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\title{Answers to exercises }

\author{}
\date{}


\begin{document}
\maketitle
\section*{Chapter 1 answers}
\section*{Exercise \(1 \mathrm{~A}\)}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-001(1)}
\end{center}

\section*{945}
10 a \(1.508 \mathrm{~g}\)
b \(0.00728 \mathrm{~g}\)

1161 minutes and 6 seconds

\begin{center}
\begin{tabular}{llll}
12 a 16500 & b 31500 &  &  \\
13 a \(\$ 480\) & b \(240 \mathrm{mins}\) & c \(240 \mathrm{~cm}\) &  \\
 & d \(\$ 218\) & e \(800 \mathrm{~mm}\) & f \(373 \frac{1}{3} \mathrm{~mm}\) \\
\end{tabular}
\end{center}

\(14 \$ 1900\)

15 a \(\$ 575\)

b \(\$ 12929\)

16 a \(100 \%\) profit

b \(17.6 \%\) loss

17 a costs \(\$ 518519\), total sales \(\$ 546519\)

b costs \(\$ 11538462\), total sales \(\$ 10788462\)

\[
\begin{array}{rrl}
18 & \text { a } \quad \text { i } \$ 0 & \text { ii } \$ 450 \\
& \text { iii } \$ 4700 & \text { iv } \$ 127700 \\
& \text { b } \quad \text { i } 0 \% & \text { ii } 3 \% \\
& \text { iii } 12.37 \% & \text { iv } 31.93 \% \\
& \text { c } \quad \text { i } \$ 22533.33 & \text { ii } \$ 34640.00 \\
& \text { iii } \$ 78571.43 & \text { iv } \$ 78000.00 \\
\text { 19 } & \text { a } \$ 1600 & \text { b } \$ 8000 \\
\text { 20 } & \text { a } \$ 720 & \text { b } \$ 2025 \\
\text { 21 } & \text { a } 7.5 \% \text { p.a. } & \text { b } 6.8 \% \text { p.a. } \\
\text { 22 } & \text { a } 50 \text { years } & \text { b } 38 \text { years } \\
\text { 23 } & \text { a } \$ 15000 & \text { b } \$ 24 \text { 790.12 }
\end{array}
\]

\section*{Exercise 1B}
\begin{center}
\begin{tabular}{lll}
1 & a \(\$ 627\) & b \(\$ 9786\) \\
 & c \(\$ 483.36\) & d \(\$ 3369.60\) \\
2 & a \(\$ 8100\) & b \(\$ 4332\) \\
 & c \(\$ 801.22\) & d \(\$ 9139.20\) \\
3 & a 35840 & b 171360 \\
4 & a \(434 \mathrm{~mm}\) & b \(88 \mathrm{~mm}\) \\
\end{tabular}
\end{center}

c 278

5 a \(6 \%\) increase

b \(88 \%\) increase

c \(22 \%\) decrease

d \(44 \%\) decrease

6 a \(\$ 630\)

b \(\$ 5160\)

7 a 200

b 7299

c 68142

8 a 4750 megalitres

b 11340 megalitres

c 1840 megalitres

d 63330 megalitres

9 a \(\$ 52\)

b \(\$ 31.20\)

c \(\$ 442\)

d \(\$ 1.04\)

10 a \(\$ 3111.11\)

b \(\$ 726.77\)

c \(\$ 54.44\)

11 a \(\$ 28000\)

d \(\$ 69565.22\)

b \(\$ 5992.56\)

c \(\$ 13806.10\)

g \(4 \%\)

e \(\$ 31890.04\)

f \(\$ 421741.12\)

h \(60.22 \%\)

i \(-17.16 \%\)

\begin{center}
\begin{tabular}{llll}
12 & a i \(\$ 3014\) & ii \(\$ 147400\) & iii \(\$ 9.02\) \\
 & b i \(\$ 3570\) & ii \(\$ 388070\) & iii \(\$ 2.90\) \\
 & c i \(\$ 4862\) & ii \(\$ 3818254\) & iii \(\$ 5.17\) \\
13 & a i \(\$ 3102\) & ii \(\$ 24.97\) & iii \(\$ 373.27\) \\
 & b i \(\$ 4061.90\) & ii \(\$ 13.83\) & iii \(\$ 73.50\) \\
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-001}
\end{center}

\[
\begin{aligned} 15 a i \(9.09 \% &\) ii \(15.25 \% \\\) iii \(78.26 \% &\) iv \(3.94 \% \\\) b i \(11.11 \% &\) ii \(21.95 \% \\\) iii \(400 \% &\) iv \(4.28 \%\end{aligned}
\])

\section*{Exercise \(1 \mathrm{C}\)}
\begin{center}
\begin{tabular}{llll}
a \(\$ 12474\) & b \(\$ 8042.64\) & c \(\$ 233436.82\) \\
\end{tabular}
\end{center}

\(2 \quad \$ 3.83\)
3 a \(\$ 7706.79\)
b \(\$ 96119.13\)
c \(\$ 128.24\)
d \(\$ 3947.50\)

4 a \(50.4 \%\)
b i \(\$ 1503.63\)
ii \(\$ 25965.05\)
iii \(\$ 324.84\)

\section*{\(525.75 \%\)}
6 a decreased by \(4 \%\)
b decreased by \(64 \%\)
7 a \(\$ 2.52\)
b \(\$ 2.19\)
c \(\$ 1.99\)

\begin{center}
\begin{tabular}{llll}
a & a i \(\$ 370.74\) & ii \(\$ 8897.76\) & iii \(\$ 4199930.76\) \\
 & b \(62.9 \%\) &  &  \\
\end{tabular}
\end{center}

\(\begin{array}{lll}10 \text { a i } 49.8^{\circ} \mathrm{C} & \text { ii } 24.8^{\circ} \mathrm{C} & \text { iii } 3.1^{\circ} \mathrm{C}\end{array}\)

b i \(20.2^{\circ} \mathrm{C}\)

ii \(34.2^{\circ} \mathrm{C}\)

iii \(68.6^{\circ} \mathrm{C}\)

11 a \(18.6 \%\)

b \(11.6 \%\)

12 a \(90.9 \%\)

b \(100.0 \%\)

c \(100.0 \%\)

13 a decrease of \(25 \%\)

b decrease of \(25 \%\)

c As multiplication is commutative, if the percentage increases and decreases are the same, but in different orders, the overall effect is the same.

\section*{Exercise 1D}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
1 & a \(\$ 107000\) & b \(\$ 114490\) & c \(\$ 140255.17\) \\
\hline
 & d \(40.26 \%\) & e \(\$ 40255.17\) & f \(\$ 35000\) \\
\hline
2 & a 59750 & b \(76 \%\) &  \\
\hline
3 & a i \(\$ 1630910.87\) & ii \(329 \%\) &  \\
\hline
 & b i \(\$ 1696688.53\) & ii \(346 \%\) &  \\
\hline
4 & a \(\$ 327000\) & b \(\$ 548411.74\) & c \(82.80 \%\) \\
\hline
 & d \(\$ 248411.74\) & e \(\$ 189000\) &  \\
\hline
5 & a \(\$ 83521265.73\) & b \(\$ 153336556.81\) & c \(\$ 12000\) \\
\hline
6 & a \(\$ 33542.00\) & b \(\$ 13542.00\) &  \\
\hline
7 & a \(\$ 58349.04\) & b \(71 \%\) &  \\
\hline
 & a \(\$ 154880.98\) & b \(61 \%\) &  \\
\hline
\end{tabular}
\end{center}

\(9 \quad\) a \(\$ 43297.57 \quad \$ 12335.56\)

\(\begin{array}{lllll}10 & \text { a } 74488 & \text { b } 69356 & \text { c } 55985 & \text { d } 39178\end{array}\)

11 a \(\$ 16282.30, \$ 13282.30\)

b \(\$ 5512011.72, \$ 2512011.72\)

c \(\$ 552.75, \$ 2447.25\)

d \(\$ 1632797.69, \$ 1367202.31\)

12 a \(\$ 5400, \$ 5724, \$ 6067.44\)

b \(\$ 5551.00, \$ 5893.38, \$ 6256.87\)

\(13 \$ 1202680.28\)

14 a \(3.72 \% \quad\) b \(7.57 \%\)

\(15 \begin{array}{lll}\text { a } 32.3 \% & \text { b } 33.1 \% & \text { c } 33.8 \%\end{array}\)

\(\begin{array}{lll}\text { d } 34.0 \% & \text { e } 34.4 \% & \text { f } 34.6 \%\end{array}\)

g The total percentage growth is approximately the same, but increases as the number of years increases.

\(1687 \%\)

\section*{Exercise 1E}
\(\begin{array}{llll}\text { a } \$ 420000 & \text { b } \$ 294000 & \text { c } \$ 70589.40\end{array}\)

d \(88.2 \%\)

e \(\$ 88235.10\)

2 a \(\$ 83886.08,73.8 \%\)

b \(\$ 507799.78\)

3 Depreciated value is \(\$ 15126\). Price obtained is better by \(\$ 4874\), to the nearest dollar.

\(4 \$ 2724000, \$ 1634400, \$ 980640, \$ 588384\); \(\$ 1816000, \$ 1089600, \$ 653760, \$ 392256\)

5 Sandra \(\$ 116731.38\), Kevin \(\$ 20971.52\)

6 a taxis \(98.4 \%\), other cars \(78.3 \%\)

b taxis \(\$ 156250\), other cars \(\$ 2166757.03\), difference \(=\$ 2010507.03\)

\(\begin{array}{llll}7 & \text { a } \$ 8387 & \text { b } \$ 10822 & \text { c } \$ 83157\end{array}\)

\(\begin{array}{ll}\text { d } 92.2 \% & \text { e } \$ 7666\end{array}\)

\(\begin{array}{lll}8 & \text { a } \$ 15523.20 & \text { b } \$ 33348.49\end{array}\)

c \(\$ 83.2 \% \quad\) d \(\$ 3964.07\)

e i \(\$ 2480.80 \quad\) ii \(\$ 5941.76\)

\(9 \quad 0.4096 \%\)

10 gain of \(58.86 \%\)

\(\begin{array}{llll}11 & \text { a i } 27.75 \% & \text { ii } 27.10 \% & \text { iii } 26.61 \%\end{array}\)

\(\begin{array}{lll}\text { iv } 26.49 \% & \text { v } 26.26 \% & \text { vi } 26.14 \%\end{array}\)

b The total depreciation is approximately the same, but decreases as the number of years increases.

\(\begin{array}{llll}12 & 92.09 \% & 13 & 2.6 \%\end{array}\)

\section*{Review exercise}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-003(1)}
\end{center}

\section*{Challenge exercise}
\begin{center}
\begin{tabular}{lll}
1 & a \(\$ 37036\) & b \(0.32 \%\) decrease \\
2 & a Bank B & b \(\$ 9653.19\) \\
3 & a \(\$ \frac{50}{n}\) & b \(\$ 50.51\) \\
 & c \(\$ 47.98\) & d \(\$ 53.03\) \\
\(\mathbf{4}\) & \(12.36 \%\) & \(\mathbf{5} 0.8 \%\) increase \\
6 & \(25 \%\) & \(\mathbf{7} 80.4 \%\) \\
\(\mathbf{8}\) & \(\$ 369814.11\) & \(\mathbf{9} r \approx 4.88\) \\
\end{tabular}
\end{center}

\section*{Chapter 2 answers}
Exercise \(2 \mathrm{~A}\)

1

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-003}
2 a \(2 \sqrt{2}\)
b \(2 \sqrt{3}\)
c \(4 \sqrt{2}\)
d \(5 \sqrt{2}\)
e \(3 \sqrt{6}\)
f \(6 \sqrt{3}\)

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
 & g \(7 \sqrt{2}\) & h \(10 \sqrt{2}\) & i \(12 \sqrt{2}\) \\
\hline
 & j \(7 \sqrt{3}\) & k \(4 \sqrt{7}\) & I \(5 \sqrt{7}\) \\
\hline
 & \(\mathbf{m} 7 \sqrt{5}\) & n \(7 \sqrt{6}\) & o 15 \\
\hline
 & p 30 & q \(15 \sqrt{2}\) & r \(20 \sqrt{2}\) \\
\hline
 & s \(10 \sqrt{10}\) & t \(24 \sqrt{3}\) &  \\
\hline
3 & a \(10 \sqrt{3}\) & b \(20 \sqrt{5}\) & c \(18 \sqrt{11}\) \\
\hline
 & d \(15 \sqrt{6}\) & e \(24 \sqrt{5}\) & f \(35 \sqrt{5}\) \\
\hline
 & g \(24 \sqrt{2}\) & h \(35 \sqrt{2}\) & i \(66 \sqrt{3}\) \\
\hline
 & j 560 & k \(35 \sqrt{3}\) & l \(12 \sqrt{11}\) \\
\hline
 & \(\mathbf{m} 8 \sqrt{13}\) & n \(25 \sqrt{11}\) & o \(40 \sqrt{3}\) \\
\hline
4 & a \(\sqrt{8}\) & b \(\sqrt{45}\) & c \(\sqrt{147}\) \\
\hline
 & d \(\sqrt{216}\) & e \(\sqrt{300}\) & f \(\sqrt{160}\) \\
\hline
 & g \(\sqrt{605}\) & h \(\sqrt{2450}\) & i \(\sqrt{108}\) \\
\hline
 & j \(\sqrt{180}\) &  &  \\
\hline
5 & a \(\sqrt{6}\) & b \(\sqrt{77}\) & c \(\sqrt{40}=2 \sqrt{10}\) \\
\hline
 & d \(\sqrt{39}\) & e \(\sqrt{48}=4 \sqrt{3}\) &  \\
\hline
6 & a \(\sqrt{5}\) & b \(\sqrt{2}\) & c \(\sqrt{6}\) \\
\hline
 & d \(\sqrt{5}\) & e \(\sqrt{3}\) & f \(\sqrt{8}=2 \sqrt{2}\) \\
\hline
7 & a \(\sqrt{14}\) & b 4 & c \(3 \sqrt{2}\) \\
\hline
 & d \(6 \sqrt{2}\) & e \(\sqrt{6}\) & f 0 \\
\hline
 & g 25 & h 3 &  \\
\hline
8 & a \(\sqrt{5} \times \sqrt{6}=\sqrt{30}\) & b \(\sqrt{12} \times\) & \(\sqrt{3}=\sqrt{36}\) \\
\hline
 & c \(\sqrt{15} \times \sqrt{3}=\sqrt{45}\) & d \(\frac{\sqrt{100}}{\sqrt{5}}=\) & \(=\sqrt{20}\) \\
\hline
 & e \(\frac{\sqrt{33}}{\sqrt{11}}=\sqrt{3}\) & f \(\sqrt{5}\) &  \\
\hline
9 & a \(\sqrt{21} \mathrm{~cm}^{2}\) & b \(\frac{1}{2} \sqrt{30} \mathrm{~cm}^{2}\) & c \(17 \mathrm{~cm}^{2}\) \\
\hline
 & d \(\sqrt{11} \mathrm{~cm}\) & e \(3 \sqrt{7} \mathrm{~cm}\) &  \\
\hline
10 & a \(3 \sqrt{2}\) & b \(\sqrt{6}\) & c \(3 \sqrt{5}\) \\
\hline
 & d \(2 \sqrt{2}\) & e \(5 \sqrt{2}\) & f \(3 \sqrt{5}\) \\
\hline
11 & a \(112 \mathrm{~cm}^{2}\) & b \(4 \sqrt{14} \mathrm{~cm}\) &  \\
\hline
12 & a \(7 \sqrt{3} \mathrm{~cm}^{2}\) & b \(\sqrt{52}=2 \sqrt{13} \mathrm{~cm}\) &  \\
\hline
\end{tabular}
\end{center}

Exercise 2B
1 a \(13 \sqrt{2}\)
b \(25 \sqrt{3}\)
c \(3 \sqrt{2}\)
d \(-27 \sqrt{5}\)
e \(-2 \sqrt{3}\)
f \(22 \sqrt{5}\)

2 a \(12 \sqrt{2}+2 \sqrt{3}\)

c \(13 \sqrt{11}-3 \sqrt{10}\)

b \(10 \sqrt{7}-10 \sqrt{5}\)

e \(6 \sqrt{15}-4 \sqrt{7}\)

d \(10 \sqrt{2}-4 \sqrt{3}\)

f \(11 \sqrt{5}-2 \sqrt{2}\)

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
3 & a \(6 \sqrt{2}\) & b \(5 \sqrt{3}\) & c \(5 \sqrt{5}\) \\
\hline
 & d \(15 \sqrt{2}\) & e \(-5 \sqrt{3}\) & f \(-2 \sqrt{5}\) \\
\hline
 & g \(3 \sqrt{3}+4 \sqrt{5}\) & h \(\sqrt{11}+8 \sqrt{5}\) & i \(3 \sqrt{3}-8 \sqrt{10}\) \\
\hline
 & j \(-4 \sqrt{5}-8 \sqrt{2}\) &  &  \\
\hline
 & a \(5 \sqrt{3}\) & b \(5 \sqrt{2}\) & c \(2 \sqrt{2}\) \\
\hline
 & d \(-3 \sqrt{5}\) & e \(27 \sqrt{2}-12 \sqrt{3}\) & f \(45 \sqrt{3}\) \\
\hline
 & g \(18 \sqrt{5}\) & h \(\sqrt{7}\) & i \(28 \sqrt{11}\) \\
\hline
 & j \(-15 \sqrt{3}\) &  &  \\
\hline
5 & a \(\sqrt{2}\) & b \(6 \sqrt{3}\) & c \(6 \sqrt{2}\) \\
\hline
 & d \(\sqrt{3}\) & e \(9 \sqrt{2}\) & f \(-7 \sqrt{5}\) \\
\hline
 & g \(5 \sqrt{6}\) & h \(4 \sqrt{3}\) & i \(2 \sqrt{5}\) \\
\hline
 & j \(11 \sqrt{2}\) & k \(\frac{5 \sqrt{5}}{2}\) & l \(15 \sqrt{2}-10 \sqrt{5}\) \\
\hline
\(P\) & a \(10 \sqrt{2}-4 \sqrt{3}\) & b \(24 \sqrt{ }\) & \(-\sqrt{5}\) \\
\hline
 & c \(6 \sqrt{3}+15 \sqrt{10}\) & d \(24 \sqrt{ }\) & \(-19 \sqrt{2}\) \\
\hline
 & a \(x=7\) & b \(x=5\) & c \(x=6\) \\
\hline
8 & a \(2 \sqrt{2}+2 \sqrt{3}, \sqrt{6}\) & b \(2 \sqrt{3}\) & \(2 \sqrt{5}, \sqrt{15}, 2 \sqrt{2}\) \\
\hline
\end{tabular}
\end{center}

\section*{Exercise 2C}
\begin{center}
\begin{tabular}{llll}
1 & a \(\sqrt{15}\) & b \(\sqrt{55}\) & c \(\sqrt{112}=4 \sqrt{7}\) \\
d \(\sqrt{65}\) & e \(\sqrt{12}=2 \sqrt{3}\) & f \(\sqrt{54}=3 \sqrt{6}\) &  \\
2 & a \(\sqrt{5}\) & b \(\sqrt{2}\) & c \(\sqrt{6}\) \\
 & d \(\sqrt{5}\) & e \(\sqrt{3}\) & f \(\sqrt{8}=2 \sqrt{2}\) \\
3 & a \(12 \sqrt{10}\) & b \(77 \sqrt{15}\) & c \(54 \sqrt{35}\) \\
 & d \(32 \sqrt{6}\) & e \(-30 \sqrt{6}\) & f \(-28 \sqrt{22}\) \\
g \(12 \sqrt{77}\) & h \(18 \sqrt{14}\) & i \(-22 \sqrt{42}\) &  \\
4 & a \(2 \sqrt{3}\) & b \(\frac{5 \sqrt{5}}{2}\) & c \(\frac{2 \sqrt{6}}{3}\) \\
 & d \(\frac{-5 \sqrt{2}}{2}\) & e \(-2 \sqrt{3}\) & f \(\frac{1}{3 \sqrt{2}}\) \\
5 & a 2 & b 11 & c 9 \\
d 40 & e 30 & f 90 &  \\
g -56 & h 24 & i -126 &  \\
6 & a \(\sqrt{3}\) & b \(\sqrt{2}\) & c \(2 \sqrt{2}\) \\
d \(3 \sqrt{3}\) & e \(3 \sqrt{7}\) & f \(6 \sqrt{5}\) &  \\
g \(6 \sqrt{2}\) & h \(12 \sqrt{3}\) & i \(10 \sqrt{5}\) &  \\
j \(8 \sqrt{2}\) & k \(32 \sqrt{2}\) & l \(50 \sqrt{5}\) &  \\
7 & a \(-3 \sqrt{3}\) & b \(-3 \sqrt{5}\) & c \(3 \sqrt{3}\) \\
d \(7 \sqrt{2}\) & e \(25 \sqrt{35}\) & f \(24 \sqrt{42}\) &  \\
g \(3 \sqrt{3}\) & h \(6 \sqrt{2}\) & i \(3 \sqrt{3}\) &  \\
j \(16 \sqrt{15}\) &  & c \(6 \sqrt{6}\) &  \\
8 \(16 \sqrt{6}\) & b \(43 \sqrt{15}\) & f \(-\sqrt{40}=-2 \sqrt{10}\) &  \\
d \(6 \sqrt{10}\) & e \(-17 \sqrt{30}\) &  &  \\
\end{tabular}
\end{center}

g \(8 \sqrt{2}\)
h \(-4 \sqrt{5}\)
i \(2 \sqrt{2}\)
j \(\sqrt{2}\)

9 a 8

b 12

c 45

d 150

e 63

f 44

g 250

h \(a^{2} b\)

10 a \(\sqrt{2}\)

b 2

c \(2 \sqrt{2}\)

d 4

e \(4 \sqrt{2}\)

f 8

g \(3 \sqrt{3}\)

h \(5 \sqrt{5}\)

i \(16 \sqrt{2}\)

j \(54 \sqrt{2}\)

k \(192 \sqrt{3}\)

l \(40 \sqrt{5}\)

m \(625 \sqrt{5}\)

n \(56 \sqrt{7}\)

o 512

p \(2187 \sqrt{3}\)

11 a \(\sqrt{6}+\sqrt{10}\)
b \(\sqrt{35}+\sqrt{42}\)
c \(\sqrt{35}-\sqrt{10}\)

d \(6 \sqrt{10}-9 \sqrt{6}\)

e \(20 \sqrt{6}+24 \sqrt{15}\)

f \(4 \sqrt{6}-4 \sqrt{3}\)

g \(6 \sqrt{15}+15\)

h \(18 \sqrt{2}+8 \sqrt{3}\)

i \(60 \sqrt{2}-32 \sqrt{5}\)

j \(24-6 \sqrt{2}\)

k \(30+24 \sqrt{5}\)

l \(36 \sqrt{2}-18\)

m \(80-24 \sqrt{10}\)

n \(105 \sqrt{5}-42 \sqrt{3}\)

o \(66 \sqrt{2}-132 \sqrt{3}\)

12 a \(\sqrt{15}+\sqrt{21}+\sqrt{10}+\sqrt{14}\)

b \(\sqrt{55}+\sqrt{30}+\sqrt{77}+\sqrt{42}\)

c \(\sqrt{15}-\sqrt{35}+\sqrt{6}-\sqrt{14}\)

d \(\sqrt{15}-\sqrt{21}+\sqrt{30}-\sqrt{42}\)

e \(5 \sqrt{6}\)

f \(45-17 \sqrt{15}\)

g \(37-10 \sqrt{5}\)

h \(10 \sqrt{6}-6 \sqrt{2}+30 \sqrt{3}-18\)

i \(-11-11 \sqrt{14}\)

j \(2 \sqrt{21}-4 \sqrt{35}+\sqrt{15}-10\)

k \(34-9 \sqrt{14}\)

l \(3 \sqrt{21}-9 \sqrt{35}-8 \sqrt{3}+24 \sqrt{5}\)

13 a \(-18 \sqrt{2}\)

b \(\frac{-3 \sqrt{2}}{2}\)

c 66

d \(\frac{11}{108}\)

e \(24 \sqrt{3}\)

f \(1296 \sqrt{3}\)

g \(\frac{9}{4 \sqrt{3}}=\frac{3 \sqrt{3}}{4}\)

h -42

14 a 4

b \(6-\sqrt{3}\)

c \(\sqrt{3}+10\)

d \(2 \sqrt{3}\)

e \(3 \sqrt{3}-2\)

f 1

g \(\sqrt{3}\)

h 1

15 a \(8 \sqrt{6}, 4 \sqrt{3}+8 \sqrt{2}\)

b \(24,12 \sqrt{3}\)

c 29,28

d \(7+3 \sqrt{5}, 6+4 \sqrt{5}\)

16 a \(\sqrt{15}\)

b \(10+5 \sqrt{3}+\sqrt{15}\) c \(6 \sqrt{5}+\sqrt{15}\)

17 a \(8+20 \sqrt{3}\)

b \(79+20 \sqrt{3}\)

\section*{Exercise 2D}
\begin{center}
\begin{tabular}{lll}
a \(7+2 \sqrt{10}\) & b \(10+2 \sqrt{21}\) & c \(5-2 \sqrt{6}\) \\
d \(13-2 \sqrt{42}\) & e \(13+4 \sqrt{3}\) & f \(22-12 \sqrt{2}\) \\
g \(14+4 \sqrt{6}\) & h \(59-24 \sqrt{6}\) & i \(83+12 \sqrt{35}\) \\
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-005}
\end{center}

\[
\begin{aligned}
& 4 \quad \text { a } \frac{1}{4}(5+5 \sqrt{3}-4 \sqrt{2}) \\
& \text { c } \frac{1}{17}(17 \sqrt{10}+4 \sqrt{2}-10) \quad \text { d }-\frac{2}{17}(3 \sqrt{3}+2 \sqrt{5}) \\
& \text { e } \frac{1}{10}(6 \sqrt{3}-5 \sqrt{2}) \quad \text { f }-10 \\
& \begin{array}{llll}
5 & \text { a } \frac{13 \sqrt{3}}{3} & \text { b }-\frac{\sqrt{3}}{3} & \text { c } 14
\end{array} \\
& \text { d } \frac{6}{7} \quad \text { e } \frac{559 \sqrt{3}}{9}
\end{aligned}
\]

\(610(\sqrt{2}+1)\)

\[
\begin{aligned}
& 7 \quad \text { a } \frac{3-\sqrt{3}}{2} \\
& \text { b } \frac{16-9 \sqrt{3}}{2} \\
& \text { c } 6 \sqrt{3}+10 \\
& \text { d } 15 \sqrt{3}+26
\end{aligned}
\]

8 a \(\frac{4}{289}(19+6 \sqrt{2})\)

b \(\frac{1}{2}(3 \sqrt{2}-1)\)

c \(\frac{1}{4}(19-6 \sqrt{2})\)

d \(\frac{1}{34}(63 \sqrt{2}-13)\)

e \(\frac{1}{578}(915 \sqrt{2}-137)\)

f \(\frac{1}{1156}(5795-1638 \sqrt{2})\)

9 i a \(\frac{1}{8}(15-5 \sqrt{5})\)

b \(\frac{1}{5}(5+\sqrt{5})\)

c \(\frac{1}{5}(6+2 \sqrt{5})\)

d \(\frac{1}{20}(45-\sqrt{5})\)

e \(\frac{1}{40}(115-17 \sqrt{5})\)

f \(\frac{1}{40}(123-9 \sqrt{5})\)

ii a \(588+240 \sqrt{6}\)

b \(\frac{1}{6}(5 \sqrt{3}-6 \sqrt{2})\)

c \(\frac{1}{36}(147-60 \sqrt{6})\)

d \(\frac{1}{6}(65 \sqrt{3}+66 \sqrt{2})\)

e \(\frac{1}{6}(3528+1440 \sqrt{6}+5 \sqrt{3}-6 \sqrt{2})\)

f \(\frac{5}{12}(1421+572 \sqrt{6})\)

\section*{Review exercise}
1 a \(4 \sqrt{2}\)
b \(\sqrt{6}\)
c \(3 \sqrt{3}-\sqrt{2}\)
d \(4 \sqrt{2}-3 \sqrt{5}\)
2 a \(5 \sqrt{2}\)
b \(\sqrt{2}\)
c \(-4 \sqrt{7}\)
d \(11 \sqrt{3}\)

3 a \(30 \sqrt{2}\)

b \(30 \sqrt{2}\)

c \(12 \sqrt{10}\)

d \(28 \sqrt{42}\)

4 a \(6 \sqrt{2}\)

b \(3 \sqrt{5}\)

c \(2 \sqrt{6}\)

d \(3 \sqrt{3}\)

e \(4 \sqrt{5}\)

f \(2 \sqrt{11}\)

g \(6 \sqrt{2}\)

h \(8 \sqrt{3}\)

j \(18 \sqrt{3}\)

k \(40 \sqrt{2}\)

5 a \(\sqrt{75}\)

b \(\sqrt{112}\)

c \(\sqrt{242}\)

d \(\sqrt{125}\)

e \(\sqrt{384}\)

f \(\sqrt{891}\)

h \(\sqrt{176}\)

g \(\sqrt{208}\)

\(\sqrt{176}\)

6 a \(9 \sqrt{2}\)

d \(19 \sqrt{7}\)

7 a \(19 \sqrt{2}\)

b \(2 \sqrt{5}+5 \sqrt{3}\)

c \(3 \sqrt{3}\)

e \(11 \sqrt{5}\)

f \(55 \sqrt{6}\)

c \(15 \sqrt{2}+\sqrt{6}\)

b \(4 \sqrt{2}\)

d \(16 \sqrt{7}-27 \sqrt{3}\)

8 a \(6 \sqrt{3}+6\)

d \(30 \sqrt{5}-50\)

b \(30-10 \sqrt{2}\)

c \(24-16 \sqrt{21}\)

9 a \(10-\sqrt{2}\)

e \(12 \sqrt{7}-21\)

f \(45-12 \sqrt{6}\)

b \(32-9 \sqrt{3}\)

c 2

d 17

e 50

f \(8+2 \sqrt{15}\)

g \(14+4 \sqrt{6}\)

h \(14-4 \sqrt{6}\)

i 10

j \(a-b\)

k \(11-4 \sqrt{7}\)

l \(28+10 \sqrt{3}\)

10 a \(\frac{5 \sqrt{3}}{3}\)

b \(\frac{7 \sqrt{3}}{6}\)

c \(\frac{2 \sqrt{2}}{3}\)

d \(\frac{3 \sqrt{2}}{7}\)

e \(\frac{5 \sqrt{3}}{9}\)

f \(\frac{\sqrt{6}}{4}\)

g \(\frac{\sqrt{42}}{7}\)

h \(\frac{7 \sqrt{42}}{12}\)

11 a \(-\frac{1}{2}(\sqrt{5}+\sqrt{7})\)

b \(\frac{1}{11}(10+3 \sqrt{5})\)

c \(\frac{1}{2}(3-\sqrt{5})\)

d \(\frac{1}{2}(3+\sqrt{5})\)

12 a \(\sqrt{2}+1\)

b \(2-\sqrt{3}\)

c \(\sqrt{3}-\sqrt{2}\)

d \(\sqrt{3}+\sqrt{2}\)

\(13 p=5, q=2\)

14 a \(5 \sqrt{5}+2\)

15 a \(2(\sqrt{3}+2)\)

c \(4 \sqrt{3}+7\)

16 a 6

b \(2 \sqrt{3}-2\)

c \(26-4 \sqrt{3}\)

d \(12 \sqrt{3}-12\)

g \(\frac{22-14 \sqrt{3}}{13}\)

e \(2-\sqrt{3}\)

f \(\frac{30-12 \sqrt{3}}{13}\)

17 a \(4 \sqrt{3}+8\)

h \(2 \sqrt{3}+5\)

b \(\frac{7}{3}\)

b 0

d 1

\(1840-20 \sqrt{3}\)

19 a \(\sqrt{14}\)

d \(\sqrt{42}\)

b 6

c \(\sqrt{118}\)

d

e \(\frac{30-10 \sqrt{3}}{3}\)

f \(20 \sqrt{2}-10 \sqrt{6}\)

20
a i \(\frac{3+\sqrt{5}}{2}\)
ii \(\frac{\sqrt{5}-1}{2}\)
iii \(\frac{1+\sqrt{5}}{2}\)

iv \(2+\sqrt{5}\)

v 2

\(2120 \sqrt{2}\)
22 a \(-4 \sqrt{3}\)
b -8
c -4
d \(-7-4 \sqrt{3}\)
\(2314 \sqrt{5}\)
24 a 6
b -11
c \(-\frac{6}{11}\)

Challenge execrise

1 a \((\sqrt{x})^{2}+2 \sqrt{x} \sqrt{y}+(\sqrt{y})^{2}=x+y+2 \sqrt{x y}\)
b i \(\sqrt{11}+\sqrt{5}\)
ii \(\sqrt{11}-\sqrt{5}\)
iii \(\sqrt{6}+\sqrt{5}\)

2 a \(a+b-a^{2}\)

b \(4\left(1+x^{2}\right)-1=3+4 x^{2}\)

c \(2 x\)

d \(\sqrt{a}+\sqrt{b}\)

e \(2 a+2 \sqrt{a^{2}-b^{2}}\)

\(\begin{array}{rlrl}\text { a } x=\frac{9}{4} & \text { b } x=9 & \text { c } x=2 \\ \text { d } x & =9 & \text { e } x=9 & \text { f } x=8\end{array}\)

\(4 x-2 \sqrt{x}-\frac{2}{\sqrt{x}}+\frac{1}{x}+3\)

\(5 \frac{5^{\frac{2}{3}}}{5}\)

\(6 \frac{\sqrt{15}-\sqrt{5}-\sqrt{3}+1}{8}\)

\(7 \frac{2 \sqrt{3}+3 \sqrt{2}-\sqrt{30}}{12}\)

\(8 a+b-\sqrt{a b}\)

9 a \(\sqrt{9 \frac{9}{80}}=\sqrt{\frac{729}{80}}=9 \sqrt{\frac{9}{80}}\)

\(\sqrt{4 \frac{4}{15}}=\sqrt{\frac{64}{15}}=4 \sqrt{\frac{4}{15}}\)

b

\[
\begin{aligned}
\sqrt{a+\frac{b}{c}} & =a \sqrt{\frac{b}{c}} \\
a+\frac{b}{c} & =a^{2} \frac{b}{c} \\
a & =\frac{b}{c}\left(a^{2}-1\right) \\
\frac{a}{a^{2}-1} & =\frac{b}{c}
\end{aligned}
\]

Mixed numbers of the form \(a \frac{a}{a^{2}-1}\).

\(10(2 \sqrt{6}+5) \mathrm{cm}\)
111
\(12 \frac{2}{5}\)

13 a 625

b 1799

14 a i \(\sqrt{5}\)

ii \(\sqrt{5}-1\)

\[
\begin{array}{cc}
\text { iii } \frac{\sqrt{5}-1}{2} & \text { iv } \frac{\sqrt{5}-1}{2} \\
\text { b }\left(\frac{\sqrt{5}-1}{2}\right)^{2}+\left(\frac{\sqrt{5}-1}{2}\right)=1 &
\end{array}
\]

\section*{Chapter 3 answers}
\section*{Exercise \(3 \mathrm{~A}\)}
1 a \(8 x\)
b \(2 x\)
c \(-6 x\)
d \(4 a+7 b\)
e \(2 t-s(=-s+2 t)\)
f \(-3 x+2 y\)
g \(4 a+b\)
h \(-4 a-4 b\)
i \(3 x-4 y\)

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\multicolumn{2}{|c|}{}\begin{tabular}{l}
2 a \(3 a b\) \\
 \\
d \(9 s t^{2}+s^{2} t\) \\
\end{tabular} & b \(6 x^{2} y\) & \begin{tabular}{l}
c \(12 c d^{3}\) \\
f \(12 x^{2} y\) \\
\end{tabular} \\
\hline
\multirow[t]{4}{*}{3} & a \(3 a+6\) & b \(7 b+21\) & c \(4 a-10\) \\
\hline
 & d \(24 p+108\) & \(-12 b-27\) & \(\mathbf{f}-20 m-25\) \\
\hline
 & g \(2-6 a\) & h \(42-24 b\) & i \(35-10 b\) \\
\hline
 & \begin{tabular}{l}
\(\mathbf{j} x+3\) \\
m \(3-5 \ell\) \\
\end{tabular} & n \(4 x+\frac{1}{2}\) & \begin{tabular}{l}
l \(4 t-3\) \\
o \(\frac{1}{5}-\frac{x}{10}\) \\
\end{tabular} \\
\hline
\multirow[t]{4}{*}{4} & a \(a^{2}+4 a\) & b \(c^{2}-5 c\) & c \(6 g^{2}-10 g\) \\
\hline
 & d \(20 h^{2}-28 h\) & \(12 j^{2}+21 j\) & f \(4 k-5 k^{2}\) \\
\hline
 & \(\mathbf{g} \ell-3 \ell^{2}\) & h \(6 a c+3 b c\) & i \(10 d^{2}-20 d e\) \\
\hline
 & j \(10 m^{2}-20 m n\) & \(10 x^{2} y+15 x\) & \(6 p-15 p^{2} q\) \\
\hline
\multirow[t]{5}{*}{5} & a \(4 a+4 b\) & \multicolumn{2}{|c|}{b \(5 a+5\)} \\
\hline
 & c \(8 p-56\) & \multicolumn{2}{|c|}{d \(15-3 p\)} \\
\hline
 & e \(a^{2}+a b\) & \multicolumn{2}{|c|}{f \(6 m^{2}+10 m\)} \\
\hline
 & g \(30-6 x\) & \multicolumn{2}{|c|}{h \(12 a^{2}-21 a\)} \\
\hline
 & i \(12 a^{2}+20 a\) & \multicolumn{2}{|c|}{j \(6 x^{2}-21 x y\)} \\
\hline
\multirow[t]{4}{*}{6} & a \(8 a+23\) & b \(3 d+3\) & c \(2 e+5\) \\
\hline
 & d \(13 g+10\) & \(3 i-30\) & f \(8 a^{2}+13 a\) \\
\hline
 & g \(10 b^{2}-9 b\) & \(2 b^{2}-15 b\) & i 10 \\
\hline
 & j \(\frac{5 x}{6}+2\) & \(\frac{9}{5}\) & l \(\frac{3 x}{4}-\frac{9}{4}\) \\
\hline
7 & a \(5 y+14\) & b \(7 a\) &  \\
\hline
 & c \(16 b-9\) & d \(a\) &  \\
\hline
 & e -5 & f \(x^{2}\) & \(x-6\) \\
\hline
 & g \(2 p^{2}+7 p+5\) & h 10 &  \\
\hline
 & i \(4 y^{2}-16 y\) & j 15 & \(-10 z\) \\
\hline
8 & a \(x^{2}+10 x+21\) & b \(a^{2}\) & \(3 a+40\) \\
\hline
 & c \(a^{2}+12 a+27\) & d \(x^{2}\) & \(x+5\) \\
\hline
 & e \(x^{2}-5 x+6\) & f \(a^{2}\) & \(1 a+28\) \\
\hline
 & g \(x^{2}-x-42\) & h \(x^{2}\) & \(x-33\) \\
\hline
 & i \(x^{2}-5 x-24\) & j \(6 x\) & \(17 x+12\) \\
\hline
 & k \(15 x^{2}+22 x+8\) & \(16 x\) & \(7 x-5\) \\
\hline
 & \(\mathbf{m} 20 x^{2}-11 x-3\) & n \(2 a\) & \(11 a+15\) \\
\hline
 & o \(6 p^{2}+11 p-10\) & p 12 & \(+13 x-14\) \\
\hline
 & q \(15 x^{2}-46 x+16\) & r \(6 x\) & \(17 x+5\) \\
\hline
3 & a \(2 a^{2}+7 a b+3 b^{2}\) & b \(6 n\) & \(7 m n-3 n^{2}\) \\
\hline
 & c \(6 p^{2}+p q-5 q^{2}\) & d \(3 x\) & \(13 a x-10 a^{2}\) \\
\hline
 & e \(6 x^{2}+13 x y-5 y^{2}\) & f \(6 c\) & \(35 c d-6 d^{2}\) \\
\hline
 & g \(8 p^{2}+2 p q-3 q^{2}\) & h \(3 b\) & \(5 a b-2 a^{2}\) \\
\hline
 & i \(15 q^{2}-4 p q-4 p^{2}\) &  &  \\
\hline
10 & a \(x^{2}+2 x+1\) & b \(9 x\) & \(12 x+4\) \\
\hline
 & c \(9 a^{2}+24 a b+16 b^{2}\) & d \(4 x\) & \(12 x y+9 y^{2}\) \\
\hline
 & e \(x^{2}-14 x+49\) & f \(x^{2}\) & \(2 x y+y^{2}\) \\
\hline
 & g \(4 a^{2}-12 a b+9 b^{2}\) & h \(9 a\) & \(24 a b+16 b^{2}\) \\
\hline
 & i \(4 x^{2}-12 x y+9 y^{2}\) &  &  \\
\hline
\end{tabular}
\end{center}

11 a \(x^{2}-9\)

c \(4 a^{2}-9\)

d \(25-x^{2}\)

e \(49-9 y^{2}\)

f \(4 m^{2}-p^{2}\)

g \(9 x^{2}-4 y^{2}\)

h \(64 m^{2}-25 n^{2}\)

i \(49 r^{2}-4 t^{2}\)

12 a \(x^{2}-64\)

b \(4 a^{2}+4 a+1\)

c \(4 a^{2}-1\)

d \(9-4 x^{2}\)

e \(9+12 a+4 a^{2}\)

f \(9-12 x+4 x^{2}\)

g \(a^{2}+6 a b+9 b^{2}\)

h \(a^{2}-9 b^{2}\)

i \(a^{2}-6 a b+9 b^{2}\)

Exercise 3B
\(1 \quad \mathbf{a} \quad a=2\)
b \(b=3\)
c \(x=\frac{9}{2}\)
d \(y=-5\)
e \(x=6\)
f \(y=\frac{22}{5}\)
g \(x=4\)
h \(y=6\)
i \(a=3\)
j \(b=-1\)
2 a \(x=-3\)
b \(x=2\)
c \(x=-3\)
d \(x=-2\)
e \(x=\frac{23}{5}\)
f \(a=\frac{26}{17}\)
g \(y=-2\)
h \(x=\frac{55}{28}\)
3 a \(x>3\)
b \(x \leq 2\)
c \(x \leq-4\)
d \(x>-8\)
e \(x \geq 1\)
f \(x<8\)
g \(x>\frac{1}{2}\)
h \(x \leq \frac{2}{3}\)
4 a \(x>-\frac{11}{2}\)
b \(x \leq 7\)
c \(x \geq 13\)
d \(x<11\)
e \(x \leq-12\)
f \(x<\frac{9}{2}\)
g \(x<4\)
h \(x \geq-\frac{29}{4}\)
i \(x \leq-20\)
j \(x<5\)

528

6 width \(5 \mathrm{~m}\), length \(13 \mathrm{~m}\)

\(7 \quad 17\) and 25

\(8 \quad \$ 90\)

\(94 \mathrm{~km} / \mathrm{h}\)

\(10 \$ 840000\)

\section*{Exercise 3C}
1 Giorgio 7 km/h, Fred \(5 \mathrm{~km} / \mathrm{h}\)

\(22 \mathrm{~L}\)

3 17 20-cent pieces, \(10 \$ 2\) coins

\(4 \quad \$ 3600\)

\(5 \quad 12 \frac{1}{2}\) hours, 7 hours

\(6 \quad 1.5 \mathrm{~L}\)

\(7 A\) to \(B 305 \mathrm{~km}, B\) to \(C 335 \mathrm{~km}\)

8 a \(\$ 129\)

b \(\$ 107, \$ 137\)

926 hours

11 a \(y \geq \frac{150-x}{2}\)
b i \(y \geq 57.5\)
ii \(y \geq 50.5\)

\section*{Exercise 3D}
\begin{center}
\begin{tabular}{llllll}
\(\mathbf{1}\) & 127 & \(\mathbf{2}\) & 43.2 & \(\mathbf{3}\) & 44.7 \\
\(\mathbf{4}\) & \(\frac{6}{5}\) & \(\mathbf{5}\) & 2.45 & \(\mathbf{6}\) & 2.92 \\
\(\mathbf{7}\) & a \(339 \mathrm{~cm}^{2}\) &  & b \(109 \mathrm{~cm}^{2}\) &  & c \(3695 \mathrm{~cm}^{2}\) \\
\(\mathbf{8}\) & a 465 & b 1275 & c 7260 & d 5985 &  \\
\end{tabular}
\end{center}

\(9 \quad 88.36\)
\(10 \quad 99 \mathrm{~cm}\)
\(1120 \mathrm{~mm}\)

\[
\begin{aligned}
& 12 \text { a } y=z-x \\
& \text { b } r=\frac{A}{\pi s} \\
& \text { c } y=\sqrt{\frac{a x}{t}} \\
& \mathbf{d} v=\sqrt{\frac{2 w}{m}} \\
& \text { e } x=\frac{m^{2}}{4} \\
& \text { f } m=\frac{k^{2}}{4 t} \\
& \text { g } p=\frac{b^{2}}{a+b} \\
& \text { h } t=\frac{x-1}{x+1} \\
& \text { i } a=\sqrt{c^{2}-b^{2}} \\
& \text { j } r=\frac{n-q}{v^{2}} \\
& \mathbf{k} q=n-v^{3} r \\
& \mathbf{l} L=g\left(\frac{t}{2 \pi}\right)^{2} \\
& \mathbf{m} k=h-g\left(\frac{t}{2 \pi}\right)^{2} \\
& \mathbf{n} a=\frac{m b c}{b c-m c-m b} \\
& \text { o } A=\frac{B^{2}+C^{2}}{2 B} \\
& \mathbf{p} s=\frac{r c}{c-r v}
\end{aligned}
\]

\section*{Exercise \(3 \mathrm{E}\)}
\[
\begin{array}{lll}
\mathbf{1} & \mathbf{a}(b+2)(b-2) & \text { b }(3+a)(3-a) \\
\text { c }(3 x+4)(3 x-4) & \text { d }(5 a+1)(5 a-1) \\
\text { e } 16(y+2)(y-2) & \text { f }(a+2 b)(a-2 b) \\
\text { g }(3 x+y)(3 x-y) & \text { h } 2(2 x+1)(2 x-1) \\
\text { i } 3(2+y)(2-y) & \text { j } 5(m+2 n)(m-2 n) \\
\text { k } 3(3 r+t)(3 r-t) & \mathbf{l}(1+3 a b)(1-3 a b) \\
\mathbf{2} & \text { a }(x+\sqrt{3})(x-\sqrt{3}) & \text { b }(x+\sqrt{7})(x-\sqrt{7}) \\
& \text { c }(x+\sqrt{13})(x-\sqrt{13}) & \mathbf{d}(x+\sqrt{6})(x-\sqrt{6}) \\
\mathbf{3} & \mathbf{a}(x+2 \sqrt{5})(x-2 \sqrt{5}) & \text { b }(x+3 \sqrt{2})(x-3 \sqrt{2}) \\
& \text { c }(x+3 \sqrt{3})(x-3 \sqrt{3}) & \text { d }(x+2 \sqrt{6})(x-2 \sqrt{6}) \\
& \text { e }(x+2 \sqrt{10})(x-2 \sqrt{10}) & \text { f }(x+2 \sqrt{7})(x-2 \sqrt{7}) \\
& \text { g }(x+5 \sqrt{5})(x-5 \sqrt{5}) & \text { h }(x+10 \sqrt{2})(x-10 \sqrt{2}) \\
& \text { i }\left(x+\frac{\sqrt{7}}{5}\right)\left(x-\frac{\sqrt{7}}{5}\right) & \text { j }\left(x+\frac{\sqrt{3}}{2}\right)\left(x-\frac{\sqrt{3}}{2}\right) \\
& \text { k }\left(x+\frac{\sqrt{5}}{3}\right)\left(x-\frac{\sqrt{5}}{3}\right) & \mathbf{l}\left(x+\frac{\sqrt{11}}{4}\right)\left(x-\frac{\sqrt{11}}{4}\right)
\end{array}
\]

4 a \((x+2+\sqrt{2})(x+2-\sqrt{2})\)

b \((x-1+\sqrt{3})(x-1-\sqrt{3})\)

c \((x-5+\sqrt{5})(x-5-\sqrt{5})\) d \(\left(x-4+\frac{\sqrt{2}}{2}\right)\left(x-4-\frac{\sqrt{2}}{2}\right)\)

e \((x-1+2 \sqrt{3})(x-1-2 \sqrt{3})\)

f \((x+3+2 \sqrt{2})(x+3-2 \sqrt{2})\)

g \(\left(x+3+\frac{\sqrt{5}}{3}\right)\left(x+3-\frac{\sqrt{5}}{3}\right)\)

h \(\left(x-2+\frac{\sqrt{7}}{3}\right)\left(x-2-\frac{\sqrt{7}}{3}\right)\)

i \((2 x+1+3 \sqrt{3})(2 x+1-3 \sqrt{3})\)

j \(\left(2 x-3+\frac{\sqrt{6}}{3}\right)\left(2 x-3-\frac{\sqrt{6}}{3}\right)\)

\(\mathbf{k}\left(4 x-1+\frac{\sqrt{15}}{5}\right)\left(4 x-1-\frac{\sqrt{15}}{5}\right)\)

l \(12 z(3 z+2)\)

5 a \(2(x+\sqrt{7})(x-\sqrt{7})\)

b \(3(x+\sqrt{2})(x-\sqrt{2})\)

c \(5(x+\sqrt{3})(x-\sqrt{3})\)

d \(4(x+\sqrt{7})(x-\sqrt{7})\)

e \(3(x+2 \sqrt{3})(x-2 \sqrt{3})\)

f \(2(x+3 \sqrt{2})(x-3 \sqrt{2})\)

g \(2(x+1+\sqrt{5})(x+1-\sqrt{5})\)

h \(3(x+2+\sqrt{2})(x+2-\sqrt{2})\)

i \(3(x-2+2 \sqrt{2})(x-2-2 \sqrt{2})\)

j \(2(x+1+2 \sqrt{3})(x+1-2 \sqrt{3})\)

\(\mathbf{k}-2(x+3+\sqrt{2})(x+3-\sqrt{2})\)

l \(-4(x-2+\sqrt{5})(x-2-\sqrt{5})\)

6 a \((\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) \quad\) b \(\sqrt{a}+\sqrt{b}\)

\section*{Exercise \(3 \mathrm{~F}\)}
\begin{center}
\begin{tabular}{ll}
\(\mathbf{a}(x+1)(x+8)\) & \(\mathbf{b}(x+2)(x+6)\) \\
\(\mathbf{c}(x+2)(x+4)\) & \(\mathbf{d}(x+6)^{2}\) \\
\(\mathbf{e}(x+3)^{2}\) & \(\mathbf{f}(x+3)(x+8)\) \\
\(\mathbf{g}(x+3)(x-2)\) & \(\mathbf{h}(x+6)(x-5)\) \\
\(\mathbf{i}(x+8)(x-5)\) & \(\mathbf{j}(x+10)(x-6)\) \\
\(\mathbf{k}(x-3)(x-4)\) & \(\mathbf{l}(x-5)^{2}\) \\
\(\mathbf{m}(x-2)(x-16)\) & \(\mathbf{n}(x+5)(x-7)\) \\
o \((x+3)(x-7)\) & \(\mathbf{p} 3(x-1)(x-5)\) \\
q \(2(x+8)(x-3)\) & \(\mathbf{r} 4(x+4)(x-6)\) \\
s \(5(x+12)(x-2)\) & \(\mathbf{t} 3(x+1)(x-3)\) \\
u \(2(x+7)(x-4)\) &  \\
a \((m+3)(x+2)\) & b \((x-4)(x-3)\) \\
c \((a-9)(b-7)\) & d \((2 x+5)(x-1)\) \\
e \((3 a+5)(b+2)\) & \(\mathbf{f}(2 x+y)(3 x-2 y)\) \\
g \((2 a-5)(b-6)\) & h \((x+2)^{2}(x-2)\) \\
i \((4 x-5)(3-2 y)\) & j \((5 x-2 y)(2 x+3 y)\) \\
\end{tabular}
\end{center}

3 a \((x-11)(x+7)\)

c \(2 x(x-3)\)

e \((x+3)(y+2)\)

g \((x+10)(x+4)\)

i \((x-\sqrt{11})(x+\sqrt{11})\)

k \((11-a)(7+a)\)

\(\mathbf{m}(z+3)^{2}\)

o \(2(x-3)(x-4)\)

q \(25(2-x)(2+x)\)

s \((x+y)(x+z)\)

\section*{Exercise 3G}
\[
\begin{aligned}
& 1 \text { a }(2 x+1)(x+1) \\
& \text { c }(3 x-2)(2 x-1) \\
& \text { b }(5 x-2)(x-1) \\
& \text { e }(5 x-4)(4 x-3) \\
& \text { d }(4 x+3)(2 x-1) \\
& \text { f } 12(x-3)(x+2) \\
& 2 \quad \mathbf{a}(2 x+3)(x+2) \\
& \text { b }(2 x+1)(x+5) \\
& \text { c }(2 x+3)(x+3) \\
& \text { d }(3 x+2)(x+3) \\
& \text { e }(3 x+1)(x+3) \\
& \text { f }(3 x+2)(x+4) \\
& \text { g }(2 x+5)(x-2) \\
& \text { h }(2 x+1)(x-7) \\
& \text { i }(3 x+5)(x-2) \\
& \text { j }(5 x-3)(x+2) \\
& \mathbf{k}(3 x-1)(x+9) \\
& \text { l }(7 x+10)(x-1) \\
& \mathbf{m}(5 x+6)(x-2) \\
& \text { n }(3 x-2)(x+2) \\
& \text { o }(4 x-1)(x+3)
\end{aligned}
\]

\section*{Exercise \(3 \mathrm{H}\)}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-009}
\end{center}

\section*{Exercise 3I}
1 a \(\frac{a+b}{b}\)

c \(\frac{a-b}{a+b}\)

e \(\frac{x+y}{x-y}\)

g \(\frac{x+4}{x-2}\)

i \(\frac{x-3}{x+7}\)

k \(\frac{2 x-1}{4(x-3)}\)

2 a \(\frac{2 y+x}{3(2 x+y)}\)

c \(\frac{x}{(x+3)}\)

e \(\frac{a(a-1)}{a+1}\)

g \(\frac{3 x-2}{9}\)

i \(\frac{x^{2}}{6 y}\)

3 a \(\frac{x^{2}-x+6}{(x-2)(x+2)}\)

c \(\frac{4 x}{x^{2}-9}\)

e \(\frac{2}{(x-2)(x+2)(x+4)}\)

g \(\frac{x+5}{(x-1)(x+3)}\)

i \(\frac{a}{(a-4)(a+1)(a-2)}\) b \(\frac{x+y}{x-y}\)

d \(\frac{b+1}{b}\)

f \(3(x-y)\)

h \(\frac{a-2}{2}\)

j \(\frac{x+2}{x+1}\)

l \(\frac{3 x-y}{4 x-y}\)

b \(\frac{(a-3)(a+2)}{a^{2}}\)

d \(\frac{(x-2)(3 x-1)}{6(x+2)}\)

f \(2 x\)

h \(2 x\)

j \(\frac{x-y}{x}\)

b \(\frac{x}{6(x-3)}\)

d \(\frac{2 x+5}{(x+2)(x+3)}\)

f \(\frac{x^{2}+2}{(x-1)(x-3)(x+2)}\)

h \(\frac{2}{2 x+3}\)
4 a \(\frac{-2(x+4)}{x+3}\)
b \(\frac{5 x+49}{49-x^{2}}\)
c \(\frac{12}{a^{2}-1}\)
d \(\frac{a^{2}+3 x^{2}}{2 a x}\)
e \(\frac{2}{(x-4)(x-6)}\)
f \(\frac{1}{(2 x+1)(x-1)(3 x-2)}\)
g \(\frac{-4(4 a-1)}{(5 a+3)(2 a-1)(a+4)}\)
h \(\frac{13 a+2}{a^{2}-9}\)
5 a \(\frac{2 x+3}{3 x+5}\)
b \(\frac{x+7}{x+13}\)
c \(\frac{x}{x+1}\)
d \(\frac{a-11}{a-2}\)
e \(\frac{4 x+3 a}{x+2}\)
f \(\frac{x+1}{x+5}\)
6 a \(x=2\)
b \(x=\frac{13}{2}\)
c \(x=-17\)
d \(x=4\)
e \(x=-\frac{1}{7}\)
f \(x=7\)
7 a \(\frac{28}{5 x}\)
b \(x=\frac{28}{5}\)
\(8 x=-18\)
\(9 x=2\)
10 a \(x=8\)
b \(x=\frac{21}{5}\)
c \(x=\frac{54}{13}\)
d \(x=6\)
e \(x=2\)
f \(x=\frac{2}{5}\)

\section*{Review exercise}
1 a \(13 x\)
b \(7 p^{3} q\)
c \(-5 a b-a b^{2}\)
d \(20 x^{3} y-10 x^{2} y\)
2 a \(-2 a-8\)
b \(-3 b-18\)
c \(-12 b+20\)
d \(-20 b+35\)
e \(8 p+4\)
f \(-4 p+5\)
3 a \(d^{2}-9 d\)
d \(-10 m^{2}+8 m\)
b \(3 e^{2}+e\)
c \(5 f^{2}+6 f\)
g \(-6 x^{2}-15 x y\)
e \(-15 n^{2}-21 n\)
f \(6 p q-10 p r\)
h \(-6 z^{2}+8 z y\)
i \(6 a+8 a^{2} b\)

4 a \(2 c+5\)

b \(7 h+4\)

c \(4+3 q\)

d \(4 a b\)

e \(\frac{19 x}{12}-\frac{10}{3}\)

f \(b-22\)

g \(12 y^{2}-11 y-15\)

h \(6 p^{2}-6 p-4\)

i \(m^{2}-m-20\)

j \(6 x^{2}+7 x-10\)

k \(7 x^{2}-36 x+5\)

l \(2 m^{2}+7 m n+3 n^{2}\)

5
a \(x=3\)
b \(x=5\)
c \(x=2\)
d \(x=15\)
e \(x=4\)
f \(x=4\)
g \(x=\frac{24}{13}\)
h \(x=20\)
i \(x=2\)

6 a \(x>4\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-010(4)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-010(7)}
\end{center}

c \(x \geq-9\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-010(5)}
\end{center}

d \(x \geq-\frac{5}{2}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-010(1)}
\end{center}

e \(x \leq-\frac{2}{7}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-010(3)}
\end{center}

f \(x \geq-\frac{4}{3}\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-010(6)}

h \(x \geq-15\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-010}
\end{center}

i \(x \geq \frac{1}{8}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-010(2)}
\end{center}

7 a \(\frac{12}{7}\)

b 3

\(8 \quad\) a 52

b \(a=\frac{p}{4}-b \quad\) c -1

\(9 \quad \mathbf{a} h=\frac{A-\pi r^{2}}{\pi r}\)

b \(h=\sqrt{\left(\frac{A}{\pi r}\right)^{2}-r^{2}}\)

c \(u=\sqrt{v^{2}-\frac{2 E}{m}}\)

d \(x=\frac{c b+b y}{c-a}=\frac{b(c+y)}{c-a}\)

10 a \((7+z)(7-z)\)

b \((3+2 a)(3-2 a)\)

c \((5+8 x)(5-8 x)\)

d \((a b+2 c d)(a b-2 c d)\)

e \((3 x+1)(3 x-1)\)

f \((6 p+7 q)(6 p-7 q)\)

g \((1+10 b)(1-10 b)\)

h \((11 a+9 b)(11 a-9 b)\)

11 a \((x-\sqrt{11})(x+\sqrt{11})\)

b \((x-\sqrt{15})(x+\sqrt{15})\)

c \((x-2 \sqrt{7})(x+2 \sqrt{7})\)

d \((x-4 \sqrt{2})(x+4 \sqrt{2})\)

\[
\begin{aligned}
& 12 \text { a }(x+4-\sqrt{2})(x+4+\sqrt{2}) \\
& \text { b }(x-3-\sqrt{5})(x-3+\sqrt{5}) \\
& \text { c }(2 x+1-2 \sqrt{5})(2 x+1+2 \sqrt{5}) \\
& \text { d }(3-2 x-\sqrt{10})(3-2 x+\sqrt{10})
\end{aligned}
\]

a \(2(x-6)(x+11)\)
b \(3(x-7)(x+11)\)
c \(3(x-7)(x+1)\)
d \(3(x-5)(x+2)\)
e \(3(x-5)(x-6)\)
f \(6(x-1)(x+2)\)
18 a \(x=4\)
b \(x=19\)
c \(x=11\)
d \(x=\frac{6}{7}\)
e \(x=10\)
f \(x=\frac{68}{11}\)
g \(x=-\frac{86}{37}\)
h \(x=33\)
a 5
b 70 bottles of each

\section*{Challenge exercise}
1 a \((x+10 y)(5 x+4 y)\)
b \((8 x+y)(2 x+3 y)\)
2 a \((a-x+2 b)(a-x-2 b)\)
b \((3 a-x+c)(3 a+x-c)\)
c \((b+d+a-c)(b+d-a+c)\)
d \((y+b+a+3 x)(y+b-a-3 x)\)
3 a \(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c\)
b \(343 a^{6}-27 x^{3}\)
4 a \(4 x^{2} y(7 x-5)(x+3)\)
b \((p+2 q)(p-2 q)(x+2 \sqrt{2} y)(x-2 \sqrt{2} y)\)

\(5 \quad x=6\)

6 a length \(18 \mathrm{~m}\), width \(12 \mathrm{~m}\) b length \(15 \mathrm{~m}\), width \(12 \mathrm{~m}\)

\(7 \quad \frac{a m}{b p} \mathrm{~km}\)

\(8 \quad \frac{c p}{p-q} \mathrm{~km}\)

\section*{Chapter 4 answers}
\section*{Exercise 4A}
1 a \(\sqrt{10} \approx 3.16\)
b 4
c 7
2 a \(\sqrt{5} \approx 2.24\)
b \(\sqrt{29} \approx 5.39\)
c \(4 \sqrt{5} \approx 8.94\)
d \(\sqrt{89} \approx 9.43\)
e \(5 \sqrt{2} \approx 7.07\)
f 10
3 a \((1,4)\)
b \(\left(\frac{1}{2}, 5\right)\)
c \((4,6)\)
d \((-3,-7)\)
e \(\left(\frac{1}{2}, 6\right)\)
f \((-5,12)\)

4 a \((6,5)\)
b i 5
ii 6

c \(A C\) and \(B C\) are equal, so \(\triangle A B C\) is isosceles.

5
a \(u=4\)
b \(v=-1\) or 9
c \(w=-5\) or 11

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-011}
\end{center}

6 a 5

b \(M\left(1 \frac{1}{2}, 2\right)\)
c i \(2 \frac{1}{2}\)
ii \(2 \frac{1}{2}\)
iii \(2 \frac{1}{2}\)

7 a \((-4,1) \quad\) b \((22,-10)\)

\(8 \quad P Q\) is perpendicular to \(P R\).

\(9 \quad X Y=X Z\)

\(10 A B=C D=D A=C A\)

\(11(-2,-3)\)

\(12 A B=C D=\sqrt{20}\) and \(B C=D A=\sqrt{50}\)

\section*{Exercise 4B}
1 a \(\frac{3}{4}\)
b 2
c -2
d \(-\frac{1}{3}\)
e \(\frac{5}{4}\)
f -1
g 1
h 0
i \(-\frac{13}{9}\)

2 both gradients \(=1\)

\(3 a=8 \quad 4 \quad b=-12\)

5 a 4

b -4

c 8
d 7
e 1
f -2

6 gradient of \(A B=8\), gradient of \(P Q^{\prime}=-\frac{1}{8}\)

7 a \(-\frac{1}{6}\)

b 2

c \(-\frac{2}{3}\)

d \(\frac{5}{4}\)

e 1

8 a 2

b 2

c \(-\frac{1}{2}\)

9

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(10)}
\end{center}

b gradient of \(A B=\) gradient of \(D C=0\); gradient of \(A D=\) gradient of \(B C=1\)

c \(\left(\frac{5}{2}, 1\right)\); Because the diagonals of a parallelogram bisect each other

\(10 \quad \mathbf{M}_{E F}=\mathrm{M}_{G H}=\frac{1}{4}\) and \(\mathrm{M}_{E H}=\mathrm{M}_{F G}=12\)

11 a \(\sqrt{2}\)

b 1

c 1

12 a gradient of \(A B=\frac{1}{2}\), gradient of \(B C=\frac{7}{5}\), points not collinear

b both gradients \(=3\), points are collinear

c both gradients \(=3\), points are collinear

d gradient of \(A B=\frac{11}{5}\), gradient of \(B C=\frac{13}{4}\), points not collinear
13 a 1
b 1
c \((6,-3)\)
d \(\left(5 \frac{1}{2}, 3 \frac{1}{2}\right)\)

\section*{Exercise 4C}
1 a gradient \(=4, y\)-intercept is 2

b gradient \(=-\frac{2}{3}, y\)-intercept is 5

c gradient \(=-7, y\)-intercept is 10

d gradient \(=-\frac{4}{11}, y\)-intercept is \(\frac{2}{3}\)

2 a \(y=8 x+3\)

b \(y=11 x+5\)

c \(y=-6 x-7\)

d \(y=-\frac{3}{4} x+\frac{2}{5}\)

3

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(12)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(7)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(6)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(1)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(8)}
\end{center}

k

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(3)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(9)}
\end{center}

\(\mathbf{f}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(11)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(5)}
\end{center}

j

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(4)}
\end{center}

I

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-012(2)}
\end{center}

\(\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}, \mathbf{i}\) and \(\mathbf{j}\) have gradient \(=0\).

\(\mathbf{e}, \mathbf{f}, \mathbf{g}, \mathbf{h}, \mathbf{k}\) and \(\mathbf{l}\) do not have a gradient.

4 a \(2 x+y-6=0\)

b \(2 x+3 y-33=0\)

c \(9 x+15 y+10=0\)

d \(-24 x+42 y-7=0\)

e \(2 x+5 y+4=0\)

f \(-4 x+10 y+3=0\)

g \(3 x-y-12=0\)

h \(9 x-16 y+36=0\)

5 a \(y=\frac{3}{2} x-3\)

b \(y=-\frac{5}{2} x-5\)

c \(y=\frac{2}{3} x-4\)

d \(y=\frac{1}{6} x-3\)

e \(y=\frac{2}{15} x-\frac{6}{5}\)

f \(y=\frac{2}{3} x+4\)

g \(y=-\frac{5}{4} x-\frac{5}{2}\)

h \(y=\frac{3}{2} x-6\)

i \(y=\frac{3}{5} x+3\)

6 a gradient \(=\frac{3}{2}, y\)-intercept is 6

b gradient \(=-4, y\)-intercept is -24

c gradient \(=-\frac{3}{8}, y\)-intercept is -6
d gradient \(=\frac{4}{7}, y\)-intercept is 8

e gradient \(=-\frac{11}{4}, y\)-intercept is 11

f gradient \(=2, y\)-intercept is 4

g gradient \(=\frac{3}{7}, y\)-intercept is 6

h gradient \(=\frac{2}{7}, y\)-intercept is -2

i gradient \(=-5, y\)-intercept is 20

7 a \(x\)-intercept is \(5, y\)-intercept is -10

b \(x\)-intercept is \(-\frac{11}{3}, y\)-intercept is 11

c \(x\)-intercept is \(16, y\)-intercept is 6

d \(x\)-intercept is \(-16, y\)-intercept is 20

e \(x\)-intercept is \(14, y\)-intercept is -6

f \(x\)-intercept is \(-\frac{11}{5}, y\)-intercept is \(\frac{11}{2}\)

a
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-013(3)}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-013}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-013(5)}
\end{center}

b
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-013(6)}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-013(8)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-013(2)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-013(1)}
\end{center}

\(9 \quad\) a \(x=1\)

b \(y=5\)

c \(x=-4\)

10
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-013(7)}

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-013(4)}
11 a \(a=-\frac{5}{2}\)
b \(b=10\)
d \(d=-14\)
e \(e=-\frac{3}{2}\)
12 a \(a=-4\)
b \(b=-8\)
d \(d=4\)
e \(e=-6\)
\(\begin{array}{ll}13 & 34\end{array}\)
\(14-\frac{19}{5}\)
\(15 \quad 12\)
\(16-6\)

c \(c=-\frac{9}{4}\)

f \(f=50\)

c \(c=-\frac{9}{2}\)

f \(f=-28\)

\section*{Exercise 4D}
1 a \(y=6 x-24\)
b \(y=-4 x+14\)
c \(y=\frac{1}{2} x+8 \frac{1}{2}\)
d \(y=-3 x+11\)
e \(y=6\)
f \(x=-7\)
2 a \(y=\frac{1}{3} x+8 \frac{2}{3}\)
b \(y=2 x\)
c \(y=-\frac{1}{2} x\)
3 a \(\left(\frac{5}{2}, \frac{5}{2}\right)\)
b 7
c \(y=7 x-15\)
d \(y=-\frac{1}{7} x+2 \frac{6}{7}\)
4 a \(y=-4 x-6\)
b \(y=-4 x+20\)
c \(y=5 x+28\)
d \(y=-2 x-12\)
5 a \(y=x-4\)
b \(y=-3 x-9\)
c \(y=-x+6\)
d \(y=3\)
e \(x=1\)
f \(y=\frac{9}{4} x-3\)
\(6 y=5 x-4\)
7 a \(y=\frac{1}{2} x+5\)
b \(y=-x+8\)
c \(y=\frac{1}{2} x+2\)
d -1
e \(2 \sqrt{2}\)
f \(2 \sqrt{2}\)

8
a i \(\left(3 \frac{1}{2}, 3 \frac{1}{2}\right)\)
ii \(\left(3 \frac{1}{2}, 3 \frac{1}{2}\right)\)
b \(m_{A C}=1, m_{B D}=-1\)

\section*{Exercise 4E}
a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-014}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-014(2)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-014(1)}
\end{center}

2
a \(x=-\frac{9}{7}, y=-\frac{27}{7}\)
b \(x=\frac{3}{2}, y=\frac{3}{4}\)
c \(x=-\frac{7}{5}, y=-\frac{9}{5}\)
d \(x=\frac{13}{5}, y=-\frac{8}{15}\)
e \(x=-\frac{1}{7}, y=\frac{9}{7}\)
f \(x=\frac{5}{2}, y=\frac{3}{2}\)

3
a \(x=4, y=1\)
b \(x=1, y=2\)
c \(x=3, y=-2\)
d \(x=-1, y=2\)

e \(x=5, y=2\)

f \(x=-3, y=7\)

g \(x=1, y=2\)

h \(x=1, y=-1\)

i \(x=-4, y=3\)

j \(x=2, y=-1\)

k \(x=\frac{60}{17}, y=\frac{10}{17}\)

l \(x=-\frac{5}{13}, y=-\frac{46}{13}\)

m \(x=3, y=5\)

o \(x=-\frac{30}{19}, y=-\frac{121}{19}\)

n \(x=-\frac{6}{7}, y=\frac{24}{7}\)

4 a \(x=-\frac{28}{33}, y=-\frac{173}{33}\)

b \(x=\frac{864}{101}, y=-\frac{35}{101}\)

c \(x=-\frac{97}{6}, y=\frac{38}{3}\)

d \(x=-\frac{63}{19}, y=\frac{30}{19}\)

e \(x=\frac{9}{13}, y=-\frac{51}{13}\)

f \(x=\frac{552}{59}, y=\frac{60}{59}\)

g \(x=\frac{385}{24}, y=-\frac{19}{12}\)

h \(x=\frac{10}{11}, y=\frac{8}{11}\)

i \(x=\frac{168}{25}, y=-\frac{24}{25}\)

j \(x=0, y=-6\)

k \(x=26, y=\frac{31}{3}\)

l \(x=-\frac{220}{17}, y=\frac{64}{17}\)

\(5 y=3 x-6\)

\(6 \begin{array}{ll}\text { a } y=0 & \text { b } x=2\end{array}\)

\(7 \quad A(-1,-1), B(3,3), C\left(\frac{11}{5}, \frac{27}{5}\right)\)

\(8 \quad y=2 x+2, x+y=9,\left(\frac{7}{3}, \frac{20}{3}\right)\)

9 a i \(y=\frac{1}{2} x+\frac{7}{2} \quad\) ii \(y=-2 x+16\)

b \((5,6)\)

10 a i \(y=x \quad\) ii \(y=-x\)

b \((0,0)\)

\(11 m=-\frac{27}{4}\)

12 a \(y=\frac{1}{2} x-1 \quad\) b \(y=-2 x+19 \quad\) c \(\left(\frac{26}{5}, \frac{43}{5}\right)\)
d \(\frac{14 \sqrt{5}}{5} \approx 6.26\)
\(\begin{array}{ll}\text { e } 5 \sqrt{5} \approx 11.18 & \text { f } 70\end{array}\)

\(13 a=4, b=-15\)

14 The point is \((11,5)\)

\(\begin{array}{ll}\text { a } y=3 x & \text { b } y=x\end{array}\)

\(16 a=9, b=-6\)

\(17\left(\frac{21}{5}, \frac{26}{5}\right)\)

18

\(B D: 6 y-x=6 ; A C: y+6 x=21\)
b \(\left(\frac{120}{37}, \frac{57}{37}\right)\)
c \(\frac{9 \sqrt{37}}{37}\)
d 9

\section*{Exercise 4F}
\(1 \quad 45\) and 67

\(2 \quad 46\) and 37

3 The daughter is 8 and the father is 36 .

\(458 \$ 2\) items, \(43 \$ 5\) items

5 Andrew is 36 and Brian is 14

610 metres and 14 metres

7 standard: \(5 \mathrm{~kg}\), jumbo: \(7.5 \mathrm{~kg}\)

\(88000 \$ 60\) tickets sold and \(2000 \$ 80\) tickets sold
\(9 \quad 1.875 \mathrm{~L}\)
105 hours
11 3:20 p.m.

1250 tonnes of the \(52 \%\) iron alloy, 150 tonnes of the \(36 \%\) iron alloy

\(13480 \mathrm{~km} / \mathrm{h}\) and \(640 \mathrm{~km} / \mathrm{h}\)

14 model horse \(\$ 30.00\); model cow \(\$ 10.00\)

\(15 \frac{2}{15}\)

\(1630 \mathrm{~km}\)

\section*{Review exercise}
1 a \(2 \sqrt{17}\)
b 6
c 12
d \(\sqrt{26}\)
e \(\sqrt{34}\)
f \(4 \sqrt{5}\)
2 a \(\left(\frac{3}{2}, 1\right)\)
b \(\left(-1,4 \frac{1}{2}\right)\)
c \(\left(-\frac{1}{2}, 0\right)\)
d \(\left(\frac{9}{2}, 5\right)\)
e \(\left(-\frac{3}{2}, \frac{13}{2}\right)\)
f \(\left(\frac{9}{2},-3\right)\)
3 a 4
b 3
c 3
d 1
e \(\frac{2}{3}\)
f \(-\frac{5}{2}\)
g -3
h 0
i 0
\(4 \quad-4\)
5 a gradient \(=2, y\)-intercept is 4
b gradient \(=1, y\)-intercept is -4
c gradient \(=\frac{1}{2}, y\)-intercept is 1
d gradient \(=-2, y\)-intercept is 5
e gradient \(=-1, y\)-intercept is 6
f gradient \(=-1, y\)-intercept is 2
g gradient \(=-3, y\)-intercept is 4
h gradient \(=-3, y\)-intercept is 4
i gradient \(=\frac{2}{3}, y\)-intercept is -2

j gradient \(=-\frac{3}{4}, y\)-intercept is 3

\(\mathbf{k}\) gradient \(=2, y\)-intercept is 0

1 gradient \(=3, y\)-intercept is 0

\(\mathbf{m}\) gradient \(=-4, y\)-intercept is 0

n gradient \(=\frac{3}{2}, y\)-intercept is 0

o gradient \(=-\frac{3}{2}, y\)-intercept is 3

6

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015(8)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015(3)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015(5)}
\end{center}

7 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015(4)}
\end{center}

c

\begin{center}
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\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015(7)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015(1)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015(10)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-015(2)}
\end{center}

f

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-015(9)}
g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-016(3)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-016(6)}
\end{center}

k

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-016(4)}
\end{center}

m

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-016(1)}
\end{center}

o

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-016(5)}
\end{center}

\(\mathbf{q}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-016(2)}
\end{center}

8 a \(6 x+y=11\)

b \(2 x-3 y+14=0\)

9 a \(4 x+9 y=74\)

b \(4 x+5 y=32\)

c \(4 x-3 y+26=0\)

10 a \(x=4 \frac{1}{8}, y=-1 \frac{7}{8}\)

h

j

I

n

p

\(\mathbf{r}\)

b \(x=1 \frac{7}{8}, y=1 \frac{7}{8}\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-016(7)}

b i After 1 hour \(52 \frac{1}{2}\) minutes, \(150 \mathrm{~km}\) from Mildura ii After \(2 \frac{1}{2}\) hours, \(200 \mathrm{~km}\) from Mildura

iii \(20 \mathrm{~km}\)

11 a \(x=-3, y=-7\)

b \(x=-2, y=8\)

c \(x=5, y=0\)

d \(x=1, y=1\)

e \(x=\frac{156}{11}, y=\frac{42}{11}\)

f \(x=\frac{138}{65}, y=-\frac{3}{13}\)

12 a i \(y=-\frac{5}{6} x+\frac{139}{12}\)

ii \(y=-\frac{3}{11} x+\frac{69}{11}\)

b \(\left(\frac{701}{74}, \frac{273}{74}\right)\)

\(13\left(\frac{57}{13},-\frac{14}{13}\right)\)

\(14 y=x+3 ; y=x-2 ; 3 x+2 y=21 ; 3 x+2 y=6\)

15 The gradient of \(A C=-1\) and the gradient of \(A B=1\).

16 b \(12 \sqrt{2}, 4 \sqrt{2}\)

\(17 \quad 113\) and 25

18 stool \(\$ 30\); chair \(\$ 100\)
19 a \(p=\frac{5}{2}\)
b \(p=-23\)

\(204 x+3 y+5=30\)

21 The point is \((4,3)\).

\section*{Challenge exercise}
1 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-016}
\end{center}

2 a \(\left(\frac{1}{4}\left(x_{1}+3 x_{2}\right), \frac{1}{4}\left(y_{1}+3 y_{2}\right)\right)\)

b \(\left(\frac{1}{5}\left(2 x_{1}+3 x_{2}\right), \frac{1}{5}\left(2 y_{1}+3 y_{2}\right)\right)\)

\(3\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)

\(4 \mathbf{a}\left(\frac{c}{2}, \frac{b}{2}\right)\)
b i \(\frac{\sqrt{c^{2}+b^{2}}}{2}\)
ii \(\frac{\sqrt{b^{2}+c^{2}}}{2}\)
iii \(\frac{\sqrt{c^{2}+b^{2}}}{2}\)

5 a i \(y=\frac{b}{a+c} x\)

ii \(y=\frac{b}{a-c}(x-c)\)

b i \(\left(\frac{a+c}{2}\right), \frac{b}{2} \quad\) ii \(\left(\frac{a+c}{2}, \frac{b}{2}\right)\)

\(6 \begin{array}{lll}\text { a i } \frac{b}{a+c} & \text { ii } \frac{b}{a-c} & \text { c } \sqrt{a^{2}+b^{2}}\end{array}\)
10 a \(\frac{p}{m} ;-\frac{p}{\ell}\)
b \(y=-\frac{m}{p} x+\frac{m \ell}{p}\)
c \(y=\frac{1}{p} x+\frac{\ell m}{p}\)

\section*{Chapter 5 answers}
\section*{Exercise 5A}
1 a \(0,-3\)
d \(3,-6\)
b 0,7
c \(0,-5\)
g \(0,-\frac{4}{5}\)
e \(-7,-9\)
f 10,7
j \(\frac{3}{2},-\frac{4}{3}\)
h \(-\frac{3}{4}, \frac{2}{3}\)
i \(-\frac{7}{2},-4\)
2 a 0,5
k \(\frac{5}{2},-4\)
l \(\frac{2}{3}, \frac{4}{3}\)
d 0,25
b \(0,-7\)
c \(0,-8\)
3 a \(-1,-8\)
e \(0,-18\)
f \(0, \frac{1}{2}\)
b \(-2,-6\)
c \(-3,-9\)
d -6
e 2,4
f \(2,-3\)
g \(5,-6\)
h \(5,-8\)
i \(6,-10\)
j 1,6
k 3,4
l 5
m 2, 16
n \(-3,7\)
o 10
4 a 0,8
d no solution
b 1,16
c 1,2
g \(0,-3\)
e 3,5
f no solution
j \(-10,1\)
h \(-5,4\)
i no solution
l \(-1,1\)
5 a \(-1,-2\)
b 3,2
c \(-7,5\)
6 a \(-\frac{3}{2},-4\)
b \(-\frac{1}{3},-4\)
c \(-\frac{3}{2},-2\)
d \(-\frac{1}{2}, 2\)
e \(\frac{3}{2}, 3\)
f \(\frac{4}{3}, 2\)
g \(-\frac{3}{2},-\frac{4}{5}\)
h \(\frac{3}{2}, \frac{4}{3}\)
i \(-\frac{1}{2}, \frac{5}{4}\)
j \(-\frac{2}{3}, \frac{3}{4}\)
k \(-\frac{1}{4}, \frac{2}{3}\)
l \(\frac{2}{3},-\frac{5}{2}\)
\(\mathbf{m}-\frac{2}{3}, 7\)
n \(-\frac{2}{5},-3\)
o \(-\frac{15}{12}, 3\)
p \(\frac{1}{15},-15\)
q \(\frac{9}{8}, \frac{8}{9}\)
r \(-\frac{2}{3}, \frac{3}{2}\)
7 a \(\frac{1}{2}, \frac{4}{3}\)
b \(-\frac{1}{6}, \frac{5}{4}\)
c \(1, \frac{3}{4}\)
d \(-2, \frac{3}{2}\)
e \(\frac{2}{3}, \frac{5}{2}\)
f \(-\frac{1}{2}, \frac{4}{3}\)
g \(\frac{5}{2}, \frac{2}{5}\)
h \(-\frac{1}{4},-\frac{3}{5}\)
i \(2, \frac{1}{2}\)
j \(3,-\frac{5}{4}\)
k \(-\frac{2}{7}, \frac{1}{3}\)
I \(\frac{5}{4},-7\)

\section*{Exercise 5B}
1 a \(9,-2\)
b \(8,-4\)
c \(-\frac{3}{2}, 4\)
d \(-5,5\)
e no solution
f \(-3,-1\)
g \(0,-5\)
h 5,1
i \(8,-8\)
j \(-\frac{4}{3}, \frac{3}{2}\)
k \(\frac{3}{2},-\frac{7}{5}\)
l 4,8
2 a \(-7,2\)
b \(5,-3\)
c \(-3,2\)
d \(2,-1\)
e 6,1
f 16,2
g \(-6,5\)
h \(-5,4\)
i \(-\frac{1}{2},-4\)
j 4
\(\mathbf{k}-13,11\)
l \(-\frac{1}{2}, 3\)
m \(-\frac{5}{4},-\frac{7}{6}\)
n \(-\frac{7}{3}, 3\)
3 a \(\frac{50}{x}\)
b \(\frac{50}{x}=x+5, x=5\)

\section*{Exercise 5C}
\(\begin{array}{llllllll}1 & 11 \text { sides } & \mathbf{2} & 12 \mathrm{~cm} & \mathbf{3} & 16 & \mathbf{4} & 5\end{array}\)

\(5 \quad 9\) and -7 or 7 and \(-9 \quad 6 \quad 8 \mathrm{~cm}\)

\(74 \mathrm{~cm}, 32 \mathrm{~cm}, 42 \mathrm{~cm} \quad \mathbf{8} 5\) and 7

\(96 \mathrm{~km} / \mathrm{h}\)

10 a \(20-x \mathrm{~cm}\) b \(14 \mathrm{~cm}, 6 \mathrm{~cm}\)

\(114 \mathrm{~m}\)

\(128 \mathrm{~cm}, 12 \mathrm{~cm}, 4 \sqrt{3} \mathrm{~cm}\)

\(1325 \mathrm{~km} / \mathrm{h}\)

\(1420 \mathrm{~cm}\) and \(25 \mathrm{~cm}\)

\section*{Exercise 5D}
\begin{center}
\begin{tabular}{ccc}
1 -1 & b -2 & c -4 \\
g \(\frac{2}{5}\) & e \(-\frac{1}{2}\) & f \(\frac{3}{2}\) \\
j \(-\frac{1}{5}\) & h \(-\frac{3}{4}\) & i \(\frac{1}{3}\) \\
\(\mathbf{m} \frac{5}{7}\) & k \(\frac{2}{5}\) & l \(-\frac{2}{7}\) \\
\end{tabular}
\end{center}

2 a parts \(\mathbf{i}, \mathbf{v}, \mathbf{i x}\)

b parts ii, iv, vii, \(\mathbf{x}\), xii
3 a 4
b 16
c 25
d 36
e 100
f \(\frac{9}{4}\)
g \(\frac{1}{4}\)
h \(\frac{49}{4}\)
i \(\frac{121}{4}\)

\[
\begin{array}{ll}
\text { a }(x+3)^{2}+1 & \text { b }(x+4)^{2}-21 \\
\text { c }(x+6)^{2}-46 & \text { d }(x-5)^{2}-19 \\
\text { e }(x-3)^{2}-17 & \text { f }(x-10)^{2}-95 \\
\text { g }\left(x+\frac{3}{2}\right)^{2}-\frac{17}{4} & \text { h }\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4} \\
\text { i }\left(x-\frac{5}{2}\right)^{2}-\frac{1}{4} & \text { j }\left(x-\frac{1}{2}\right)^{2}-\frac{41}{4} \\
\text { k }\left(x+\frac{3}{2}\right)^{2}+\frac{19}{4} & \text { l }\left(x-\frac{11}{2}\right)^{2}-\frac{117}{4} \\
\text { 5 } 3(x+1)^{2}+9 & \text { b } 5(x+3)^{2}-35 \\
\text { c } 3(x-2)^{2}+3 & \text { d }-(x+1)^{2}+5 \\
\text { e }-(x-4)^{2}+6 & \text { f }-(x+3)^{2}+13 \\
\text { g } 3(x-1)^{2}-4 & \text { h } 2(x-3)^{2}+15 \\
\text { i } 4(x-6)^{2}-45 & \text { j } 2\left(x+\frac{3}{4}\right)^{2}+\frac{7}{8} \\
\text { k } 4\left(x-\frac{1}{8}\right)^{2}-\frac{65}{16} & \text { l } 3\left(x-\frac{4}{3}\right)^{2}+\frac{11}{3} \\
\text { m } 5\left(x-\frac{1}{10}\right)^{2}+\frac{19}{20} & \text { n } 2\left(x-\frac{5}{4}\right)^{2}-\frac{81}{8} \\
\text { o }-3\left(x+\frac{1}{6}\right)^{2}+\frac{49}{12} &
\end{array}
\]

\section*{Exercise 5E}
1 a \(\pm \sqrt{5}\)
b \(\pm \sqrt{11}\)
c \(\pm \sqrt{3}\)
d \(\pm \sqrt{2}\)
e \(\pm \sqrt{10}\)
f \(\pm \sqrt{5}\)
2 a \(-1 \pm \sqrt{2}\)
b \(-2 \pm \sqrt{3}\)
c \(6 \pm \sqrt{13}\)
d \(-3 \pm \sqrt{2}\)
e \(4 \pm \sqrt{17}\)
f \(-4 \pm 2 \sqrt{5}\)
g \(-5 \pm 2 \sqrt{6}\)
h \(-6 \pm \sqrt{41}\)
i \(5 \pm 5 \sqrt{3}\)
j \(-10 \pm \sqrt{95}\)
k \(50 \pm 2 \sqrt{645}\)
l \(25 \pm \sqrt{615}\)
3 a \(\frac{-1 \pm \sqrt{5}}{2}\)
b \(\frac{3 \pm \sqrt{5}}{2}\)
c \(\frac{5 \pm \sqrt{29}}{2}\)
d \(\frac{-3 \pm \sqrt{17}}{2}\)
e \(\frac{-5 \pm \sqrt{41}}{2}\)
f \(\frac{3 \pm \sqrt{29}}{2}\)
g \(\frac{7 \pm \sqrt{449}}{2}\)
h \(\frac{3 \pm \sqrt{33}}{2}\)
i \(\frac{9 \pm \sqrt{101}}{2}\)
j \(\frac{1 \pm \sqrt{21}}{2}\)
k \(\frac{3 \pm \sqrt{5}}{2}\)
1 \(\frac{5 \pm \sqrt{13}}{2}\)
b \(-1 \pm \sqrt{5}\)
c \(-1 \pm \sqrt{5}\)
d \(4 \pm \sqrt{6}\)
e \(-3 \pm \sqrt{21}\)
f \(-4 \pm 2 \sqrt{5}\)
g \(2,-\frac{1}{2}\)
h \(\frac{4 \pm \sqrt{34}}{3}\)
i \(\frac{1 \pm \sqrt{65}}{8}\)
j \(\frac{-1 \pm \sqrt{21}}{10}\)
k \(3,-\frac{1}{2}\)
l \(\frac{-5 \pm \sqrt{70}}{3}\)
\(\mathbf{m}-1 \pm \sqrt{7}\)
n \(1 \pm \sqrt{5}\)
o \(\sqrt{2} \pm 2\)

\begin{center}
\begin{tabular}{lll}
5 \(-3 \pm \sqrt{17}\) & b \(5,-2\) & c \(1,-7\) \\
d \(2 \pm \sqrt{7}\) & e \(2,-3\) & f \(\frac{1 \pm \sqrt{13}}{2}\) \\
g \(-2,-\frac{1}{2}\) & h \(1,-\frac{1}{3}\) & i \(-1 \pm \sqrt{6}\) \\
j \(-3 \pm \sqrt{14}\) & k no solution & l no solution \\
\(\mathbf{m} \pm \frac{5}{2}\) & \(\mathbf{n} \pm \frac{1}{3}\) & o \(-7,5\) \\
p \(-3,4\) & q \(\pm \frac{\sqrt{5}}{2}\) & r no solution \\
s \(-\frac{3}{2}, \frac{4}{3}\) & t \(-\frac{5}{3},-\frac{1}{4}\) & u -3 \\
v \(-1 \pm \frac{\sqrt{3}}{3}\) & w \(2 \pm \frac{\sqrt{6}}{2}\) & x \(-\frac{1 \pm \sqrt{26}}{5}\) \\
y \(-\frac{2}{3}, \frac{1}{4}\) & z no solution &  \\
a \(-1 \pm \sqrt{6}\) & b \(1 \pm \sqrt{2}\) & c \(-2 \pm \sqrt{11}\) \\
d \(-3 \pm \sqrt{5}\) & e \(\frac{1 \pm \sqrt{5}}{2}\) & f \(3, \frac{-3}{2}\) \\
\end{tabular}
\end{center}

\section*{Exercise 5F}
1 - \(4 \pm \sqrt{15}\)

b \(-2,4\)

c \(\frac{3 \pm \sqrt{13}}{2}\)

d \(-2,6\)

e \(\frac{-5 \pm \sqrt{17}}{2}\)

f \(\frac{-9 \pm \sqrt{69}}{2}\)

g \(4 \pm \sqrt{14}\)

h \(-1 \pm \sqrt{5}\)

i \(-6 \pm \sqrt{33}\)

2 a \(\frac{-1 \pm \sqrt{22}}{3}\)

b \(\frac{-3 \pm \sqrt{29}}{10}\)

c \(\frac{3 \pm \sqrt{5}}{4}\)

d \(1, \frac{2}{7}\)

e \(-1, \frac{2}{5}\)

f \(\frac{1 \pm \sqrt{29}}{14}\)

g \(\frac{-6 \pm \sqrt{38}}{2}\)

h \(\frac{10 \pm \sqrt{106}}{3}\)

i \(\frac{2 \pm \sqrt{19}}{3}\)

3 a \(1.53,-0.13\)

b \(7.87,0.13\)

c \(-2,5\)

d \(-0.20,-14.80\)

e 2,3

f \(-0.41,3.41\)

g \(-0.35,2.85\)

h \(-0.24,0.84\)

i \(0.15,3.35\)

4 a i no solutions

ii one solution
b i 2 , irrational

iii 1 , rational

v 0

vii 2 , irrational
5 a \(\frac{1+\sqrt{5}}{2}\)
b \(3.52 \mathrm{~cm}\)
\(6 \quad A P=12+4 \sqrt{5} \mathrm{~cm}, B P=4+4 \sqrt{5} \mathrm{~cm}\)
\(\begin{array}{ll}7 & 50\end{array}\)
\(8 \frac{5+5 \sqrt{7}}{2}\) and \(\frac{-5+5 \sqrt{7}}{2}\) or \(\frac{5-5 \sqrt{7}}{2}\) and \(\frac{-5-5 \sqrt{7}}{2}\)
\(9 \quad 4.47\)
\(105(\sqrt{201}+1) \mathrm{km} / \mathrm{h} \approx 75.89 \mathrm{~km} / \mathrm{h}\)
\(1150-\sqrt{2095}\) and \(50+\sqrt{2095} \approx 4.23 \mathrm{~m}\) and \(95.77 \mathrm{~m}\)
12 a \(x=4\) or \(x=7\)
b \(x=6-2 \sqrt{10}\) or \(x=6+2 \sqrt{10}\)
c \(x=\frac{-1-\sqrt{19}}{3}\) or \(x=\frac{-1+\sqrt{19}}{3}\)

ii 2 , rational

iv 1 , rational

vi 0

viii 2 , irrational

\section*{Review exercise}
1 a \(6,-3\)
b \(-2, \frac{1}{3}\)
c \(-1, \frac{1}{2}\)
d \(-\frac{3}{2}, \frac{1}{3}\)
e \(-\frac{3}{2}, 4\)
f \(-\frac{1}{2},-4\)
g \(-2,4\)
h \(-1,-3\)
i \(\frac{5}{2},-1\)
j \(1,-\frac{5}{7}\)
k \(\frac{1}{2}, 5\)
l \(-\frac{1}{2},-5\)
m \(-\frac{1}{2}, 3\)
n \(\frac{1}{3}, 3\)
o \(\frac{1}{3}\)
p \(\frac{7}{3}\)
q \(\frac{7}{3},-\frac{7}{2}\)
2 a 3,5
b 5,6
c \(2 \pm \sqrt{3}\)
d \(-1 \pm \sqrt{2}\)
e \(\frac{-1 \pm \sqrt{13}}{2}\)
f \(10 \pm \sqrt{107}\)
g \(\frac{3 \pm \sqrt{37}}{2}\)
h \(3,-1\)
i \(-1 \pm \sqrt{33}\)
j \(2, \frac{1}{2}\)
k \(\frac{-1 \pm \sqrt{37}}{6}\)
1 \(\frac{3 \pm \sqrt{41}}{8}\)

3 a \(6,-4 \quad\) b \(-2, \frac{1}{2}\)

c \(\frac{-1 \pm \sqrt{5}}{2} \quad\) d \(\frac{-5 \pm \sqrt{17}}{4}\)

e \(\frac{-1 \pm \sqrt{7}}{2} \quad\) f \(\frac{1 \pm \sqrt{13}}{6}\)

\(4 \begin{array}{ll}\text { a } 1 \pm \sqrt{2} & \text { b } \frac{-5 \pm \sqrt{57}}{4}\end{array}\)

c \(\frac{1 \pm \sqrt{17}}{8} \quad\) d \(\frac{1 \pm \sqrt{17}}{2}\)

e \(1,-\frac{7}{6} \quad\) f \(\frac{1}{3},-1\)

5 a \(\frac{7+\sqrt{13}}{2}, \frac{7-\sqrt{13}}{2}\)

b \(\frac{-7-\sqrt{73}}{4}, \frac{-7+\sqrt{73}}{4}\)

c \(\frac{1+\sqrt{51}}{10}, \frac{1-\sqrt{51}}{10} \quad\) d \(\frac{4+\sqrt{6}}{5}, \frac{4-\sqrt{6}}{5}\)

e \(\frac{3+\sqrt{21}}{4}, \frac{3-\sqrt{21}}{4} \quad\) f \(\frac{9+\sqrt{113}}{4}, \frac{9-\sqrt{113}}{4}\)

g \(\frac{5+\sqrt{33}}{4}, \frac{5-\sqrt{33}}{4} \quad\) h \(\frac{-2+\sqrt{10}}{3}, \frac{-2-\sqrt{10}}{3}\)

6 a -7 and 9 or 7 and -9

b length \(16 \mathrm{~cm}\), width \(9 \mathrm{~cm}\)

c 3,5 or \(-3,-5\)

\(7 \quad x=2 \sqrt{5}\)

8 train \(28 \mathrm{~km} / \mathrm{h}\), car \(49 \mathrm{~km} / \mathrm{h}\)

\(9(10+5 \sqrt{2}) \mathrm{cm}\) and \((10-5 \sqrt{2}) \mathrm{cm}\)

\(106 \mathrm{~cm}^{2}\) or \(\frac{28}{27} \mathrm{~cm}^{2}\)

\(11 A P=4 \mathrm{~cm}\)

\section*{Challenge exercise}
\(1 \pm 2, \pm 3\)
\(2-1,-2,3,4\)
\(4 x^{2}-4 x+1=0\)
\(5-\frac{7}{2}, \frac{17}{2}\)
\(\begin{array}{ll}6 & 7\end{array}\)
\(7 \quad 10+2 \sqrt{21}\)
8 a AA
b \(\frac{-1+\sqrt{5}}{2}\)
d \(\frac{\sqrt{2(5+\sqrt{5})}}{4}\)
9 a \(2 x^{2}+6 x-3=0\)
b \(\frac{-3+\sqrt{15}}{2}\)
c \(\frac{-1+\sqrt{5}}{2}\)
\(100,5 a\)

\section*{Chapter 6 answers}
\section*{Exercise 6A}
1 a \(318 \mathrm{~cm}^{2}\)

c \(192 \pi \mathrm{cm}^{2} \approx 603.2 \mathrm{~cm}^{2}\)

b \(448 \mathrm{~cm}^{2}\)

e \(4140 \mathrm{~cm}^{2}\)

d \(54+36 \pi \mathrm{cm}^{2} \approx 167.1 \mathrm{~cm}^{2}\)

g \(1000 \mathrm{~cm}^{2}\)

f \(100+37.5 \pi \mathrm{m}^{2} \approx 217.8 \mathrm{~m}^{2}\)

i \(194 \mathrm{~cm}^{2}\)

h \(596 \mathrm{~cm}^{2}\)

j \(144+40 \sqrt{2} \approx 200.6 \mathrm{~cm}^{2}\)

2 a \(3 \sqrt{3} \approx 5.1962 \mathrm{~cm}\)

b \(216+18 \sqrt{3} \approx 247.1769 \mathrm{~cm}^{2}\)

\(39 \mathrm{~m}\)

\(4 \quad 4.5 \mathrm{~cm}\)

5 a \(1425 \pi \mathrm{cm}^{2} \approx 4476.8 \mathrm{~cm}^{2}\)

b \(\frac{350}{3 \pi} \mathrm{cm} \approx 37.1 \mathrm{~cm}\)

6 a \(280 \mathrm{~cm}^{3}\)

b \(216 \mathrm{~m}^{3}\)

c \(144 \mathrm{~m}^{3}\)

d \(28000 \mathrm{~mm}^{3}\)

e \(300 \pi \mathrm{cm}^{3} \approx 942.5 \mathrm{~cm}^{3}\)

f \(1215 \pi \mathrm{cm}^{3} \approx 3817.0 \mathrm{~cm}^{3} \quad\) g \(23040 \mathrm{~cm}^{3}\)

h \(375 \pi \mathrm{cm}^{3} \approx 1178.1 \mathrm{~cm}^{3} \quad\) i \(3240 \mathrm{~cm}^{3}\)

\(\begin{array}{lll}\text { j } 1080 \mathrm{~cm}^{3} & \text { k } 480 \mathrm{~cm}^{3} & \text { l } 360 \mathrm{~cm}^{3}\end{array}\)

\(\begin{array}{lll}7 & \text { a } 5.29 \mathrm{~cm} & \text { b } 761.98 \mathrm{~cm}^{3}\end{array}\)

\(\begin{array}{lll}8 & \text { a } \quad 19.9 \mathrm{~cm} & \text { b } 6.31 \mathrm{~cm}\end{array}\)

9 a \(4 \mathrm{~cm} \quad\) b \(7.4 \mathrm{~cm}^{3}\)

\(10 \quad 16.25 \mathrm{~m}^{2}\)

\(114 \mathrm{~cm}\)

\(12 \frac{1500}{2+3 \sqrt{2}} \mathrm{~cm} \approx 240.28 \mathrm{~cm}\)

13 a \(1099.56 \mathrm{~cm}^{3}\)

b \(1400 \mathrm{~cm}^{3}\)

c \(300.44 \mathrm{~cm}^{3}\)

\section*{Exercise 6B}
\(160 \mathrm{~cm}^{3}\)

2 a \(80 \mathrm{~cm}^{3}\)

b \(100 \mathrm{~cm}^{3}\)

c \(216 \mathrm{~cm}^{3}\)

d \(180 \mathrm{~cm}^{3}\)

3 a \(400 \mathrm{~cm}^{3}\)

b \(53 \frac{1}{3} \mathrm{~cm}^{3}\)

c \(80 \mathrm{~cm}^{3}\)

d \(96 \mathrm{~cm}^{3}\)

\(4 \quad 2547758.6 \mathrm{~m}^{3}\)

5 a \(800 \mathrm{~cm}^{3}\)

b \(2333.33 \mathrm{~cm}^{3}\)

6 a \(5 \mathrm{~cm} \quad\) b \(13 \mathrm{~cm} \quad\) c \(65 \mathrm{~cm}^{2} \quad\) d \(360 \mathrm{~cm}^{2}\)

7 a \(64+32 \sqrt{13} \approx 179.378 \mathrm{~cm}^{2}\)

b \(36+12 \sqrt{34} \approx 105.971 \mathrm{~cm}^{2}\)

8 a \(\sqrt{61} \mathrm{~cm} \approx 7.810 \mathrm{~cm} \quad\) b \(2 \sqrt{13} \mathrm{~cm} \approx 7.211 \mathrm{~cm}\)

c \((80+8 \sqrt{61}+20 \sqrt{13}) \mathrm{cm} \approx 214.593 \mathrm{~cm}^{2}\)

9 a \(\triangle V D A, \triangle V A B, \triangle V C B, \triangle V D C\) : all are right-angled triangles.
b i \(10 \mathrm{~cm}\)
ii \(2 \sqrt{41} \mathrm{~cm} \approx 12.806 \mathrm{~cm}\)

\section*{Exercise 6C}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-020}
\end{center}

\section*{Exercise 6D}
1 a \(804.25 \mathrm{~cm}^{2}\)
b \(2827.43 \mathrm{~cm}^{2}\)
c \(615.75 \mathrm{~cm}^{2}\)
d \(1385.44 \mathrm{~cm}^{2}\)

\(\begin{array}{ll}\text { a } 64 \pi \mathrm{cm}^{2} \approx 201.06 \mathrm{~cm}^{2} & \text { b } 128 \pi \mathrm{cm}^{2} \approx 402.12 \mathrm{~cm}^{2}\end{array}\)

c \(192 \pi \mathrm{cm}^{2} \approx 603.19 \mathrm{~cm}^{2}\)

\(3 \quad 6.3078 \mathrm{~cm}\)

\(\begin{array}{lll}4 & \text { a } 2.11 \mathrm{~m} & \text { b } 1.72 \mathrm{~m}\end{array}\)

\(5 \begin{array}{lll}\text { a } 195 \pi \mathrm{cm}^{2} & \text { b } 115 \pi \mathrm{cm}^{2} & \text { c }(440+16 \pi) \mathrm{cm}^{2}\end{array}\)

\(\begin{array}{llll}\text { 6 } & \text { a } 2144.66 & \text { b } 4188.79 & \text { c } 904.78\end{array}\)

\(\begin{array}{lll}\text { d } 452.39 \mathrm{~cm}^{3} & \text { e } 3619.11 \mathrm{~cm}^{3} & \text { f } 883.57 \mathrm{~cm}^{3}\end{array}\)

\(\begin{array}{llll}7 & \text { a } 6.20 \mathrm{~cm} & \text { b } 179 \mathrm{~mm} & \text { c } 9.8 \mathrm{~cm}\end{array}\)

\(\begin{array}{lll}\text { a } 368.61 \mathrm{~cm}^{3} & \text { b } 575.96 \mathrm{~cm}^{3} & \text { c } 329.87 \mathrm{~cm}^{3}\end{array}\)

\(9 \quad 8249 \mathrm{~cm}^{3}\)

\(10 \quad 545131.78 \mathrm{~mm}^{3}\)

\section*{Exercise 6E}
1 a i \(9 \quad\) ii 9

b i \(\frac{36}{25} \quad\) ii \(\frac{36}{25}\)

c i \(\frac{9}{4} \quad\) ii \(\frac{9}{4}\)

c \(192 \mathrm{~cm}^{2}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-021(3)}
\end{center}

9 a The area of the triangle increases by a factor of 7.
b 36
c 9
d 81

\section*{Review exercise}
1 a i \(1000 \mathrm{~cm}^{3}\)
ii \(600 \mathrm{~cm}^{2}\)
b i \(600 \mathrm{~cm}^{3}\)
ii \(460 \mathrm{~cm}^{2}\)
c i \(480 \mathrm{~cm}^{3}\)
ii \(528 \mathrm{~cm}^{2}\)
d i \(168 \pi \mathrm{cm}^{3}\)
ii \((36 \pi+12 \sqrt{58} \pi) \mathrm{cm}^{2}\)
e i \(90 \pi \mathrm{m}^{3}\)
ii \(78 \pi \mathrm{m}^{2}\)
f i \(\frac{2048}{3} \pi \mathrm{m}^{3}\)
ii \(256 \pi \mathrm{m}^{2}\)
2 a i \(550 \mathrm{~m}^{3}\)
ii \((375+25 \sqrt{17}) \mathrm{cm}^{2} \approx 478.1 \mathrm{~cm}^{2}\)
b i \(200 \pi \mathrm{cm}^{3} \approx 628.3 \mathrm{~cm}^{3}\)

\[
\text { ii }\left(\frac{340 \pi}{3}+120\right) \mathrm{cm}^{2} \approx 476.0 \mathrm{~cm}
\]

\(3 \quad 2 \mathrm{~cm}\)
4
a \(1500000 \mathrm{~m}^{3}\)
b \(80 \mathrm{~cm}^{3}\)
c \(\frac{250 \pi}{3} \mathrm{~cm}^{3} \approx 261.8 \mathrm{~cm}^{3}\)
d \(156 \pi \mathrm{cm}^{3} \approx 490.1 \mathrm{~cm}^{3}\)
\(5 \quad 8 \mathrm{~cm}\)
6 a i \(\frac{592 \pi}{3} \mathrm{~cm}^{3} \approx 619.94 \mathrm{~cm}^{3}\)
ii \(168 \pi \mathrm{cm}^{2} \approx 527.8 \mathrm{~cm}^{2}\)
b i \(\left(4+\frac{2 \pi}{3}\right) \mathrm{m}^{3} \approx 6.09 \mathrm{~m}^{3}\)
ii \((16+\pi) \mathrm{m}^{2} \approx 19.1 \mathrm{~m}^{2}\)
7
\(1.2 \mathrm{~m}\)

\[
\begin{aligned}
& 8 \text { a } 662.5 \mathrm{~cm}^{2} \\
& \text { b } 937.5 \mathrm{~cm}^{3} \\
& 9 \text { a } 80 \mathrm{~m}^{3} \\
& \text { b }(63+8 \sqrt{34}+3 \sqrt{89}) \mathrm{m}^{3} \approx 137.95 \mathrm{~m}^{2} \\
& 10 \text { a } V=\frac{250000}{3} \pi \mathrm{mm}^{3}, S=\left(20000+\frac{12500}{3} \pi\right) \mathrm{mm}^{2} \\
& \text { b } V=2250 \mathrm{~cm}^{3}, S=675 \pi \mathrm{m}^{2} \\
& \text { c } V=4 \pi \mathrm{m}^{3}, S=(6+\sqrt{13}) \pi \mathrm{m}^{2}
\end{aligned}
\]

\[
\text { d } V=\frac{875}{3} \pi \mathrm{m}^{3}, S=150 \pi \mathrm{m}^{2}
\]

\section*{Challenge exercise}
\(14 \mathrm{~cm}\)

\(36 \mathrm{~cm}\)

6 c Substitute the result of part \(\mathbf{b}\) into that of part \(\mathbf{a}\).

7 b \(7: 8\)

9 a Use similar triangles.

c Show \(s^{2}=h^{2}+(R-r)^{2}\)

10 a Use Pythagoras' theorem

c cross-section area \(=\) area of cylinder cross-section area of cone cross-section

d \(\pi r^{3}-\frac{1}{3} \pi r^{3}=\frac{2}{3} \pi r^{3}\)

e Volume of a sphere is twice the result in part \(\mathbf{d}\).

\section*{Chapter 7 answers}
\section*{Exercise 7A}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-021(2)}
\end{center}

3 a
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-021(1)}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-021}
\end{center}

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-021(4)}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(8)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(11)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(9)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(6)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(13)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(3)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(15)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(4)}
\end{center}

5 a \(y=-(x-3)^{2}\)

c \(y=-x^{2}-6\)

b \(y=-(x+b)^{2}\)

6 a \(y=(x+4)^{2}+3\)

d \(y=-x^{2}+c\)

c \(y=(x-a)^{2}+b\)

b \(y=(x-6)^{2}-5\)

d \(y=(x+d)^{2}-c\)

7

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-022(14)}
a \(y=x^{2}\)
b \(y=(x+5)^{2}-11\)

c \(y=(x+3-a)^{2}-8+b\)

\(\begin{array}{lll}\text { a } a=0 & \text { b } a=2 \quad \text { c } a=-1 \quad \text { d } a=-8\end{array}\)

9 a \(b=5\)

b \(b=7\)

c \(b=0\)

d \(b=-3\)

10 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(5)}
\end{center}

\section*{Exercise 7B}
1 a i 3

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(2)}
\end{center}

b i -7

iii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(7)}
\end{center}

c i 8

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(12)}
\end{center}

d i 2

iii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(10)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-022(1)}
\end{center}

ii \(x=0,(0,3)\)

iv no \(x\)-intercepts

ii \(x=0,(0,-7)\)

iv two \(x\)-intercepts

ii \(x=2,(2,4)\)

iv no \(x\)-intercepts

ii \(x=3,(3,-7)\)

iv two \(x\)-intercepts
e i

25

iii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(9)}
\end{center}

f i 48

iii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(11)}
\end{center}

g i \(\quad-9\)

iii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(7)}
\end{center}

h i 4

iii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(8)}
\end{center}

i i -5

iii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(10)}
\end{center}

2 a i 0

iii \(x=3,(3,-9)\)

v two \(x\)-intercepts

b i 0

iii \(x=-3,(-3,-9)\)

v two \(x\)-intercepts ii \(y=(x-3)^{2}-9\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(6)}
\end{center}

ii \(x=-5,(-5,0)\)

iv one \(x\)-intercept

ii \(x=7,(7,-1)\)

iv two \(x\)-intercepts

ii \(x=3,(3,0)\)

iv one \(x\)-intercept

ii \(x=-1,(-1,5)\)

iv two \(x\)-intercepts

ii \(x=2,(2,-1)\)

iv no \(x\)-intercepts

ii \(y=(x+3)^{2}-9\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(4)}
\end{center}

c i -4

iii \(x=-1,(-1,-5)\)

v two \(x\)-intercepts

d i -1

iii \(x=-2,(-2,-5)\)

v two \(x\)-intercepts

e i -3

iii \(x=-3,(-3,-12)\)

v two \(x\)-intercepts

\section*{f i -4}
iii \(x=-6,(-6,-40)\)

v two \(x\)-intercepts

g i 5

iii \(x=\frac{3}{2},\left(\frac{3}{2}, \frac{11}{4}\right)\)

v two \(x\)-intercepts

h i 2

iii \(x=\frac{5}{2},\left(\frac{5}{2},-\frac{17}{4}\right)\)

v two \(x\)-intercepts

i i 0

iii \(x=-\frac{7}{2},\left(-\frac{7}{2},-\frac{49}{4}\right)\)

v two \(x\)-intercepts ii \(y=(x+1)^{2}-5\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(1)}
\end{center}

ii \(y=(x+2)^{2}-5\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023}
\end{center}

ii \(y=(x+3)^{2}-12\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(5)}
\end{center}

ii \(y=(x+6)^{2}-40\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(3)}
\end{center}

ii \(y=\left(x-\frac{3}{2}\right)^{2}+\frac{11}{4}\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(13)}
\end{center}

ii \(y=\left(x-\frac{5}{2}\right)^{2}-\frac{17}{4}\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-023(2)}
\end{center}

ii \(y=\left(x+\frac{7}{2}\right)^{2}-\frac{49}{4}\)

iv

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-023(12)}
j i 0

ii \(y=-(x+1)^{2}+1\)

iii \(x=-1,(-1,1)\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-024(8)}
\end{center}

v two \(x\)-intercepts

k i 7

ii \(y=-(x-4)^{2}+23\)

iii \(x=4,(4,23)\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-024(3)}
\end{center}

v two \(x\)-intercepts
l i -15
ii \(y=-\left(x-\frac{5}{2}\right)^{2}-\frac{3}{4}\)

iii \(x=\frac{5}{2},\left(\frac{5}{2}, \frac{-3}{4}\right)\)

iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-024(7)}
\end{center}

v no \(x\)-intercepts

3
a \(-3,1\)
b \(-5,1\)

c \(-2,8\)

d \(4-\sqrt{7}, 4+\sqrt{7}\)

e \(-3-\sqrt{11},-3+\sqrt{11}\)

f \(1,-3\)

g \(2+2 \sqrt{2}, 2-2 \sqrt{2}\)

h \(5+3 \sqrt{2}, 5-3 \sqrt{2}\)

i \(3+5 \sqrt{2}, 3-5 \sqrt{2}\)

j \(-3+2 \sqrt{5},-3-2 \sqrt{5}\)

k \(-2+4 \sqrt{2},-2-4 \sqrt{2}\)

l \(3,-5\)

4 a \(-1-\sqrt{5},-1+\sqrt{5}\)

b \(3-\sqrt{2}, 3+\sqrt{2}\)

c \(4-\sqrt{3}, 4+\sqrt{3}\)

d \(-2-2 \sqrt{2},-2+2 \sqrt{2}\)

e \(-11,1\)

f \(10-5 \sqrt{6}, 10+5 \sqrt{6}\)

g \(-1,13\)

h \(3+\sqrt{5}, 3-\sqrt{5}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-024(1)}
\end{center}

ii \(y=(x+2)^{2}-9\)

iii \(x=-2,(-2,-9)\)

iv \(-5,1\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-024(2)}
\end{center}

b i 5

ii \(y=(x+2)^{2}+1\)

iii \(x=-2,(-2,1)\)

iv no \(x\)-intercepts

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-024(6)}
\end{center}

c i 9

ii \(y=(x+3)^{2}\)

iii \(x=-3,(-3,0)\)

iv -3

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-024}
d i -7
ii \(y=(x-3)^{2}-16\)

iii \(x=3,(3,-16)\)

iv \(-1,7\)

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-024(5)}
e i -3
ii \(y=(x+4)^{2}-19\)
iii \(x=-4,(-4,-19)\)
iv \(-4-\sqrt{19},-4+\sqrt{19}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-024(4)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-025(3)}
\end{center}

g i 10

iii \(x=-\frac{5}{2},\left(-\frac{5}{2}, \frac{15}{4}\right)\)

ii \(y=\left(x+\frac{5}{2}\right)^{2}+\frac{15}{4}\)

ii \(y=\left(x-\frac{1}{2}\right)^{2}-\frac{9}{4}\)

iv \(-1,2\)

\(\mathbf{v}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-025(6)}
\end{center}

iv no \(x\)-intercepts h i -3

iii \(x=-\frac{7}{2},\left(-\frac{7}{2},-\frac{61}{4}\right)\)

ii \(y=\left(x+\frac{7}{2}\right)^{2}-\frac{61}{4}\)

iv \(\frac{-7+\sqrt{61}}{2}, \frac{-7-\sqrt{61}}{2}\)

v

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-025(7)}
\end{center}

i i 4

ii \(y=(x-1)^{2}+3\)

iii \(x=1,(1,3)\)

iv no \(x\)-intercepts

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-025(4)}
\end{center}

j i 3

iii \(x=2,(2,7)\)

ii \(y=-(x-2)^{2}+7\)

\(\mathbf{v}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-025}
\end{center}

k i 6

ii \(y=-(x-6)^{2}+40\)

iii \(x=6,(6,40)\)

iv \(6-2 \sqrt{10}, 6+2 \sqrt{10}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-025(5)}
\end{center}

l i -2

ii \(y=-(x-1)^{2}-1\)

iii \(x=1,(1,-1)\)

iv no \(x\)-intercepts

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-025(2)}
\end{center}

mi 1

ii \(y=-\left(x-\frac{1}{2}\right)^{2}+\frac{5}{4}\)

iii \(x=\frac{1}{2},\left(\frac{1}{2}, \frac{5}{4}\right)\)

iv \(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\)

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-025(1)}
n i

\(-1\)

iii \(x=\frac{-5}{2},\left(\frac{-5}{2}, \frac{21}{4}\right)\)

ii \(-\left(x+\frac{5}{2}\right)^{2}+\frac{21}{4}\)

v

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-026(8)}
\end{center}

o i 20

ii \(y=-\left(x-\frac{11}{2}\right)^{2}+\frac{201}{4}\)

iii \(x=\frac{11}{2},\left(\frac{11}{2}, \frac{201}{4}\right)\)

iv \(\frac{11-\sqrt{201}}{2}, \frac{11+\sqrt{201}}{2}\)

\(\mathbf{v}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-026(9)}
\end{center}

\section*{Exercise 7C}
1 a

\begin{center}
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b

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c

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2 a

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b

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c

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d

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e

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\(3 \mathbf{a}, \mathbf{d}, \mathbf{e}, \mathbf{g}\) are congruent; \(\mathbf{b}, \mathbf{f}, \mathbf{l}\) are congruent; \(\mathbf{c}, \mathbf{h}\) are congruent; \(\mathbf{j}\) and \(\mathbf{k}\) are congruent

4 a \(y=3(x+1)^{2}\)

b \(y=3\left(x-\frac{3}{2}\right)^{2}+\frac{41}{4}\)

c \(y=2\left(x+\frac{5}{2}\right)^{2}-\frac{51}{2}\)

d \(y=2\left(x+\frac{7}{4}\right)^{2}-\frac{25}{8}\)

e \(y=3\left(x+\frac{5}{6}\right)^{2}+\frac{59}{12}\)

f \(y=-5(x-2)^{2}+57\)

ICE-EM MATHEMATICS YEAR 10 ICE-EM Mathematics 103 d

ISBNAR Photocopying is restricted under law and this material must not be transferred to another party

\section*{\(\begin{array}{lll}5 & \mathbf{a} & \mathbf{i} \\ & 3\end{array}\)}
iii \(x=-1,(-1,1)\)

v

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b i 7

iii \(x=-1,(-1,9)\) ii \(y=-2(x+1)^{2}+9\)

iv \(\frac{-2+3 \sqrt{2}}{2}, \frac{-2-3 \sqrt{2}}{2}\)

ii \(y=2(x+1)^{2}+1\)

iv no \(x\)-intercepts v

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c i -19

ii \(y=3(x+2)^{2}-31\)

iii \(x=-2,(-2,-31)\)

iv \(\frac{-6+\sqrt{93}}{3}, \frac{-6-\sqrt{93}}{3}\)

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d i -2

ii \(y=3\left(x+\frac{4}{3}\right)^{2}-\frac{22}{3}\)

iii \(x=-\frac{4}{3},\left(-\frac{4}{3},-\frac{22}{3}\right)\)

iv \(\frac{-4+\sqrt{22}}{3}, \frac{-4-\sqrt{22}}{3}\)

v

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e i -15

ii \(y=2\left(x+\frac{1}{4}\right)^{2}-\frac{121}{8}\)

iii \(x=-\frac{1}{4},\left(-\frac{1}{4},-\frac{121}{8}\right)\) iv \(-3, \frac{5}{2}\)

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f i 5

ii \(y=2\left(x-\frac{1}{4}\right)^{2}+\frac{39}{8}\)

iii \(x=\frac{1}{4},\left(\frac{1}{4}, \frac{39}{8}\right)\)

iv no \(x\)-intercepts

v

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\(6 \quad \mathbf{a} \quad x=\frac{1}{2},\left(\frac{1}{2}, \frac{11}{4}\right)\), no \(x\)-intercepts

b \(x=-\frac{5}{6},\left(-\frac{5}{6},-\frac{181}{12}\right)\), has \(x\)-intercepts

c \(x=-\frac{1}{4},\left(-\frac{1}{4}, \frac{7}{8}\right)\), no \(x\)-intercepts

d \(x=-\frac{3}{10},\left(-\frac{3}{10}, \frac{131}{20}\right)\), no \(x\)-intercepts

e \(x=\frac{5}{6},\left(\frac{5}{6},-\frac{59}{12}\right)\), no \(x\)-intercepts

f \(x=\frac{3}{2},\left(\frac{3}{2}, \frac{17}{4}\right)\), has \(x\)-intercepts

7 a 6

b 8

c -8

d -2

8 a 3

b 1

c \(\frac{1}{2}\)

d -3

\(9 \quad \mathbf{a} \pm 1\)

b \(\pm 3\)

c \(\pm \frac{\sqrt{2}}{2}\)

d \(\pm \sqrt{3}\)

\(10 y=(x-1)^{2}-2\) or \(y=x^{2}-2 x-1\)

\(11 y=3(x+2)^{2}-1\) or \(y=3 x^{2}+12 x+11\)

\(12 y=-2(x-1)^{2}+6\) or \(y=-2 x^{2}+4 x+4\)

13 a

\begin{center}
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axis: \(x=1\)

b

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vertex: \((1,8)\)

\(y\)-intercept: 6

\(x\)-intercepts: \(-1,3\)

axis: \(x=1\)

c

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d

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e

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f

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g

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h
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-028(5)}

\section*{Exercise 7D}
14

21

ICE-EM MATHEMATICS YEAR 10 ICE-EM Mathematics
a
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e

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f

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d
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g
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i

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4 a \(-5,-1\)

b \(-4,-3\)

d \(-5,3\)

g \(-\frac{1}{2}, \frac{1}{2}\)

e \(-\frac{1}{2}, 10\)

h 0,3

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e

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g

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i

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k

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\(9 y=6 x^{2}-15 x+6\) d

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f

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h

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j

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l

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\(6 \begin{array}{ll}\text { a } y=x^{2}-4 x & \text { b } y=9-x^{2}\end{array}\)

c \(y=-3 x^{2}+6 x+9\)

\(7 \quad y=x^{2}-x-2\)

d \(y=-\frac{3 x^{2}}{5}-\frac{9 x}{5}+6\)

\(8 y=\frac{1}{2} x^{2}-\frac{x}{2}-3\)

\section*{Exercise 7E}
a

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b

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c

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d
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f

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g

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h

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i

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j

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ICE-EM MATHEMATICS YEAR 10 Photocopying is restricted under law and this material must not be transferred to another party Updated June 2022
k
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-031}

\section*{Exercise \(7 \mathrm{~F}\)}
\(180 \mathrm{~m}, 130 \mathrm{~m}\)

\(2 \quad 60 \mathrm{~cm}\)

325,26

\(4 \quad 14,16\)

513,15

63

78 years

\(8 \quad \$ 6\)

\(98 \mathrm{~cm}, 15 \mathrm{~cm}, 17 \mathrm{~cm}\)

\(1020 \mathrm{~cm}, 21 \mathrm{~cm}, 29 \mathrm{~cm}\)

\(115 \mathrm{~cm}, 30 \mathrm{~cm}, 40 \mathrm{~cm}\)

1316

\(14 \quad 1.84\) seconds

\(\begin{array}{ll}15 & -7\end{array}\)

\(16 \frac{5}{4}\)

17 a \(\max =4, \min =\frac{-73}{12}\)

b \(\max =46, \min =-2\)

\(1825 \mathrm{~cm}\) by \(25 \mathrm{~cm}, 17 \mathrm{~cm}\)

\(1975 \mathrm{~m}\) by \(150 \mathrm{~m}\)

\(2031.888 \mathrm{~m}, 5.102\) seconds

\(2162.5 \mathrm{~m}\) by \(80 \mathrm{~m}\), or \(160 \mathrm{~m}\) by \(31.25 \mathrm{~m}\)

\(223 \frac{3}{8}\) square units

\section*{Exercise 7G}
1 a \(-3<x<5\)

\[
\begin{aligned}
& \text { b } x<-4 \text { or } x>1 \\
& \text { d } x \leq-1 \text { or } x \geq 2 \\
& \text { f } x<-\frac{3}{2} \text { or } x>\frac{3}{2}
\end{aligned}
\]

\[
\text { c } x=-2
\]

e all values of \(x\)

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
1 & a 2 & b 24 & c 0 \\
\hline
 & d 0 & e 4 & f -16 \\
\hline
2 & a 0 & b \(\pm 4\) &  \\
\hline
 & \(\mathbf{c} \pm 3\) & \(\mathbf{d} \pm 2\) &  \\
\hline
3 & a \(-4,1\) & b \(-6,-\frac{1}{2}\) & c \(-\frac{3}{4}, \frac{3}{2}\) \\
\hline
 & \(\mathrm{d}-\frac{1}{2}, \frac{5}{2}\) & e \(-7,7\) & f 0,5 \\
\hline
4 & a \(-2-\sqrt{5},-2+\sqrt{5}\) &  & b \(3-\sqrt{2}, 3+\sqrt{2}\) \\
\hline
 & c \(-1-\sqrt{5},-1+\sqrt{5}\) &  & d \(2-\sqrt{5}, 2+\sqrt{5}\) \\
\hline
 & e \(3+\frac{\sqrt{35}}{5}, 3-\frac{\sqrt{35}}{5}\) &  & f \(2+\sqrt{2}, 2-\sqrt{2}\) \\
\hline
\end{tabular}
\end{center}

\[
\begin{array}{lll}
2 & \text { a } x<-2 \text { or } x>3 & \text { b }-4 \leq x \leq-1 \\
& \text { c } x \leq 2 \text { or } x \geq 5 & \text { d }-3<x<0 \\
3 & \text { a } x<-10 \text { or } x>7 & \text { b }-3<x<8 \\
& \text { c } x \leq-5 \text { or } x \geq-4 & \text { d } 3 \leq x \leq 4 \\
4 & \text { a }-8<x<5 & \text { b }-3 \leq x \leq 8 \\
& \text { c } x \leq 5 \text { or } x \geq 7 & \text { d } x<0 \text { or } x>11
\end{array}
\]

\section*{Review exercise}
5 a \(-2-\sqrt{6},-2+\sqrt{6}\)

b \(3-2 \sqrt{2}, 3+2 \sqrt{2}\)

c \(-\frac{5}{2}-\frac{\sqrt{19}}{2},-\frac{5}{2}+\frac{\sqrt{19}}{2}\)

d \(-2-\frac{\sqrt{26}}{2},-2+\frac{\sqrt{26}}{2}\)

6 a minimum \(\quad\) b maximum

c maximum d minimum

\(7 y=x^{2}\) is congruent to \(y=x^{2}-x\),

\(y=-2 x^{2}\) is congruent to \(y=3-2 x^{2}\),

\(y=3 x^{2}\) is congruent to \(y=3 x^{2}+1\),

\(y=2+3 x-4 x^{2}\) is congruent to \(y=1+4 x^{2}\)

8 a translate 1 unit down

b translate 2 units up

c reflect in the \(x\)-axis then translate 4 units up

d reflect in the \(x\)-axis then translate 1 unit up

(There are other correct answers to this question.)

9 a translate 2 units left

b translate 1 unit right

c reflect in the \(x\)-axis then translate 1 unit left

d translate 1 unit left and 3 units down

e translate 2 units right and 3 units down

f reflect in the \(x\)-axis then translate 3 units right and 1 unit up

(There are other correct answers to this question.)
10 a \(y=(x+2)^{2}\)
b \(y=(x-3)^{2}+1\)
c \(y=(x-5)^{2}-2\)
d \(y=(x+3)^{2}-2\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-032(1)}
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ICE-EM MATHEMATICS YEAR 10 ICE-EM MathematicS 10 MEd

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-032}
d

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\(\mathbf{e}\)

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f

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22 The side lengths are \(5 \mathrm{~cm}, 12 \mathrm{~cm}\) and \(13 \mathrm{~cm}\).

\(2352 \mathrm{~cm}, 42 \mathrm{~cm}\) and \(4 \mathrm{~cm}\)

\(24125 \mathrm{~m}\) by \(250 \mathrm{~m}\)

25 a \(-3<x<5\)

b \(-5 \leq x \leq-2\)

c \(x<0\) or \(x>2\)

\section*{Challenge exercise}
1 a -5
b -5
2 a -21
b 4
c \(c>4\)
3 d \(x=80, t=\frac{3}{2} ; x=90, t=\frac{4}{3}\)
\(4 \quad 40 \mathrm{~km} / \mathrm{h}\)
545 cattle
\(632 \mathrm{~km} / \mathrm{h}, 40 \mathrm{~km} / \mathrm{h}\)
\(7 \quad 75\) runs
\(8 \quad 4.844 \mathrm{~m} / \mathrm{s}, 5.844 \mathrm{~m} / \mathrm{s}\)
\(940 \mathrm{~km} / \mathrm{h}, 60 \mathrm{~km} / \mathrm{h}\)
\(10 a=\frac{24}{5}\)
\(11 \mathbf{a}\left(x-\frac{5}{2}\right)^{2}+\left(y-\frac{7}{2}\right)^{2}=\frac{25}{2}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-033(3)}
\end{center}

b Let \(A(0,1)\) and \(B(g, h)\) be the given points and \(P(x, y)\) be any point. \(P\) is on the circle with

diameter \(A B\) if \(\angle A P B=90^{\circ}\) (angle in semicircle).

If gradient of \(A P \times\) gradient of \(B P=-1\), then

\(\frac{y-1}{x} \times \frac{y-h}{x-g}=1\).

So \((y-1)(y-h)+x(x-g)=0\) is the equation of the circle. This meets the \(x\)-axis when \(y=0\). That is when \(x^{2}-g x+h=0\).

12 a \(\frac{x^{2}}{4 \pi}+\frac{(100-x)^{2}}{16}\)

b The minimum area is \(\frac{2500}{\pi+4} \mathrm{~cm}^{2}\) when \(x=\frac{100 \pi}{\pi+4}\)

\(13 \mathbf{d}\) If \((x, y)\) goes to \((x, 5 y)\), then \(\left(a, a^{2}\right)\) goes to \(\left(a, 5 a^{2}\right)\).

The graph of \(y=x^{2}\) goes to the graph of \(y=5 x^{2} .(0,0)\) goes to \((0,0),(1,0)\), goes to \((1,0)\), and \((0,1)\) goes to \((0,5)\). It is not a similarity transformation since some distances are unchanged and others multiplied by 5 . Note that 5 can be replaced by any \(a>0\).

e The enlargement with centre at \((0,0)\) and enlargement factor \(\frac{1}{5}\) maps \(A\left(a, a^{2}\right)\) to \(B\left(\frac{a}{5}, \frac{a^{2}}{5}\right)\). Since \(\frac{a^{2}}{5}=5\left(\frac{a}{5}\right)^{2}\), \(B\) lies on \(y=5 x^{2}\). That is \(y=x^{2}\) is similar to \(y=5 x^{2}\).

f By c \(y= \pm a x^{2}+b x+c\) is congruent to \(y=a x^{2}, a>0\). By \(\mathbf{d} y=a x^{2}\) is similar to \(y=x^{2}\). That is, all parabolas are similar.

14

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-033}
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b

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\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-033(5)}
\end{center}

\section*{Chapter 8 answers}
\section*{Exercise 8A}
1 a \(y=6, \beta=45^{\circ}\)
b \(x=5, \alpha=30^{\circ}\)
c \(\alpha=\gamma=60^{\circ}, \beta=120^{\circ}, x=6\)
d \(\alpha=90^{\circ}, \beta=60^{\circ}\)
e \(x=4, \alpha=45^{\circ}\)
f \(\alpha=60^{\circ}, x=5\)

2 a \(\alpha=\beta=\gamma=60^{\circ}\) (equilateral \(\triangle R S T\) )

b \(\theta=20^{\circ}\) (angle sum of \(\triangle W X Y\) ),

\(x=12 \mathrm{~cm}\) (isosceles \(\triangle W X Y\) )

c \(\angle R Q P=60^{\circ}\) (base angle of isosceles \(\triangle P Q R\) ),

\(\theta=60^{\circ}\) (angle sum of \(\triangle P Q R\) ),

\(y=7 \mathrm{~cm}\) (equilateral \(\triangle R P Q\) )

d \(y=5, \alpha=60^{\circ}\) (equilateral \(\triangle L N Q\) ),

\(\alpha=2 \beta\) (exterior angle of isosceles \(\triangle L Q M\) ), \(\beta=30^{\circ}\)

e \(\angle A B D=55^{\circ}\) (straight angle at \(B\) ),

\(\angle B A D=55^{\circ}\) (base angle of isosceles \(\triangle A B D\) ),

\(\angle A D B=70^{\circ}\) (angle sum of \(\left.\triangle A B D\right)\),

\(\theta=70^{\circ}\) (vertically opposite angles at \(D\) ),

\(\angle A G E=55^{\circ}\) (alternate angles, \(A C \| H E\) ),

\(\alpha=125^{\circ}\) (straight angle at \(G\) )

f \(\angle D C G=55^{\circ}\) (vertically opposite angles at \(C\) ), \(\angle C D G=55^{\circ}\) (base angle of isosceles \(\triangle C D G\) ), \(\alpha=70^{\circ}\) (angle sum of \(\triangle C D G\) ),

\(\angle A G E=55^{\circ}\) (corresponding angles, \(B D \| H E\) ),

\(\angle E G D=55^{\circ}+70^{\circ}=125^{\circ}\),

\(\beta=125^{\circ}\) (vertically opposite angles at \(G\) )

g \(\angle D A B=50^{\circ}\) (complementary angles),

\(\alpha=50^{\circ}\) (alternate angles, \(A B \| C E\) ),

\(\angle A F D=40^{\circ}\) (base angle of isosceles \(\triangle A D F\) ),

\(\alpha+\angle F D C=100^{\circ}\) (angle sum of \(\left.\triangle A D F\right)\),

\(\angle F D C=50^{\circ}, \theta=130^{\circ}\) (straight angle at \(D\) )

h \(y=2 \mathrm{~m}\) (radii of circle),

\(\beta=25^{\circ}\) (base angle of isosceles \(\triangle A B O\) ),

\(\alpha=60^{\circ}\) (base angle of isosceles

\(\triangle B C O\), angle sum of \(\triangle B C O)\)

i \(3 \alpha+\beta=180^{\circ}\) (co-interior angles, \(A D \| B C\) ),

\(\alpha+2 \beta=180^{\circ}\) (angle sum of isosceles triangle),

\(\alpha=36^{\circ}, \beta=72^{\circ}\)

j \(\angle C B O=2 \alpha(\triangle B O C\) isosceles \()\), \(\angle O B A=90-\frac{3 \alpha}{2}\) (angle sum of isosceles triangle), \(\angle O B C+\angle A B O=95^{\circ}\) (alternate angles, \(A B \| C D\) ), \(\alpha=10^{\circ}\)

\(\mathbf{k} \alpha=130^{\circ}\) (angle sum of quadrilateral)

l \(\alpha=108^{\circ}\) (angle sum of regular pentagon)

m \(\alpha=120^{\circ}\) (co-interior angles, \(A B \| D C\) ),

\(\beta=60^{\circ}\) (co-interior angles, \(B C \| A D\) )

36 sides

\(436^{\circ}\)

\section*{Exercise 8B}
1 a \(\triangle B A C \equiv \triangle \mathrm{YXZ}\) (SAS)
b \(\triangle A B C \equiv \triangle X Z Y(\mathrm{ASA})\)
c \(\triangle A B C \equiv \triangle P Q R(\mathrm{SSS})\)
d \(\triangle A B C \equiv \triangle X Y Z(\mathrm{ASA})\)
e \(\triangle P Q R \equiv \triangle X Y Z\) (ASA)
2
a \(\angle B C A=40^{\circ}, \angle F E D=100^{\circ}, \angle F D E=40^{\circ}\),
\(A B=B C=3 \mathrm{~cm}, F D=4.6 \mathrm{~cm}\)
b \(\angle A B C=10^{\circ}, \angle E F D=130^{\circ}, C B=4.2 \mathrm{~cm}\),
\(A C=1.1 \mathrm{~cm}, E D=5 \mathrm{~cm}\)
c \(\angle F E D=32^{\circ}, \angle C A B=111^{\circ}, \angle B C A=37^{\circ}\),
\(A B=9.1 \mathrm{~cm}, \angle A C=8 \mathrm{~cm}, F E=14.1 \mathrm{~cm}\)

d \(\angle F E D=67^{\circ}, \angle C A B=67^{\circ}, \angle A B C=67^{\circ}\), \(\angle B C A=46^{\circ}, C B=38.3 \mathrm{~mm}, A C=38.3 \mathrm{~mm}\), \(F E=38.3 \mathrm{~mm}, E D=30 \mathrm{~mm}\)

3 a \(A D=B C\) (equal side lengths of square), \(D E=C E\) (given), \(\angle A D E=\angle B C E\) (given), \(\triangle A D E \equiv \triangle B C E(\mathrm{SAS})\)

b \(A E=B E\) (matching sides in congruent triangles)

4 a \(P R=R S\) (equal side lengths of square), \(P X=S T\)

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-034(2)}
(half side lengths of square),

\(\angle R P X=\angle R S T=90^{\circ}\),

\(\triangle R P X \equiv \triangle R S T\) (SAS),

\(R X=R T\) (matching sides in congruent triangles),

b \(R X=R T\) (from a),

\(V R=V R\) (common side),

\(V T=V X\) (half side lengths of square)

\(\triangle R X V \equiv \triangle R T V(\mathrm{SSS})\),

\(\angle T R V=\angle X R V\) (matching angles of congruent triangles)

5 a Diagonals of a rectangle are equal and bisect each other.
b i \(\angle A O D=56^{\circ}\)
ii \(\angle A O B=124^{\circ}\)
iii \(\angle O B C=62^{\circ}\)
iv \(\angle A B O=28^{\circ}\)

\(6 \angle Y D O=\angle O B X\) (alternate angles, \(A B \| C D\) ), \(\angle B O X=\angle D O Y\) (vertically opposite),

\(O B=O D\) (diagonals of a parallelogram bisect each other), \(\triangle B O X \equiv \triangle D O Y(\mathrm{AAS})\)

\(O X=O Y\) (matching sides of congruent triangles)

\(7 \quad B P=P C(P\) is the midpoint of \(B C)\),

\(\angle Q P B=\angle C P D\) (vertically opposite),

\(\angle P Q B=\angle C D P\) (alternate angles, \(A Q \| D C\) ),

\(\triangle B Q P \equiv \triangle C D P(\mathrm{AAS})\),

\(B Q=C D\) (matching sides of congruent triangles),

\(C D=A B\) (opposite sides of parallelogram), \(A Q=2 A B\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-034(1)}
\end{center}

\(8 D C=A B=Y X\) (opposite sides of parallelogram),

\(D C\|A B\| Y X\) (opposite sides of parallelogram)

Therefore, \(D C X Y\) is a parallelogram (opposite sides equal and parallel).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-034(3)}
\end{center}

\(9 \angle A K O=\angle A O K\) (isosceles triangle \(A K O\) ),

\(\angle K A O=45^{\circ}\) (diagonals of a square bisect the vertex angles), \(\angle A K O=\angle A O K=67 \frac{1}{2}^{\circ}\) angle sum of triangle),

\(\angle B O A=90^{\circ}\) (diagonals of a square intersect at right angles),

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-034}
\(\angle K B O=45^{\circ}\) (diagonals of a square bisect the vertex angles), \(\angle B O K=22 \frac{1}{2}^{\circ}, \angle A O K=3 \angle B O K\)
a \(\triangle D B C \equiv \triangle D B A(\mathrm{AAS})\)

\(A B=C B\) (matching sides of congruent triangles)

\(D A=D C\) (matching sides of congruent triangles)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-035(3)}
\end{center}

b \(\triangle A X B \equiv \triangle C X B\) (SAS)

So \(A B=C B\) (matching sides of congruent triangles)

\(\triangle C D X \equiv \triangle A D X(\mathrm{SAS})\)

\(A D=C D\) (matching sides of congruent triangles)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-035}
\end{center}

\(11 \angle D A C=\angle B C A\) (alternate angles, \(A D \| B C\) )

\(\angle A C D=\angle C A B\) (alternate angles, \(A B \| C D\) )

\(\triangle A B C=\triangle C D A(\mathrm{SAS})\)

\(C D=B A\) (matching sides of congruent triangles)

\(A D=B C\) (matching sides of congruent triangles)

Thus, opposite sides are equal.

\(\angle A B C=\angle C D A\) (matching angles of congruent triangles)

\(\angle B A D=\alpha+\beta=\angle O C B\)

Thus, opposite angles are equal.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-035(2)}
\end{center}

12 Let \(X\) be the point of intersection of diagonals \(B D\) and \(A C\). We use the results of Question 11.

\(A D=C B\)

\(\angle D A C=\angle B C A(\) alternate angles, \(B C \| A D)\)

\(\angle B X C=\angle D X A\) (vertically opposite)

\(\angle B X C \equiv \angle D A X(\) AAS \()\)

\(C X=A X\) (matching sides of congruent triangles)

\(D X=B X\) (matching sides of congruent triangles)

Thus, the diagonals of a parallelogram bisect each other.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-035(5)}
\end{center}

13 A rhombus is a parallelogram.

\(B C=C D\) (all sides of a rhombus are equal)

\(B X=D X\) (diagonals of a parallelogram bisect each other)

\(\triangle B X C \equiv \triangle D X C(\mathrm{SSS})\)

\(\angle B X C=\angle D X C\) (matching angles of congruent triangles)

\(\angle B X C+\angle D X C=180^{\circ}\)

So \(\angle B X C=\angle D X C=90^{\circ}\)

Thus, diagonals of a rhombus are perpendicular.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-035(1)}
\end{center}

\(14 \quad \triangle B A D \equiv \triangle D C B(\mathrm{SAS})\)

\(B D=A C\) (matching sides of congruent triangles)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-035(6)}
\end{center}

15 Let \(\angle A B K=\angle D B K=\alpha\)

Then \(\angle C B D=2 \alpha\)

(diagonals of rhombus bisect the vertex angles) \(\angle B A K=180^{\circ}-4 \alpha\) (co-interior angles \(A D \| B C\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-035(4)}
\end{center}

\(\angle B K A=180^{\circ}-\left(180^{\circ}-4 \alpha\right)-\alpha(\) angle sum of triangle \()\) \(=3 \alpha\) \(=3 \angle \mathrm{ABK}\)

16

\(\angle C B X=\angle C D Y=90^{\circ}+60^{\circ}=150^{\circ}\)

\(\angle X A Y=360^{\circ}-\left(90^{\circ}+60^{\circ}+60^{\circ}\right)=150^{\circ}\)

\(\triangle X B C \equiv \triangle C D Y \equiv \triangle X A Y\) (SAS congruence test)

Hence, \(X C=C Y=Y X\) (matching sides of congruent triangles)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-036(2)}
\end{center}

\(172 \alpha+2 \beta=360^{\circ}\) (angle sum of quadrilateral \(\left.A B C D\right)\) \(\alpha+\beta=180^{\circ}\)

so \(\angle A B C+\angle B A D=180^{\circ}\)

so \(\angle B C D+\angle A D C=180^{\circ}\)

Thus, \(A B \| O C\) and \(B C \| A D\) (co-interior angles are supplementary).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-036(1)}
\end{center}

18

\(B C=A D\) and \(A B=D C\)

\(\triangle B C D \equiv \triangle D A B(\mathrm{SSS})\)

Hence,

\(\angle A B D=\angle C D B\) (matching angles of congruent triangles)

\(\angle A D B=\angle C B D\) (matching angles of congruent triangles)

Hence, \(A B \| C D\) (alternate angles are equal)

and \(B C \| A D\) (alternate angles are equal)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-036}
\end{center}

\section*{Exercise 8C}
1 a \(\triangle A B C\) is similar to \(\triangle R Q P\) (SSS)

b \(\triangle L M N\) is similar to \(\triangle Q R P\) (AAA)

c \(\triangle S T U\) is similar to \(\triangle X Z Y\) (AAA)

d \(\triangle L M N\) is similar to \(\triangle T S U\) (SAS)

e \(\triangle B C A\) is similar to \(\triangle E C D\) (AAA)

f \(\triangle G H F\) is similar to \(\triangle J H I\) (SAS)

g \(\triangle O M K\) is similar to \(\triangle L M N\) (AAA)

h \(\triangle S T R\) is similar to \(\triangle S U Q\) (AAA)

i \(\triangle E D C\) is similar to \(\triangle A D B\) (SAS)

j \(\triangle I F G\) is similar to \(\triangle I H F\) is similar to \(\triangle F H G\) (AAA)

2 a AAA

b \(x=3.5\)

3 a SAS

b \(y=14.4\)

4 a \(\triangle J K L\) is similar to \(\triangle I G H\) (SSS)

b \(\alpha=49^{\circ} a=\beta, b=131-\beta\)

5 a \(\triangle D E C\) is similar to \(\triangle D B A\) (SAS)

b \(\frac{E C}{B A}=\frac{E D}{D B}=\frac{D C}{D A}\)

c \(x=2\)

6 a \(\angle A C B=\angle D C E, \angle C A B=\angle C E D, \angle A B C=\angle E D C\)
b \(3.75 \mathrm{~cm}\)
c \(7.2 \mathrm{~cm}\)

7 a \(\triangle A D E\) is similar to \(\triangle A C B\) (AAA)

b \(8 \frac{4}{7} \mathrm{~cm}\)

8 a AAA b \(L M=16 \frac{2}{3} \mathrm{~cm}\)

\(9 \quad 18.75 \mathrm{~m}\)

\(10130 \mathrm{~m}\)

\(1133 \mathrm{~m}\)

\(12 \angle D O C=\angle A O B\) (vertically opposite), \(\angle A B O=\angle C D O\) (alternate angles, \(A B \| C D)\), \(\triangle A O B\) is similar to \(\triangle C O D\) (AAA)

\(\frac{O B}{O D}=\frac{O A}{O C}=2, O B=2 O D\)

13 a \(\angle B A D=\angle B A C\) (common angle),

\(\angle A D B=\angle A B C=90^{\circ}\),

\(\triangle A B D\) is similar to \(\triangle A C B\) (AAA)

b \(\angle B C D=\angle B C A\) (common angle), \(\angle A B C=\angle B D C=90^{\circ}, \triangle B C D\) is similar to \(\triangle A C B\) (AAA)

14 a \(\angle A E P=\angle C B P\) (alternate angles, \(C B \| A D\) ), \(\angle E P A=\angle B P C\) (vertically opposite),

\(\triangle A P E\) is similar to \(\triangle C P B\) (AAA)

b \(\frac{A P}{P C}=\frac{A E}{A D}=\frac{A E}{C B}\) (ratios of matching sides of similar triangles) and \(A D=C B\) (opposite sides of a parallelogram) \(\frac{A D}{P C}=\frac{A E}{C D}=\frac{A E}{C B}=\frac{1}{2}\),

\(A P=2 P C\) and \(A C=3 A P\)

\(\angle E D B=180^{\circ}-\angle A C B\) (given),

\(\angle A D E=\angle A C B\) (supplementary)

\(\angle B A C=\angle D A E\) (common),

\(\triangle A D E\) is similar to \(\triangle A C B\) (AAA)

b \(\frac{A E}{A B}=\frac{A D}{A C}\) (equal ratios of matching sides of similar triangles)
\(\angle P D B=90^{\circ}\) (given), \(\angle A F B=90^{\circ}\) (given)

\(\angle D B P=\angle A B F\) (common)

\(\triangle P B D\) is similar to \(\triangle A B F\) (AAA)

\(\triangle A B F \equiv \triangle A C F(\mathrm{RHS})\)

So \(\triangle P D B\) is similar to \(\triangle A C F\)

b \(\frac{F C}{D B}=\frac{A C}{P B}\) (matching sides of similar triangles)

\(17 L M=\frac{1}{2} A C\) and \(N M=\frac{1}{2} A B\). But \(A C=A B\). Therefore, \(L M=N M\).

18 a \(\angle A B C=\angle A D B=90^{\circ}, \angle B A C=\angle B A D\) (common), \(\triangle A B C\) is similar to \(\triangle A D B\) (AAA)

b \(\frac{b}{x}=\frac{a}{c}=\frac{c}{a-y}\). Therefore, \(a(a-y)=c^{2}(1)\)

c \(\angle A B C=\angle B D C=90^{\circ}, \angle B C A=\angle B C D\) (common), \(\triangle A B C\) is similar to \(\triangle B D C\) (AAA)

d \(\frac{c}{x}=\frac{b}{y}=\frac{a}{b}\), so \(a y=b^{2}(2)\)

e From (1) \(\quad a(a-y)=c^{2}\) and \(a^{2}=c^{2}+a y\)

From (2) \(a y=b^{2}\)

\(a^{2}=b^{2}+c^{2}\)

\section*{Exercise 8D}
1 Let \(M\) be the midpoint of \(A B\).

Let \(N\) be the midpoint of \(B C\).

\(M B=\frac{1}{2} A B\) and \(N B=\frac{1}{2} C B\)

Hence, \(\frac{M B}{A B}=\frac{1}{2}\) and \(\frac{N B}{C B}=\frac{1}{2}\)

\(\triangle M B N\) is similar to \(\triangle A B C\) (SAS)

Hence, \(\angle B A C=\angle B M N\) (matching angles of similar triangles)

\(M N \| A C\) (corresponding angles equal)

Also, \(M N=\frac{1}{2} A C\) (similarity factor of \(\frac{1}{2}\) )

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-037(1)}
\end{center}

2 Let \(M\) be the midpoint of \(A B\) and \(M N \| A C\).

\(\angle B M N=\angle B A C\) (corresponding angles, \(M N \| A C\) )

\(\triangle B M N\) is similar to \(\triangle B A C\) (AAA)

\(B M=\frac{1}{2} B A \quad\) (matching sides of similar angles)
\(\frac{B M}{B A}=\frac{B N}{B C}=\frac{1}{2}\)
Thus, \(N\) is the midpoint of \(B C\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-037(3)}
\end{center}

\(3 \quad M\) is the midpoint of \(A B\).

\(N\) is the midpoint of \(B C\).

\(P\) is the midpoint of \(A C\).

From question 1,

\(M N=\frac{1}{2} A C, M P=\frac{1}{2} B C, P N=\frac{1}{2} A B\)

Therefore,

\(N P=M B=A M\)

\(M N=A P=P C\)

\(M P=B N=N C\)

Hence, all the triangles are congruent by the SSS test.

All triangles have angles \(\alpha, \beta\) and \(\gamma\).

Therefore, each of the small triangles is similar to the large triangle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-037(2)}
\end{center}

4 Let \(M, N, Q\) and \(P\) be the midpoints of \(A B, B C, C D\) and \(D A\) respectively.

Join vertices \(B\) and \(D\).

In \(\triangle C B D, N Q \| B D\) (Question 1)

In \(\triangle A B D, M P \| B D\) (Question 1)

Therefore, \(N Q \| M P\).

In a similar way, \(M N \| P Q\).

Thus, \(M N Q P\) is a parallelogram.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-037}
\end{center}

5 In \(\triangle P A Q\) and \(\triangle B A C\)

\(\angle P A Q=\angle B A C\) (common)

\[
\frac{B A}{P A}=\frac{A C}{A Q}=\frac{7}{4}
\]

\(\triangle P A Q\) is similar to \(\triangle B A C\) (SAS)

\(\angle A P Q=\angle A B C\) (matching angles of similar triangles)
\(P Q \| B C\) (corresponding angles are equal)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-038}
\end{center}

6

a \(\triangle S A C\) is similar to \(\triangle S B D\)

\[
\begin{aligned}
\frac{S A}{S A+A B} & =\frac{S C}{S C+C D} \\
S A \times(S C+C D) & =S C \times(S A+A B) \\
S A \times S C+S A \times C D & =S C \times S A+S C \times A B \\
S A \times C D & =S C \times A B \\
\frac{S A}{A B} & =\frac{S C}{C D} \\
S A: A B & =S C: C D
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-038(1)}
\end{center}

b \(\triangle S A C\) is similar to \(\triangle S B D\)

\[
\begin{aligned}
\frac{S B}{S B-A B} & =\frac{S D}{S D-C D} \\
S B \times(S D-C D) & =S D \times(S B-A B) \\
S B \times S D-S B \times C D & =S D \times S B-S D \times A B \\
S B \times C D & =S D \times A B \\
\frac{S B}{A B} & =\frac{S D}{C D}
\end{aligned}
\]

That is, \(S B: A B=S D: C D\).

c \(\frac{S A}{S B}=\frac{S C}{S D}\) or \(\triangle S A C\) is similar to \(\triangle S B D\).

\(S A: S B=S C: S D\)

7

\[
\begin{aligned}
\frac{S A}{A B} & =\frac{S C}{C D} \\
S A \times C D & =A B \times S C
\end{aligned}
\]

\[
\begin{aligned}
S A \times S C+S A \times C D & =A B \times S C+S C \times S A \\
S A \times(S C+C D) & =S C \times(A B+S A) \\
S A \times S D & =S C \times S B \\
\frac{S A}{S B} & =\frac{S C}{S D}
\end{aligned}
\]

Therefore, \(\triangle S A C\) is similar to \(\triangle S B D\) (SAS).

Hence, \(A C \| B D\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-038(3)}
\end{center}

\section*{Review exercise}
\(1 \alpha=37^{\circ}, \beta=80^{\circ}, \gamma=20^{\circ}\)

2 a \(\triangle P C Q\) is similar to \(\triangle A C B\)

b \(x=5\)

3 a In \(\triangle B E C\) and

\(\triangle F E D, \angle C B E=\angle D F E\) (alternate angles, \(B C \| A D\) ) \(\angle C E B=\angle F E D\) (vertically opposite),

\(\triangle B E C\) is similar to \(\triangle F E D\) (AAA)
b i 2
ii 4

4 a \(\alpha=112^{\circ}\)

b \(\alpha=78^{\circ}, \beta=58^{\circ}\)

c \(\alpha=20^{\circ}\)

d \(\beta=104^{\circ}, \alpha=76^{\circ}, \gamma=52^{\circ}\)

e \(\alpha=36^{\circ}, \beta=144^{\circ}\)

f \(\alpha=130^{\circ}, \beta=50^{\circ}, \gamma=130^{\circ}\)

5 a \(\angle B A C=70^{\circ}, \angle A C B=55^{\circ}\)

b \(\angle F I C=55^{\circ}\) (corresponding angles, \(F I \| A B\) ). \(\angle F C I=55^{\circ}\). So \(\triangle F I C\) is isosceles.

c \(\angle D I B=55^{\circ}\) (corresponding angles, \(A C \| D I\) ). \(\triangle D B I\) is isosceles (equal base angles), \(D B=D I\)

d \(\angle G I F=53^{\circ}, \angle E F G=17^{\circ}\)

6 a \(\triangle O N M \equiv \triangle Z Y X\) (AAS)

b \(\triangle A B C \equiv \triangle R T S\) (RHS)

\(7 \triangle A O B\) is similar to \(\triangle C O D\) (AAA).

Ratio of matching sides gives: \(\frac{B O}{O D}=\frac{A O}{O C}\)

8 In \(\triangle C F A\) and \(\triangle B E A\),

\(\angle C F A=\angle B E A=90^{\circ}\) (altitudes),

\(\angle F A C=\angle E A B\) (common),

\(\triangle C F A\) is similar to \(\triangle B E A\) (AAA)

\(\frac{B E}{C F}=\frac{A B}{A C}\) (ratios of matching sides)

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-038(2)}
\(9 \quad \mathbf{a} \quad x=\frac{102}{11}\)
b \(x=\frac{21}{4}\)
10 a \(x=13\)
b \(E F=\frac{119}{13}\)

\(11 \quad 20 \mathrm{~cm}\)

\section*{Challenge exercise}
Only outlines of proofs are given in the challenge exercise
1 a \(272 \mathrm{~cm}\)
b \(50 \sqrt{55} \mathrm{~cm}\)
c \(16 \sqrt{55} \mathrm{~cm}\)
d approximately \(67 \mathrm{~cm}\)

2 Produce \(A Y\) to \(Z\). Let \(\angle B A Y=\alpha \angle A Z C=\alpha\) (alternate angles, \(B A \| Z C\) )

Then \(\angle Y A C=\alpha\) (angle bisector), and \(\triangle Z C A\) is isosceles So \(Z C=C A=2 O A\)

Next, \(\triangle A O X\) is similar to \(\triangle Z C Y\) (AAA)
\(\frac{C Y}{O X}=\frac{Z C}{O A}=2\) or \(C Y=2 O X\)

\(3 P Q=B C=A B=C D=x\) and \(B P=C Q=2 x\)

\(\triangle A X B \equiv Q X P(\mathrm{AAS})\)

\(\therefore P X=X B=x=Q Y=Y C\)

\(P Q Y X\) is a rhombus and \(X Q\) and \(P Y\) are diagonals of a rhombus. Hence, \(A Q \perp D P\).
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-039(7)}

4 Show that \(\angle B A D=\angle B D A=\angle A C F=54^{\circ}\).

Hence, \(\triangle A B D\) is isosceles with \(A B=B D\).

Draw the perpendicular from \(B\) to \(A D\). Let \(X\) be the point where the perpendicular meets \(A D\). \(X\) is the midpoint of \(A D\). \(\triangle A F C \equiv \triangle B X D(\mathrm{AAS})\). Hence, \(F C=X D\) and \(A D=2 F C\).

\(\triangle S X Y\) is similar to \(\triangle S P R\)

\(S X=\frac{1}{2} P S, S Y=\frac{1}{2} S R\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-039(4)}
\end{center}

Therefore, area of \(\Delta S X Y=\frac{1}{4}\) area of \(\triangle S P R\)

Therefore, area of \(\triangle S P R=\frac{1}{2}\) area of parallelogram \(P Q R S\)

Therefore, area of \(\triangle X Y S=\frac{1}{8}\) area of parallelogram \(P Q R S\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-039(9)}
\end{center}

a \(\angle B A D=\angle B E C\) (corresponding angles, \(A D \| E C\) ) \(\angle D A C=\angle A C E(\) alternate angles, \(A D \| E C)\)

\(\triangle B A D\) is similar to \(\triangle B E C\) (AAA)

\(\triangle A E C\) is isosceles; \(A C=A E\) (base angles of \(\triangle A E C\) )

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-039(6)}
\end{center}

b From a, \(\frac{B C}{C D}=\frac{B E}{E D}\) and \(\frac{B A}{A D}=\frac{B E}{E D}\).

Therefore, \(\frac{B C}{C D}=\frac{B A}{A D}\)

Therefore, \(\frac{A D}{A B}=\frac{C D}{C B}\)

c From a, \(\frac{A F}{F B}=\frac{A C}{B C}\)

\[
\begin{aligned}
& \frac{B D}{D C}=\frac{B A}{C A} \\
& \frac{A E}{E C}=\frac{A B}{B C}
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-039(5)}
\end{center}

\[
\frac{B D}{D C} \times \frac{C E}{E A} \times \frac{A F}{F B}=\frac{B A}{C A} \times \frac{B C}{A B} \times \frac{A C}{B C}=1
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-039(8)}
\end{center}

a \(B^{\prime}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-039(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-039}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-039(2)}
\end{center}

d \(\angle B^{\prime \prime} A^{\prime \prime} A^{\prime}\) is a right angle. The new triangle is similar to triangle \(A B C\).

e \(c\)

\[
\mathbf{f} a^{2}+b^{2}=c^{2}
\]

\(8 \quad X Y=\frac{1}{2} A C=W Z\) and \(Y Z=\frac{1}{2} B D=X W\).

Therefore, \(X Y Z W\) is a parallelogram. Diagonals bisect each other.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-039(3)}
\end{center}

\section*{Chapter 9 answers}
\section*{Exercise 9A}
1 a 16

b 125

d 27

e 10000

c 64

2 a \(2^{3}\)

b \(2^{6}\)

f 216

d \(2^{5}\)

e \(5^{4}\)

c \(3^{4}\)

3 a \(13^{-1}\)

b \(7^{-2}\)

f \(3^{5}\)

d \(2^{-10}\)

e \(3^{-4}\)

4 a \(a^{15}\)

b \(a^{11}\)

e \(a^{6} b^{7}\)

h \(15 x^{5} y^{4}\)

g \(8 a^{5} b^{5}\)

5 a \(\frac{x}{y}\)

b \(\frac{3 x y^{2}}{2}\)

d \(\frac{5 x y}{4}\)

e \(\frac{2 c^{2}}{3 b^{2}}\)

b \(x^{21} y^{35}\)

e \(9 a^{4} b^{8}\)

b \(p^{4}\)

e \(a^{5} b^{2}\)

b \(\frac{1}{4}\)

e \(\frac{1}{1000}\)

h \(\frac{25}{9}\)

k \(\frac{1}{81}\)

n 1

b \(\frac{1}{9 x^{4} y^{4}}\)

e \(\frac{x^{6} y^{6}}{27}\)

b \(\frac{8}{a^{4} b^{3}}\)

e \(\frac{y}{2 x^{6}}\)

h \(m^{7} n^{6}\)

k \(\frac{4 a b^{5}}{3}\)

n \(a^{3} b^{11}\)

q \(\frac{a^{8}}{b^{7}}\)

b \(\frac{1}{64}\)

\(13 \frac{13}{3}\)

\(12 \frac{7}{208}\)

14 a \(-\frac{x}{y}\)

c \(\frac{3 x^{2} y^{2}}{x+y}\)

d \(\frac{x^{2} y+x y^{2}}{x^{2}+y^{2}}\) e \(\frac{x^{4}}{1+2 x^{2} y+x^{4} y^{2}}\)

f \(\frac{x^{2} y}{x^{2}+y}\)

\section*{Exercise 9B}
1 a \(6.3 \times 10^{1}\)

b \(4 \times 10^{-1}\)

c \(6.2 \times 10^{-1}\)

d \(7.4 \times 10^{3}\)

e \(2.1 \times 10^{7}\)

f \(2.6 \times 10^{-4}\)

g \(-8.6 \times 10^{-2}\)

h \(2 \times 10^{12}\)

i \(9.1345 \times 10^{-5}\)

j \(5.732 \times 10^{4}\)

k \(3.012 \times 10^{-3}\)

l \(1.00051 \times 10^{-1}\)

2 a \(2.25 \times 10^{7}\)

b \(6.700 \times 10^{-7} \mathrm{~m}\)

c \(1.5 \times 10^{11} \mathrm{~m}\)

3 a 6400

b 92000

c 0.048

d 0.0087

e 7412000

f -402

g -0.004657

h 47.26

4 a \(8 \times 10^{5}\)

b \(1 \times 10^{6}\)

c \(1.26 \times 10^{8}\)

d \(2.04 \times 10^{6}\)

e \(2 \times 10^{-4}\)

f \(4 \times 10^{-6}\)

g \(1.21 \times 10^{-16}\)

h \(2 \times 10^{-21}\)

i \(2.5 \times 10^{1}\)

j \(3 \times 10^{-2}\)

k \(5 \times 10^{6}\)

l \(5 \times 10^{0}\)

5 a \(1.026 \times 10^{9}\)

b \(5.83 \times 10^{1}\)

c \(6 \times 10^{-7}\)

d \(4.34 \times 10^{9}\)

e \(1.2 \times 10^{2}\)

f \(1.6 \times 10^{2}\)

\(6 \quad 5.8473 \times 10^{9} \mathrm{~kg}=5.8473 \times 10^{6}\) tonnes

7 a \(2.18 \times 10^{18} \mathrm{~km}\)

b \(1.23 \times 10^{23} \mathrm{~km}\)

c approximately 8 minutes 18 seconds

81840 electrons

9 a \(5.766 \times 10^{2}\)

b \(4.73 \times 10^{2}\)

c \(4.7 \times 10^{2}\)

d \(5 \times 10^{2}\)

e \(5.124 \times 10^{-2}\)

f \(5.12 \times 10^{-2}\)

g \(5.1 \times 10^{-2}\)

h \(5 \times 10^{-2}\)

i \(1.603 \times 10^{3}\)

j \(1.60 \times 10^{3}\)

k \(1.6 \times 10^{3}\)

l \(2 \times 10^{3}\)

m \(2.994 \times 10^{27}\)

n \(2.99 \times 10^{27} \quad\) o \(3.0 \times 10^{27}\)

p \(3 \times 10^{27}\)

q \(5.73 \times 10^{5}\)

r \(7 \times 10^{-3}\)

10 a 5.60

b 538

c 9670

d 732000

e 0.00351

f 0.0142

g 372

h 478000

\(112.93 \times 10^{-8} \mathrm{~m}^{3}\)

\(\begin{array}{ll}12 \text { a } 4.14 \times 10^{4} & \text { b } 2.97 \times 10^{9}\end{array}\)

13 a between \(14.5 \mathrm{~cm}\) and \(15.5 \mathrm{~cm}\)

b between \(1.995 \times 10^{3} \mathrm{~kg}\) and \(2.005 \times 10^{3} \mathrm{~kg}\)

c between \(18.665 \mathrm{~m}\) and \(18.675 \mathrm{~m}\)

d between \(4.87445 \times 10^{7} \mathrm{~mL}\) and \(4.87455 \times 10^{7} \mathrm{~mL}\)

\section*{Exercise \(9 \mathrm{C}\)}
1 a 2
b 7
c 3
d 2
e 10
f 5
2 a \(\frac{1}{5}\)
b \(\frac{1}{8}\)
c \(\frac{1}{2}\)
d \(\frac{1}{100}\)
e \(\frac{1}{10}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-041(5)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-041(4)}
\end{center}

\section*{Exercise 9D}
1 a

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|}
\hline
\(\boldsymbol{x}\) & -2 & -1 & 0 & 1 & 2 \\
\hline
\(\mathbf{2}^{\boldsymbol{x}}\) & 0.25 & 0.5 & 1 & 2 & 4 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-041(2)}
\end{center}

b

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(\boldsymbol{x}\) & -2 & -1 & 0 & 1 & 2 \\
\hline
\(\mathbf{2}^{-\boldsymbol{x}}\) & 4 & 2 & 1 & 0.5 & 0.25 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-041}
\end{center}

c

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(\boldsymbol{x}\) & -2 & -1 & 0 & 1 & 2 \\
\hline
\(\mathbf{4}^{\boldsymbol{x}}\) & 0.0625 & 0.25 & 1 & 4 & 16 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-041(1)}
\end{center}

d

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(\boldsymbol{x}\) & -2 & -1 & 0 & 1 & 2 \\
\hline
\(5^{-\boldsymbol{x}}\) & 25 & 5 & 1 & 0.2 & 0.04 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-041(3)}
\end{center}

2

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-042(1)}
\end{center}

3

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-042}
\end{center}

4

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-042(2)}
\end{center}

\section*{Exercise 9E}
1 a 3

b 9

c 5

d 2

e 3

f 2

g 3

h 5

i 3

j 6

k 4

l 5

2 a -4

b -4

c 0

d -3

e -5

f -3

g -5

h -10

3 a \(\frac{1}{2}\)

b \(\frac{3}{2}\)

c \(\frac{3}{2}\)

d \(\frac{2}{3}\)

e \(\frac{3}{2}\)

f \(\frac{2}{3}\)

g \(\frac{2}{3}\)

h \(\frac{3}{4}\)

4 a

b \(\frac{7}{2}\)

c \(\frac{5}{7}\)

d \(\frac{3}{4}\)

e \(\frac{1}{3}\)

f \(-\frac{2}{3}\)

g \(-\frac{1}{3}\)

h \(-\frac{3}{2}\)

5 a 5

b -2

c \(\frac{4}{3}\)

d \(\frac{2}{15}\)

e \(\frac{3}{2}\)

f 6

g \(-\frac{5}{4}\)

h \(\frac{3}{2}\)

6

\begin{center}
\begin{tabular}{lll}
a 4 and 5 & b 2 and 3 & c 5 and 6 \\
d 2 and 3 & e 4 and 5 & f 2 and 3 \\
g 1 and 2 & h 2 and 3 & i 3 and 4 \\
j -1 and 0 & k 0 and 1 & l 1 and 2 \\
\end{tabular}
\end{center}

\section*{Exercise 9F}
1 a

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
\(\boldsymbol{t}\) & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
\(\boldsymbol{y}\) & 200 & 400 & 800 & 1600 & 3200 & 6400 \\
\hline
\end{tabular}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-042(3)}
\end{center}

\(\begin{array}{ll}\text { c i } 303.14 & \text { ii } 918.96\end{array}\)

iii 2262.74

2 a

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
\(\boldsymbol{t}\) & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
\(\boldsymbol{y}\) & 200 & 100 & 50 & 25 & 12.5 & 6.25 \\
\hline
\end{tabular}
\end{center}

b

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-042(4)}
c i 131.95
ii 21.76
iii 8.25

\(\begin{array}{llll}3 & \mathbf{a} & \mathbf{i} & 60\end{array}\)

ii 3840

iii 10861

b i 1000

ii 100

iii 1

iv 0.1

4 a \(8.2007 \times 10^{9}\)

b \(1.360 \times 10^{11}\)

5 a \(63.2^{\circ}\)

b \(42.0^{\circ}\)
6 a \(0.8 \%\)
b 80000
c 86635

7 a 1000
b i 250
ii 125
iii 31

\section*{Exercise 9C}
\(1 \begin{array}{lll}\text { a } \log _{2} 8=3 & \text { b } \log _{10} 100=2 & \text { c } \log _{7} 49=2\end{array}\)

d \(3^{4}=81\) is equivalent to \(\log _{3} 81=4\)

e \(5^{3}=125\) is equivalent to \(\log _{5} 125=3\)

f \(7^{3}=343\) is equivalent to \(\log _{7} 343=3\)

g \(2^{5}=32\) is equivalent to \(\log _{2} 32=5\)

h \(10^{4}=10000\) is equivalent to \(\log _{10} 10000=4\)

i \(10^{-3}=0.001\) is equivalent to \(\log _{10} 0.001=-3\)

j \(2^{-1}=0.5\) is equivalent to \(\log _{2} 0.5=-1\)

2 a 2

b 6

c 7

d 12

e 0

f 8

g 3

h 2

3 a

b 4

c 3

d 2

e 3

f 0

g 4

h 3

\begin{center}
\begin{tabular}{lllll}
\(\mathbf{4}\) & \(\mathbf{a}-2\) & b -1 & c -2 & d -2 \\
 & e -3 & f -5 & g -4 & \(\mathbf{h}-3\) \\
\(\mathbf{5}\) & \(\mathbf{a} 1\) & b 0 & c 3 & d 5 \\
 & e 100 & f -2 & g -7 & \(\mathbf{h}-13\) \\
\end{tabular}
\end{center}

\(\begin{array}{ll}6 & 1\end{array}\)

\(\begin{array}{ll}7 & 0\end{array}\)

\(8 \quad\) a 2.8971

d -2.2680

b -3.5229

c 7.8573

\(9 \quad\) a 6

d -3

g \(\frac{1}{8}\)

j 4

e 56.8028

b \(5 \quad\) c -4

e 27

f 25

i 2

l 2 f -3.9085

\section*{Review exercise}
1 a \(a^{17}\)
b \(1296 m^{16}\)
c \(a^{6} b^{2}\)
2 a \(\frac{1}{16}\)
b 6
c \(\frac{1}{10000}\)
d \(\frac{81}{16}\)
3 a \(\frac{1}{a^{8}}\)
b \(\frac{14}{a^{3}}\)
c \(\frac{4}{a^{2}}\)
d \(\frac{1}{4 a^{6}}\)
4 a \(\frac{1}{b^{3}}\)
b \(\frac{2}{x^{4}}\)
c \(\frac{5}{x^{3}}\)
d \(\frac{1}{2 x^{3}}\)
e \(\frac{1}{5 a^{4}}\)
f \(2 x^{2}\)
g \(4 x^{3}\)
h \(\frac{4 b^{3}}{a^{2}}\)
i \(\frac{5 m^{3}}{6}\)
5 a \(\frac{1}{36}\)
b \(\frac{1}{64}\)
c \(\frac{1}{16}\)
d \(\frac{1}{5}\)
e \(\frac{1}{100}\)
6 a 1
b 5
c 1
d 7
e 1
g 1
h \(\frac{1}{3}\)
f 5
i 4
7 a \(8 a b\)
b \(a b^{5}\)
c \(2 b\)
d \(\frac{1}{32 a}\)
e \(\frac{9}{a^{2}}\)
f \(\frac{a^{4}}{2}\)

\(82^{2 n-4}\)

\(96^{3 x}\)
10 a \(2^{-\frac{1}{6}}\)
b \(a^{\frac{11}{20}}\)
c \(2^{\frac{7}{3}}\)
d \(2^{\frac{3}{5}}\)
11 a \(4.2 \times 10^{3}\)
b \(6.2 \times 10^{-3}\)
c \(7.4 \times 10^{8}\)
d \(2 \times 10^{-7}\)

b 112000

d 0.0097

f 0.000064

h 402

13

\(\begin{array}{lll}\text { a } 20 & \text { b } 500 & \text { c } 420\end{array}\)

\(\begin{array}{lll}\text { d } 34000 & \text { e } 0.0068 & \text { f } 0.0492\end{array}\)

\(\begin{array}{llll}\text { g } 480 & \text { h } 600 & \text { i } & 0.007\end{array}\)

14 a 3

b 4

c -2

d 0

e -2

f -3

g -4

h -3

15 a 1

b 5

c 15

d -1

e -2

f -6

16 a \(-\frac{5}{3}\)

b \(-\frac{7}{3}\)

c \(-\frac{1}{5}\)

d -2

e -5

f 6

17 a 12

b \(\frac{1}{2}\)

c 4

d \(\frac{1}{128}\)

e \(\frac{1}{2}\)

f \(7^{\frac{1}{2}}\)

18 a \(\frac{3^{7} b^{2}}{3 a^{4}}\)

b \(\frac{1}{x^{10}}\)

c \(\frac{3^{7} a^{8}}{2^{7} b^{9}}\)

d \(a b^{4}\)

\(19 \quad\) a 32

b \(3^{7}=2187\)

c 1

d \(7^{2}=49\)

e \(\frac{1}{10}\)

f \(\frac{\sqrt{5}}{5}\)

g 5

h 3

i 10

20 a 8400

b 9261

c \(8000 \times(1.05)^{n}\)

\section*{Challenge exercise}
\(\begin{array}{llllll}\mathbf{1} & \text { a } 120 x & & \text { b } 128 & & \text { c } \frac{x}{16} \\ \mathbf{2} & -8 & \mathbf{3} & \frac{1}{6} & \mathbf{4} & -1\end{array}\)

5 a \(2306,2308,2312,2320,2336,2368,2432,2560\)

b \(n=12\)

\(\begin{array}{ll}\text { a } x=3, y=2 & \text { b } x=\frac{1}{2}, y=2\end{array}\)

\(\begin{array}{ll}\text { c } x=5, y=3 & \mathbf{d} x=3, y=7\end{array}\)

8 a \(a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}} \quad\) b \(x+x^{-1}+3^{\frac{1}{2}}\)

9 a \(3 a-2 a^{-\frac{1}{3}} b^{\frac{1}{2}}-a^{\frac{1}{3}} b^{-\frac{1}{2}}-6 a^{\frac{2}{3}} b+4 a^{-\frac{2}{3}} b^{\frac{3}{2}}+2 b^{\frac{1}{2}}\)

b \(a-b^{2}\)

\(103^{\frac{1}{3}}>2^{\frac{1}{2}}>5^{\frac{1}{5}} \quad 11 x y\)

\(\begin{array}{lll}12 & 13 & 18063\end{array}\)

\section*{Chapter 10 answers}
10A Review

\section*{Chapter 1: Consumer arithmetic}
\(1 \$ 4500\)

23 years

\(3 \quad 8.7 \%\) p.a.

\(4 \quad\) a \(\$ 239.40\)

b \(\$ 72\)

\(520 \%\)

6 a \(\$ 24\)

b \(\$ 120\)

7 a 108

b 103.5

c 36

d 20

e \(25 \%\) decrease

f \(150 \%\) increase

g 253.75

8
a \(4 \%\) decrease
b \(15.5 \%\) increase
c \(4.32 \%\) decrease
d \(19 \%\) decrease

\(9 \quad 11.384 \%\) increase

10 a \(20 \%\) increase

b \(5 \%\) increase

c \(20 \%\) decrease

d \(15 \%\) decrease

\(11 \$ 56314.12\)

12 a \(\$ 19042.49\)

b \(\$ 19301.25\)

c \(\$ 19362.03\)

\section*{Chapter 2: Review of surds}
1 a \(5 \sqrt{2}+2 \sqrt{3}\)
b \(7 \sqrt{5}+4\)
2 a \(3 \sqrt{2}\)
b \(8 \sqrt{2}\)
c \(24 \sqrt{2}\)
d \(9 \sqrt{3}\)
3 a \(-\sqrt{2}\)
b \(26 \sqrt{3}\)
c \(16 \sqrt{3}\)
d \(8 \sqrt{7}\)
e \(7 \sqrt{5}+18 \sqrt{2}\)
f \(2 \sqrt{a}\)
4 a \(4 \sqrt{3}-4\)
b \(2 \sqrt{2}-13\)
c 13
d \(5-2 \sqrt{6}\)

5 a 8

b \(a-b\)

c \(a+2 \sqrt{a b}+b\)

6 a \(\sqrt{6}\)

b \(\sqrt{2}+\sqrt{3}\)

c \(\frac{3+5 \sqrt{3}}{3}\)

d \(\frac{15-\sqrt{5}}{5}\)

e \(2 \sqrt{15}+2 \sqrt{2}\)

f \(\frac{2 \sqrt{2}-1}{3}\)

7 a \(\frac{7+2 \sqrt{10}}{3}\)

b \(\frac{13+5 \sqrt{6}}{19}\)

c \(\frac{4 \sqrt{3}+4 \sqrt{2}+\sqrt{6}+2}{2}\)

d \(\frac{14 \sqrt{2}-\sqrt{5}}{9}\)

\section*{Chapter 3: Algebra review}
1 a \(9 m n^{2}-3 m\)

b 0

c \(7 a b^{2}+a^{2}\)

d \(10 p^{2} q^{2}-5 p^{3}\)

2 a \(5 a+4\)

b \(2 b+21\)

c \(6 x^{2}-2 x\)

d \(17 y-5 y^{2}\)

e \(2 x^{2}+11 x+5\)

f \(6 a^{2}+17 a-14\)

g \(y^{2}+5 y+9\)

h \(b^{2}+23 b-1\)

\(3 \begin{array}{ll}\text { a } 6 x^{2}+33 x+45 & \text { b } 12 a^{2}-10 a-8\end{array}\)

c \(10 y^{2}+7 y-14 \quad\) d \(7 b^{2}+37 b+1\)

4 a \(\frac{5}{4} a+1\)

b \(\frac{7}{6} b+\frac{7}{3}\)

c \(\frac{11 x^{2}}{12}+\frac{31 x^{2}}{12}\)

d \(-\frac{7 y^{2}}{10}-\frac{y}{10}\)

e \(\frac{2 x^{2}}{3}+\frac{7 x}{6}+\frac{1}{4}\)

f \(-\frac{4 y^{2}}{9}+\frac{15 y}{4}+\frac{73}{12}\)

5 a \(x=\frac{17}{2}\)

b \(x=-\frac{10}{3}\)

c \(x=-\frac{11}{3}\)

d \(y=-4\)

e \(x=\frac{38}{3}\)

f \(y=-\frac{10}{3}\)

g \(x=16\)

h \(x=\frac{79}{7}\)

i \(y=-9\)

j \(x=\frac{65}{2}\)

6 a \((x-5)\) metres

b length \(17.5 \mathrm{~m}\), width \(12.5 \mathrm{~m}\)

\(724 \mathrm{~km} / \mathrm{h}\)

84 litres

9 a \(x>4\)

b \(x \geq \frac{11}{6}\)

c \(x<37\)

d \(x \leq-\frac{17}{7}\)

e \(x \leq-5\)

f \(x>3\)

\(109.61 \times 10^{6}\) watts

11 a \(s=42.225\)

b \(a=\frac{2(s-u t)}{t^{2}}\)

12 a \(g=\frac{4 \pi^{2} p}{T^{2}}\)

b \(g=9.8\)

13 a \(x=\frac{c-b}{a}\)

b \(x=\frac{c-a b}{a}\)

c \(x=\frac{c d-b}{a}\)

d \(x=\frac{c-b}{r-t}\)

e \(x=y a^{2}\)

f \(x=\frac{y c}{y-c}\)

g \(x=\frac{n-p}{m^{2}}\)

h \(x=\frac{-(a+b)}{a}\)

i \(x=\frac{3 y-1}{2}\)

14 a \(x^{2}-25\)

b \(x^{2}-4\)

d \(25 x^{2}-4 y^{2}\)

e \(\frac{1}{4} a^{2}-1\)

c \(9 x^{2}-1\)

f \(\frac{1}{16} x^{2}-\frac{4}{9} y^{2}\)

15 a \((x-6)(x+6)\)

b \((a-8)(a+8)\)

c \((9 b-1)(9 b+1)\)

d \((3 x-2 y)(3 x+2 y)\)

\[
\begin{aligned}
& 16 \text { a } x(x-18) \\
& \text { b } 3 x(x-6) \\
& \text { c } 2(3 b-5)(3 b+5) \\
& \text { e } 7(2 x-3 y)(2 x+3 y) \\
& \text { d } 3(2 b-3)(2 b+3) \\
& \mathbf{g}\left(\frac{1}{2} x-y\right)\left(\frac{1}{2} x+y\right) \\
& \text { f } 6(3 a-2 b)(3 a+2 b) \\
& \text { h } 3\left(\frac{1}{2} x-\frac{2}{5} y\right)\left(\frac{1}{2} x+\frac{2}{5} y\right) \\
& 17 \text { a }(x+2)(x+3) \\
& \text { b }(x+2)(x+6) \\
& \text { c }(x-1)(x-2) \\
& \text { d }(x-1)(x-5) \\
& \text { e }(x-3)(x-6) \\
& \text { f }(x-6)(x+1) \\
& \mathbf{g}(x-5)(x+2) \\
& \text { h }(x-4)(x+2) \\
& \text { i }(x-7)(x+3)
\end{aligned}
\]

a \(2(x+3)(x-7)\)

\section*{Chapter 4: Lines and linear equations}
1 a \(2 \sqrt{13}\)

b \(2 \sqrt{2}\)

c \(\sqrt{34}\)

d \(\sqrt{34}\)

2 a \((3,2)\)

b \((4,3)\)

c \(\left(-\frac{1}{2}, \frac{7}{2}\right)\)

d \(\left(-\frac{5}{2},-\frac{1}{2}\right)\)

3 a \(\frac{2}{3}\)

b 1

c \(\frac{3}{5}\)

d \(\frac{5}{3}\)
4 a \(m=2, c=-1\)

b \(m=1, c=3\)

c \(m=-1, c=7\)

d \(m=-1, c=4\)

e \(m=-\frac{2}{3}, c=\frac{1}{3}\)

f \(m=\frac{3}{4}, c=-\frac{1}{2}\)

5 a i 3

ii \(-\frac{1}{3}\)

c i \(\frac{1}{2}\)

ii -2

b i -2

d i \(-\frac{2}{3}\)

ii \(\frac{1}{2}\)

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-045(2)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-045}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-045(3)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-045(4)}
\end{center}

7 a \(y=3 x+2\)

b \(y=2 x+1\)

c \(y=-x-3\)

d \(x+2 y=8\)

8 a \(3 x+2 y=6\)

b \(2 y-x=2\)

c \(y=3\)

d \(x=5\)

e \(y=2 x-1\)

f \(y=1-x\)

9 a \(x=2, y=3\)

b \(x=1, y=1\)

c \(x=6, y=-5\)

d \(x=\frac{1}{2}, y=\frac{2}{3}\)

e \(x=6, y=-5\)

f \(x=\frac{1}{13}, y=\frac{44}{13}\)

\(10 a=3, b=2\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-045(1)}
\end{center}

11 a 3
b i \(a-2\)
ii \((a-2,1)\)
c i \(\frac{2}{a}\)
ii \(\frac{2}{4-a}\)
d i \(\frac{2}{5}\)
ii \(y=\frac{2}{5} x+1\)
iii \(\left(\frac{5}{2}, 2\right)\)
e i \(\sqrt{a^{2}+4}\)
ii \(3 \sqrt{5}\)

12 a \(31000 \mathrm{~L}\)

b \(40 \mathrm{~L} / \mathrm{h}\)

c \(V=31000-40 t\)

d after 775 hours

13 a i A started in Melbourne, B started \(100 \mathrm{~km}\) from Melbourne.

ii A finished \(160 \mathrm{~km}\) from Melbourne, B finished in Melbourne.

iii A took 2 hours 20 minutes, B took 1 hour 36 minutes.

iv A averaged \(68 \frac{4}{7} \mathrm{~km} / \mathrm{h}\), B averaged \(62.5 \mathrm{~km} / \mathrm{h}\).

b A: \(d=\frac{480}{7} t\), B: \(d=-62.5 t+100\)

c They passed after 46 minutes of travelling.

\section*{Chapter 5: Quadratic equations}
1 a \(-4,4\)
b \(-2,2\)
c \(-7,7\)
d \(-\frac{5}{2}, \frac{5}{2}\)
e \(\frac{1}{2},-\frac{1}{2}\)
f \(\frac{5}{2},-\frac{5}{2}\)
2 a \(0, \frac{3}{2}\)
b \(0,-\frac{1}{3}\)
c \(0, \frac{3}{5}\)
d \(0, \frac{1}{2}\)
e \(0, \frac{2}{7}\)
f \(0,-5\)
g 0,7
h 0,12
i \(0,-\frac{1}{4}\)

3
a \(a=4\) or -3
b \(t=-3\) or -5
c \(m=-7\) or 3
d \(m=4\) or -1
e \(x=4\)
f \(b=9\) or -3
b \(\frac{1}{3}, \frac{11}{3}\)
c \(\frac{1}{3},-8\)
d \(-\frac{7}{6},-\frac{3}{2}\)
e \(-\frac{1}{3}, 1\)
f \(-\frac{3}{2}, \frac{5}{6}\)
g \(\frac{3}{2},-4\)
h 3

4 a \(\frac{5}{2}, 7\)

5 a 3

b -5

d 4

e \(-\frac{3}{2}\)

c -1

f 5

6 a \((x+\sqrt{5})(x-\sqrt{5})\)

b \((x+2-2 \sqrt{2})(x+2+2 \sqrt{2})\)

c \(2(x-3-\sqrt{5})(x-3+\sqrt{5})\)

7
a \(y=-1 \pm \sqrt{5}\)
b \(a=2 \pm \sqrt{6}\)
c \(x=1 \pm \frac{\sqrt{41}}{2}\)
d \(x=\frac{7 \pm \sqrt{41}}{2}\)
e \(y=-\frac{1}{4} \pm \frac{\sqrt{2}}{4}\)
f \(x=\frac{1 \pm \sqrt{33}}{4}\)

g \(n=\frac{5 \pm \sqrt{41}}{2}\)

h \(x=\frac{-1+\sqrt{2}}{4}\)

8 a \(d= \pm \sqrt{2}\)

b \(y= \pm \frac{\sqrt{30}}{2}\)

c \(x=10 \pm 2 \sqrt{5}\)

d \(y=4 \pm \sqrt{13}\)

e \(m=\frac{1}{2} \pm \frac{\sqrt{5}}{2}\)

f \(n=\frac{3 \pm \sqrt{21}}{2}\)

\(9 \quad \mathbf{a}(x+7)\) metres, \((x+8)\) metres

b \(5 \mathrm{~cm}, 12 \mathrm{~cm}, 13 \mathrm{~cm}\)

\(\begin{array}{lll}10 & \text { a } 381.6 \mathrm{~m} & \text { b } 102 \text { seconds }\end{array}\)

11 a length \(=24-2 x \mathrm{~cm}\), width \(=17-2 x \mathrm{~cm}\)

b \(x=2.5\)

c \(19 \mathrm{~cm} \times 12 \mathrm{~cm} \times 2.5 \mathrm{~cm}\)
12 a \(x^{2} \mathrm{~m}^{2}\)
b \((8 x+16) \mathrm{m}^{2}\)
c \(8 x+16=\frac{5 x^{2}}{4}\)
d \(8 \mathrm{~m} \times 8 \mathrm{~m}\)
13 a \(b= \pm 4\)
b \(b<-4\) or \(b>4\)
c \(-4<b<4\)
14 a \(a=\frac{4}{3}\)
b \(a<\frac{4}{3}\)
c \(a>\frac{4}{3}\)

\section*{Chapter 6: Surface area and volume}
1 a \(2200 \mathrm{~cm}^{2}\)
b \(6000 \mathrm{~cm}^{3}\)

\(2 \quad 5 \mathrm{~cm}\)

\(3 \quad 5 \mathrm{~cm}\)

4 a \(44 \mathrm{~cm}^{2}\)

b \(728 \mathrm{~cm}^{2}\)

c \(880 \mathrm{~cm}^{3}\)

5 a 4712 litres

b \(64 \mathrm{~cm}\)

6 a i \(360 \mathrm{~cm}^{2}\)

ii \(400 \mathrm{~cm}^{3}\)

b i \(96 \pi \mathrm{cm}^{2}\)

ii \(96 \pi \mathrm{cm}^{3}\)

\(\begin{array}{lll}\text { a } 254.6 \mathrm{~m}^{2} & \text { b } 160 \mathrm{~m}^{3}\end{array}\)

\(\begin{array}{lll}\text { a } 4 \mathrm{~cm} & \text { b } 8 \sqrt{6} \mathrm{~cm} & \text { c } 328 \mathrm{~cm}^{3}\end{array}\)

\(9 \begin{array}{ll}\text { a } 18 \pi \mathrm{m}^{2} & \text { b } \frac{45}{4} \pi \mathrm{m}^{3}\end{array}\)
10 a 9,27
b \(2.25,3.375\)
c 2,8

d 6,216

e 5,25

f 9,81

\section*{Chapter 7: The parabola}
1 a \(-1,-3\)

b \(-\frac{1}{2}, 6\)

c \(-4+\sqrt{3},-4-\sqrt{3}\)

d \(2+\sqrt{2}, 2-\sqrt{2}\)

2 a \(y=(x+3)^{2}-6,(-3,-6)\)

b \(y=(x-2)^{2}-2,(2,-2)\)

c \(y=2\left(x+\frac{3}{2}\right)^{2}-\frac{7}{2},\left(-\frac{3}{2},-\frac{7}{2}\right)\)

d \(y=3\left(x+\frac{4}{3}\right)^{2}-\frac{10}{3},\left(-\frac{4}{3},-\frac{10}{3}\right)\)

3 a \(y=-2 x^{2}+6 x+8\)

b \(y=x^{2}-8 x+15\)

4 a \(y=2 x^{2}-12 x+16\)

b \(y=-3 x^{2}+12 x-7\)

5 a \(y=3 x^{2}-12 x+12\)

b \(y=-x^{2}-2 x+2\)

c \(y=x^{2}-10 x+21\)

6 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(1)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(3)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(2)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(10)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(9)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(6)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(5)}
\end{center}

k
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-047(7)}

I

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(12)}
\end{center}

7 a \(\left(\frac{1}{3},-\frac{4}{3}\right)\)

b \(\left(-\frac{1}{3}, 0\right),(1,0)\)

8 a \(12-2 x\)

b \(A=x(12-2 x)\)

c \(3 \mathrm{~m}\) by \(6 \mathrm{~m}\)

9 a \((-1,-8)\)

b \((0,-7),(-1-2 \sqrt{2}, 0),(-1+2 \sqrt{2}, 0)\)

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(11)}
\end{center}

d \(-2 \sqrt{2}-1 \leq x \leq 2 \sqrt{2}-1\)

10 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-047(8)}
\end{center}

b \(1-\frac{\sqrt{3}}{2}<x<1+\frac{\sqrt{3}}{2}\)
11 a \(-6<x<5\)
b \(x \leq-3\) or \(x \geq-2\)
c \(x \leq-6\) or \(x \geq 10\)

\section*{Chapter 8: Review of congruence and similarity}
Note: Proofs are not given in answers for this section.

\(1 \begin{array}{ll}\text { a } a=2, b=2.5 & \text { b } x=2, y=6\end{array}\)

\(2 \quad 16 \frac{2}{3} \mathrm{~cm}\)

3 a i AAA

ii \(x=3.75\)

b RHS

\(\begin{array}{ll}\text { a i AAA } & \text { ii } x=\frac{12}{5} \\ \text { b ASA }\end{array}\)

6 a \(\angle D E C=\angle A B C\) (given both \(\left.90^{\circ}\right), \angle E C D=\angle B C A\) (common), \(\triangle D E C\) is similar to \(\triangle A B C\) (AAA)
b \(x=5\)
c \(\alpha=126.87^{\circ}\)

7 a \(\triangle P Q L\) is similar to \(\triangle L T P\) is similar to \(\triangle N M L\) (AAA)

b \(L T=75 \mathrm{~m}\)

\(12 x=\frac{a d}{b}\)

\section*{Chapter 9: Indices, exponentials and}
 logarithms - part 11 a \(a^{4}\)
b \(24 x^{7} y^{5}\)
c \(a^{2} b\)
d \(\frac{1}{a^{4}}\)
e \(2 x y\)
f \(2 y\)
g \(\frac{2 y^{3}}{x^{4}}\)
h \(\frac{b^{8}}{a}\)
i \(\frac{y^{4}}{x^{2}}\)

2 a \(3.2 \times 10^{3}\)

b \(5.76 \times 10^{5}\)

c \(2.67 \times 10^{-4}\)

d \(2.5 \times 10^{-2}\)
3 a \(9.017 \times 10^{4}\)
b \(4.552 \times 10^{-4}\)
c \(6.516 \times 10^{-6}\)
d \(9.468 \times 10^{-20}\)
4 a 3
b 3
c 2
d 2 e 3
f 4
5 a 4
b 8
c 9
d 8

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-048(1)}
\end{center}

6 a \(b^{4}\)

b \(\frac{a^{2}}{b}\)

c \(\frac{a^{2}}{b}\)

d \(a^{\frac{1}{3}} b^{\frac{1}{3}}\)

e \(\frac{a}{b^{2}} \quad \mathbf{f} \frac{3 a^{\frac{2}{3}}}{b}\)

g \(\frac{4 a^{\frac{3}{2}}}{b}\)

h \(a^{2}\)

7 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-048(3)}
\end{center}

c

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-048}
8 a \(x=\frac{1}{5}\)
b \(x=\frac{1}{2}\)
d \(x=\frac{3}{4}\)
e \(x=-\frac{3}{4}\)
c \(x=-2\)
9 a \(x=5\)
b \(x=1\)
c \(x=\frac{11}{4}\)
d \(x=\frac{19}{18}\)
e \(x=-\frac{17}{15}\)
f \(x=6\)

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-048(6)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-048(5)}
\end{center}

10 a 5400

b 5832

c \(5000 \times 1.08^{x}\)

d 23305

11 a i \(4116 \quad\) ii 4037

iii \(4200 \times 0.98^{x}\)

b Population

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-048(4)}
\end{center}

c 14 years
12 a 4
b 2
c \(\frac{1}{2}\)
d \(\frac{3}{2}\)

\section*{10B Miscellaneous questions}
1 a \(140 \mathrm{~km} \quad\) b \(x \mathrm{~km}\)

c \(\frac{x}{35}\) hours, \(\frac{140}{x}\) hours

d \(70 \mathrm{~km} / \mathrm{h}\)

2 a i \(1 \frac{1}{8}\) hours ii \(\frac{5}{8}\) hours \(\quad\) iii \(144 \mathrm{~km} / \mathrm{h}\)

b i \(\quad(d-m) \mathrm{km}\) ii \(T=\frac{d-m}{v}\)

iii \(t=\frac{d-m}{v}-n\)

iv \(w=\frac{v(d-m)}{d-m-n v}\)

\(\mathbf{v} v=\frac{w(d-m)}{w n+d-m}\)

c \(56 \frac{16}{19} \mathrm{~km} / \mathrm{h}\)

3 a \(X\) takes \(2 \frac{1}{2}\) hours, \(Y\) takes \(1 \frac{2}{3}\) hours

b \(X\) averages \(24 \mathrm{~km} / \mathrm{h}, Y\) averages \(36 \mathrm{~km} / \mathrm{h}\)

c They pass \(31.2 \mathrm{~km}\) from \(A\).

4 a \(2 \mathrm{~km}\)

b Shen finishes first by approximately 10 minutes.

5 a \(30 \mathrm{~m} / \mathrm{s}\)

b \(V \mathrm{~m} / \mathrm{s} \uparrow\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-048(2)}
\end{center}

c 7 seconds

d \(5 \mathrm{~m} / \mathrm{s}^{2}\)
e i \(180 \mathrm{~m}\)
ii \(20 t\) metres
f i \(90 \mathrm{~m}\)
ii \(180 \mathrm{~m}\)

iii \((30 t-180)\) metres

g i 18 seconds ii \(360 \mathrm{~m}\)

6 a \(\mathbf{i} \frac{1}{3}\)

ii \(\frac{4}{3}\)

b i \(y=\frac{1}{3} x+2\)

ii \(y=\frac{4}{3} x-4\)
c \((6,4)\)

d 27

e 18

f \(\left(\frac{6 b^{2} 12 b}{4-b^{2}}, \frac{8 b+4 b^{2}}{4-b^{2}}\right)\); when \(b=0, E=(0,0)\); when \(b=2\) the lines are parallel so \(E\) does not exist; when \(b=-2\) the lines are the same.

7 a \(20 \mathrm{~km}\)
b i \(10 \sqrt{2} \mathrm{~km}\)
ii \(25 \sqrt{2} \mathrm{~km}\)

c \((15 \sqrt{2}, 10)\)

d i \(P=(0,-20), R=(25 \sqrt{2}, 0)\)

ii \(\frac{2 \sqrt{5}}{5}\)

iii \(y=\frac{2 \sqrt{2}}{5} x-20\)

8
b i \(7.55 \mathrm{~cm}\)
ii \(62.9 \mathrm{~cm}^{3}\)

ii \(V=\pi r^{3}\)

iii \(A=\sqrt{10} \pi r^{2}\)

9 a \(\angle A G B=\angle C G F\) (vertically opposite angles at \(G\) ), \(\angle G A B=\angle G C F\) (or \(\angle A B G=\angle C F G\) ) (alternate angles in parallel lines), so \(\triangle A G B\) is similar to \(\triangle C G F\) (AAA)

b \(\triangle E B A, \triangle B F C\)
c i \(3: 2\)
ii \(2: 3\)

10
a i AAA
ii \(x=5 \frac{1}{7}\)
iii \(\frac{24}{49}\)
b i \(x=\frac{3 y}{2}\)
ii \(\frac{4}{9}\)

11 a i 3

ii \(3 \sqrt{5}\)

iii \(\sqrt{6}\)

iv \(1+\sqrt{3}\)

b \(y=\frac{1}{x}\)

12 a \(\angle A B E=\angle A C D\) (given), \(\angle B A E=\angle C A D\) (common), so \(\triangle A B E\) is similar to \(\triangle A C D\) (AAA)

b i \((x+10)\) metres \(\quad\) ii \(\frac{x}{x+10}=\frac{2}{3}\)

c 20 metres

d i \(\frac{1}{2} x^{2} \mathrm{~m}^{2} \quad\) ii \(\left(450-\frac{1}{2} x^{2}\right) \mathrm{m}^{2}\)

e i \(\frac{1}{2} x^{2}=450-\frac{1}{2} x^{2} \quad\) ii \(15 \sqrt{2}\) metres

13 a \(A=(15-x)^{2}-(15-2 x)^{2}\)

b \(75 \mathrm{~m}^{2}\)

14 a \(10 \mathrm{~km}\)

b \(0.5 \mathrm{~km}\)

c \(A=(1.13,0.2), B=(8.87,0.2)\)

d \(7.75 \mathrm{~km}\)

15

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-049(1)}
\end{center}

b i \(y=(x-2)^{2}+5\)

ii 9

iii see graph

c \(11 \mathrm{~km}\)

d i \(y=2 a-2\)

ii \(y=a^{2}-4 a+9\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-049(2)}
\end{center}

g \(2 \mathrm{~km}, 3 \mathrm{~km}\) east of origin

16 b i \((-1,-10)\) ii \(\left(-1+\frac{\sqrt{30}}{3}, 0\right),\left(-1-\frac{\sqrt{30}}{3}, 0\right)\)

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-049}
\end{center}

d \(200 \mathrm{~km}\)

e \(3 x^{2}+6 x-7=5 x+3\)

\(3 x^{2}+x-10=0\)

f \(\left(\frac{5}{3}, \frac{34}{3}\right),(-2,-7)\)

17 a i \(h=r \tan \theta\)

ii \(s=\frac{r}{\cos \theta}\)

b i Area \((A)=\frac{1}{2} \pi r^{2}\)

ii \(\operatorname{Area}(B)=r^{2} \tan \theta\)
d \(57.5^{\circ}\)
e \(4 \pi \approx 12.57 \mathrm{~mm}^{2}\)

\section*{10C Problem-solving}
\(1 a=\frac{2 r}{5}, b=\frac{6 r}{5}-3\)

\(2 x=12\) and \(y=2\), or \(x=12\) and \(y=-2\)

or \(x=-12\) and \(y=2\), or \(x=-12\) and \(x=-2\)

or \(x=36\) and \(y=34\), or \(x=36\) and \(y=-34\),

or \(x=-36\) and \(y=34\), or \(x=-36\) and \(y=-34\)

3 a i \(2 \sqrt{259} \mathrm{~cm} \quad\) ii \(2 \sqrt{257} \mathrm{~cm}\)

b \(5 \mathrm{~cm}\)

\(4(4-2 \sqrt{3}) \mathrm{m}\)

5 a \(16 \pi \mathrm{cm}^{3}\)

b \(\frac{16 \pi}{3} \mathrm{~cm}^{3}\)

c \(\frac{32 \pi}{3} \mathrm{~cm}^{3}\)

d \(8 \pi \mathrm{cm}^{3}\)

6 Midpoint of \(A B=X\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

Midpoint of \(B C=Y\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)\)

Midpoint of \(C D=Z\left(\frac{x_{3}+x_{4}}{2}, \frac{y_{3}+y_{4}}{2}\right)\)

Midpoint of \(D A=W\left(\frac{x_{4}+x_{1}}{2}, \frac{y_{4}+y_{1}}{2}\right)\)

Gradient of \(X Y=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}\)

Gradient of \(W Z=\frac{y_{3}-y_{1}}{x_{3}-x}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-050(3)}
\end{center}

Therefore, \(X Y \| W Z\)

\(X Y=\frac{1}{2} \sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}}\)

\(W Z=\frac{1}{2} \sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}}\)

so \(X Y=W Z\)

\(X Y Z W\) is a parallelogram.

\((P A)^{2}+(P C)^{2}=x^{2}+y^{2}+(x-a)^{2}+(y-b)^{2}\)

\((P B)^{2}+(P D)^{2}=(x-a)^{2}+y^{2}+x^{2}+(y-b)^{2}\)

so \((P A)^{2}+(P C)^{2}=(P B)^{2}+(P D)^{2}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-050(2)}
\end{center}

\(\triangle A Q X \equiv \triangle C R Z \equiv \triangle B P Y\)

Quad \(A Q R Z \equiv\) Quad \(B P Q X \equiv \operatorname{Quad} C R P Y\)

\(\triangle P Q R\) is equilateral.

\(\triangle B P Y\) is similar to \(\triangle B C Z\) (AAA)

Assume \(A B=B C=C A=1\).

Then \(C Z=A X=Y B=\frac{1}{3}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-050}
\end{center}

Apply Pythagoras' theorem twice to find \(Z B=\frac{\sqrt{7}}{3}\).

The enlargement factor from \(\triangle B P Y\) to \(\triangle B C Z\) is \(\sqrt{7}\). \(\therefore\) area of \(\triangle B P Y=\frac{1}{7}\) area of \(\triangle B C Z\)

\(=\frac{1}{21}\) area of \(\triangle A B C\)

area of \(\triangle P Q R=3 \times\) area of \(\triangle P B Y\)

\[
=\frac{1}{7} \times \text { area of } \triangle A B C
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-050(4)}
\end{center}

\(9 \quad \triangle A R D\) is similar to \(\triangle P R B\)

Therefore \(\frac{A R}{R P}=\frac{R D}{R B}\)

\(\triangle B R A\) is similar to \(\triangle D R Q\)

\(\therefore \frac{R D}{R B}=\frac{R Q}{A R}\)

\(\therefore \frac{A R}{R P}=\frac{R Q}{A R}\)

\(\therefore A R^{2}=R P \times R Q\)

10 Draw \(M X \| A B\)

\(\angle A B C=\angle M X C\) (corresponding angles \(A B \| M X\) )

\(\angle A B C=\angle M C X(\triangle A B C\) isosceles \()\)

\(\therefore \angle M X C=\angle M C X\) ( \(\triangle M X C\) isosceles)

\(\therefore M X=M C\)

\(\triangle L N B\) is similar to \(\triangle L M X\)

\(N M=M L\) (given)

\(\therefore L N=2 L M\) and \(N B=2 M X=2 M C\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-050(1)}
\end{center}

\(11 \angle A E B=60^{\circ}(\triangle A E B\) is isosceles \()\)

\(\angle D E A=30^{\circ}\) (complementary to \(\left.\angle E A B\right)\)

\(\angle C B E=30^{\circ}\) (complementary to \(\angle E B C\) )

\(\triangle B E C=\triangle A E D(\mathrm{SAS})\)

\(\angle A D E=\angle A E D=\angle B E C=\angle E C B=75^{\circ}\)

\((\triangle B E C\) and \(\triangle A E D\) are isosceles)

\(\angle E B F=30^{\circ}+60^{\circ}\) (square and equilateral triangle)

\(\triangle B E F\) is isosceles

\(\angle B E F=45^{\circ}\) (angle sum of \(\left.\triangle B E F\right)\)

Therefore,

\(\angle A E D+\angle A E B+\angle A E B+\angle B E F=75^{\circ}+60^{\circ}+45^{\circ}=180^{\circ}\)

Points \(A, E\) and \(F\) are collinear.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-050(5)}
\end{center}

\section*{Chapter 11 answers}
\section*{Exercise 11A}
1 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(11)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(3)}
\end{center}

2 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(1)}
\end{center}

3 a yes

b no

d yes

e yes

4 a yes

b no

d yes

e no

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(10)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(4)}
\end{center}

b
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-051(9)}

6 a centre \((-2,-3)\), radius 2

b centre \((1,-4)\), radius \(\sqrt{13}\)

c centre \((3,4)\), radius 8

d centre \((7,4)\), radius 5

e centre \((4,3)\), radius \(\sqrt{10}\)

f centre \((4,2)\), radius \(\sqrt{10}\)

\(7 \quad \mathbf{a}(x-1)^{2}+(y-3)^{2}=9 \quad\) b \((x+2)^{2}+(y-1)^{2}=16\)

c \((x-4)^{2}+(y+1)^{2}=1 \quad \mathbf{d}(x-2)^{2}+y^{2}=4\)

\(8 \sqrt{(17-5)^{2}+(17-12)^{2}}=13\), so the point \((17,17)\) is 13 units from the centre at \((5,12)\). The equation of the circle is \((x-5)^{2}+(y-12)^{2}=169\).

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(5)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(8)}
\end{center}

c no

f yes

c yes

f yes

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(2)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(6)}
\end{center}

b \(y=-\frac{1}{2} \quad\) c \(y=2\)

e \(y=-\frac{3}{2}\)
2 a \(y=4\)
b \(y=3\)
c \(y=-24\)
d \(y=-8\)
e \(y=16\)

3 a \(y=1\)

b \(y=\frac{1}{2}\)

c \(y=2\)
d \(y=-\frac{2}{3}\)
e \(y=\frac{2}{3}\)

4 a \(y=1\)

b \(y=-1\)

d \(y=2\)

e \(y=-4\)

c \(y=\frac{1}{2}\)

5 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-051(7)}
\end{center}

b \(-1,-2,-4,4,2,1\)

6

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(3)}
\end{center}

b \(-\frac{1}{2},-1,-2,2,1, \frac{1}{2}\)

7
a -12000
b -60
c 4
d \(\frac{1}{2}\)
e \(\frac{1}{12}\)

8
a -2
b -3
c -600
d 600
e \(\frac{6}{997}\)
b 120
c 1200
d -1200
e -12000

\(9 \quad \mathbf{a} 4\)

10 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(11)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(4)}
\end{center}

11 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(7)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(8)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(5)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(9)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(10)}
\end{center}

12 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(6)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(2)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-052(12)}
\end{center}

ICE-EM MATHEMATICS YEAR 10 ICE-EM Mathematics 103 d Photocopying is restricted under law and this material must not be transferred to another party

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(3)}
\end{center}

ii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(5)}
\end{center}

b i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(11)}
\end{center}

ii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(8)}
\end{center}

c i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(10)}
\end{center}

ii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053}
\end{center}

d i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(9)}
\end{center}

ii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(7)}
\end{center}

14 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(6)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(4)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-053(1)}
\end{center}

\section*{Exercise 11C}
1 a \((2,4),(-2,4)\)

b \((1,1),(-1,1)\)

c \((2,1)\)

d \((3,9),(4,16)\)

2 a \((-1,1)\)

b \((-5,2),(1,8)\)

c \((-2,4),(1,7)\)

d \(\left(-\frac{1}{2}, 0\right),(0,1) \quad \mathbf{e}\left(-\frac{1}{3}, 2\right),(1,6)\)

f \(\left(-\frac{2}{3}, \frac{5}{3}\right),\left(-\frac{1}{2}, 2\right)\)

3
a \((2,0)\)
b \((-3,0),(3,0)\)
c \((4,4),(-4,-4)\)
d \((-3,-6 \sqrt{2}),(3,6 \sqrt{2})\)

4

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054(11)}
\end{center}

intersection points:

\((1,3),(-3,-1)\)

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054(10)}
\end{center}

intersection points:

\((4,-1),(-1,4)\)

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054(6)}
\end{center}

intersection points:

\((-2,-4),(2,4)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054(5)}
\end{center}

intersection points:

\((-1,5),(5,-1)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054(2)}
\end{center}

intersection points: \(\left(\frac{2}{5},-\frac{11}{5}\right),(2,1)\)

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054(4)}
\end{center}

intersection point:

\((-2,2)\)

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054(3)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054}
\end{center}

intersection point:

\((3,3)\)

intersection point:

\((3,4)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054(1)}
\end{center}

no intersection point

j

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-054(8)}
\end{center}

no intersection point

5 a \((3,1),(-1,-3)\)

b \((1,1),\left(-\frac{1}{2},-2\right)\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-054(9)}

c \((1,2),\left(-\frac{2}{3},-3\right) \quad\) d \((-1,1),(1,-1)\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-054(7)}

6 a \(A=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), B=(1,1)\)

b area of \(\triangle O A X\) is \(\frac{1}{4}\), area of \(\triangle O B Y\) is \(\frac{1}{2}\)

\(7 \quad(2,3)\)

\(8(x-5)^{2}+4 x^{2}=4,(x-1)^{2}=-\frac{16}{5}\), so this quadratic equation has no solution and the line does not meet the circle.

\(9 \begin{array}{ll}\text { a } a=3 \sqrt{2} \text { or }-3 \sqrt{2} & \text { b }-3 \sqrt{2}<a<3 \sqrt{2}\end{array}\)

c \(a<-3 \sqrt{2}\) or \(a>3 \sqrt{2}\)

\(10(1,2 \sqrt{2}),(1,-2 \sqrt{2})\)

ICE-EM MATHEMATICS YEAR 10 ICE-EM Mathematics 10 3ed Photocopying is restricted under law and this material must not be transferred to another party

\section*{Exercise 11D}
1 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(8)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(14)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(1)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(18)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(16)}
\end{center}

\(\mathbf{k}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(7)}
\end{center}

m

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(13)}
\end{center}

o

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(9)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(12)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(5)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(24)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(20)}
\end{center}

j
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-055(19)}

n
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-055(17)}

\(\mathbf{p}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(11)}
\end{center}

q
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-055(22)}

\[
\begin{array}{ll}
\text { a } y \geq 2 x+4 & \text { b } y<-x+2 \\
\text { c } x+y<3 & \text { d } y \geq 2 x+1
\end{array}
\]

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(15)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(10)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(4)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(6)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(2)}
\end{center}

k

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(21)}
\end{center}

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-055(23)}
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-055(3)}

1

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-055(25)}
\end{center}

4 a \(x+y \leq 2, y-x \leq 2\)

b \(y \geq x-1, y \leq 3 x+3\)

c \(y \geq 2, x<2\)

d \(y \geq x, y<2\)

e \(y \leq x, x \leq 2, y \geq 0\)

f \(x \geq 0 y \geq 0, x \leq 3, x+y \leq 6\)

5 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(6)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(1)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(10)}
\end{center}

6

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(8)}
\end{center}

Review exercise

1 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(7)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(4)}
\end{center}

2 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(9)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-056(5)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(8)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(2)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(1)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057}
\end{center}

3 a centre \((-2,-4)\), radius \(2 \sqrt{5}\)

b centre \((-2,-1)\), radius \(\sqrt{10}\)

c centre \(\left(2,-\frac{5}{4}\right)\), radius \(\frac{\sqrt{65}}{4}\)

d centre \((2,-3)\), radius \(5 \sqrt{2}\)

4 a \(x^{2}+y^{2}=9\)

b \((x+1)^{2}+(y-4)^{2}=36\)

c \((x-2)^{2}+(y-5)^{2}=1\)

d \((x+2)^{2}+(y+6)^{2}=16\)

5 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(3)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(9)}
\end{center}

6 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(10)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(7)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-057(5)}
\end{center}

7 a \((-2,18),(2,6)\)

8 a \((3,0)\)

c \((-2,-2 \sqrt{3}),(2,2 \sqrt{3})\)

9
\((4,3)\)

10 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(5)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(1)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(4)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(15)}
\end{center}

11 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(3)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(9)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058}
\end{center}

b \(\left(-\frac{7}{5},-\frac{26}{5}\right)\),

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(18)}
\end{center}

b \((0,2),(2,-12)\)

b \((-4,0),(4,0)\)

b
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-058(2)}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(13)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(17)}
\end{center}

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-058(14)}

12 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(16)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(10)}
\end{center}

13 a \((4,3),(-3,-4)\)

c \((3,2)\)

d no points of intersection

\section*{Challenge exercise}
1 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(7)}
\end{center}

2 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(12)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-058(11)}
\end{center}

b

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-058(8)}

b \(\left(\frac{1}{2},-6\right),(3,-1)\)

3

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-059(2)}
\end{center}

b
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-059}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-059(4)}
\end{center}

4 a i \(y=-\frac{1}{\sqrt{3}} x+\frac{a}{3}\)

ii \(y=\frac{1}{\sqrt{3}} x+\frac{a}{3}\)
c \(\frac{2}{3} a\)
d \(x^{2}+\left(y-\frac{a}{3}\right)^{2}=\frac{4 a^{2}}{9}\)

6 Centre \(\left(\frac{a}{2}, \frac{a}{2}\right)\), radius \(\frac{\sqrt{2}}{2} a\), expand

\[
\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{a}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2} a\right)^{2}
\]

7 a \(\mathbf{i} x=\frac{a}{2}\)

ii \(y=\frac{b}{2}\)

b \(\left(\frac{a}{2}, \frac{b}{2}\right)\)

d \(\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{4}\)

\(8\left(x-\frac{a^{2}+b^{2}}{2 b}\right)^{2}+y^{2}=\frac{(a-b)^{4}+4 b^{4}}{4 b^{2}}\)

\(9\left(x-\frac{5}{2}\right)^{2}+y^{2}=\frac{5}{4}\)
10 a \(y-3=\frac{2}{x-4}\)

\[
y=3+\frac{2}{x-4}
\]

b \(y-3=\frac{2}{x-2}\)

\[
y=3+\frac{2}{x-2}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-059(5)}
\end{center}

11 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-059(7)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-059(6)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-059(3)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-059(1)}
\end{center}

\(12(0,3)\) and \((2,0)\)

\(13\left(-\frac{9}{5}, \frac{12}{5}\right)\) and \((-1,3)\)

\(14 x^{2}+y^{2}-x-5 y+4=0\)

\(15 \quad a x+b=\frac{1}{x}\), eliminating \(y\).

\[
\begin{aligned}
& a x^{2}+b x=1 \\
& a x^{2}+b x-1=0 \\
& x=\frac{-b \pm \sqrt{b^{2}+4 a}}{2 a} \text { and } b^{2}+4 a>0, \text { since } a>0
\end{aligned}
\]

So there are always two points of intersection between the line and the hyperbola.

\section*{Chapter 12 answers}
\section*{Exercise \(12 \mathrm{~A}\)}
\begin{center}
\begin{tabular}{llll}
1 & a \(a=7.42\) & b \(b=3.71\) & c \(c=6.17\) \\
 & d \(d=2.29\) & e \(i=3.51\) & f \(j=13.59\) \\
2 & a \(a=14.97\) & b \(b=9.01\) & c \(e=11.75\) \\
 & d \(h=38.70\) &  &  \\
3 \(\quad\) a \(25.4^{\circ}\) & b \(56.3^{\circ}\) & c \(60.3^{\circ}\) &  \\
 & d \(70.9^{\circ}\) & e \(39.6^{\circ}\) & f \(70.8^{\circ}\) \\
4 & a \(x=7.10\) & b \(\theta=44.9^{\circ}\) & c \(a=8.08\) \\
 & d \(a=2.26\) & e \(\theta=61.9^{\circ}\) & f \(\theta=52.1^{\circ}\) \\
g \(x=10.15\) & h \(y=10.63\) & i \(a=9.39\) &  \\
\end{tabular}
\end{center}

5 a \(3.92 \mathrm{~cm}, 2.52 \mathrm{~cm}, 50^{\circ} \quad\) b \(6.97 \mathrm{~cm}, 40.7^{\circ}, 49.3^{\circ}\)

c \(9.78 \mathrm{~cm}, 30.8^{\circ}, 59.2^{\circ}\)

\section*{Exercise 12B}
1 a 5
b \(6 \sqrt{2}\)
c \(6 \sqrt{3}\)
d \(4 \sqrt{3}\)
e \(4 \sqrt{3}\)
2 a 1
b -1
c \(2-\sqrt{3}\)
d \(\frac{\sqrt{2}+\sqrt{6}}{4}\)
e \(\frac{\sqrt{3}}{2}\)
f 0

\(310 \mathrm{~cm}\) and \(10 \sqrt{3} \mathrm{~cm}\)
4 a \(16 \sqrt{3} \mathrm{~cm}\)
b \(12 \sqrt{3} \mathrm{~cm}\)
c \(8 \sqrt{3} \mathrm{~cm}\)
d \(4 \sqrt{3} \mathrm{~cm}\)
e \(12 \mathrm{~cm}\)

\(5 \quad a=\frac{40 \sqrt{3}}{3}\) and \(x=\frac{40}{3} \quad 6 \quad 50(\sqrt{3}+1)\)

\(75 \sqrt{3} \mathrm{~cm}\)

\section*{Exercise 12C}
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-060(1)}
6 a \(130.6 \mathrm{~m}\)
b \(96.2 \mathrm{~m}\)
c \(162.2 \mathrm{~m}\)
d \(216.4^{\circ} \mathrm{T}\)

\(7 \quad 81 \mathrm{~m}\)
8 a \(73 \mathrm{~m}\)
b \(260 \mathrm{~m}\)
c \(16^{\circ}\)
\(9 \quad 42 \mathrm{~m}\)

\section*{Exercise 12D}
\begin{center}
\begin{tabular}{llll}
\(\mathbf{1}\) & a 10.46 & b 6.88 & c 8.69 \\
\(\mathbf{2}\) & a 8.63 & b 10.89 & c 9.34 \\
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{llll}
\(\mathbf{3}\) & \(\mathbf{a} 7.33\) & b 10.63 &  \\
\(\mathbf{4}\) & \(\mathbf{a} 54^{\circ}\) & b \(34^{\circ}\) & c \(66^{\circ}\) \\
\(\mathbf{5}\) & \(\mathbf{a} 47.34^{\circ}\) & b \(61.66^{\circ}\) & c \(16.76 \mathrm{~cm}\) \\
\(\mathbf{6}\) & \(\angle A C D \approx 35.07,26.01 \mathrm{~m}\) &  &  \\
\(\mathbf{7}\) & a \(708 \mathrm{~m}\) & b \(547 \mathrm{~m}\) &  \\
\(\mathbf{8}\) & a \(27^{\circ}\) & b \(132.56 \mathrm{~m}\) & c \(119 \mathrm{~m}\) \\
\(\mathbf{9}\) & \(16 \mathrm{~m}\) & \(\mathbf{1 0} 7.11 \mathrm{~m}\) &  \\
\end{tabular}
\end{center}

\section*{Exercise 12E}
\begin{center}
\begin{tabular}{lllll}
a \(65^{\circ}\) & b \(57^{\circ}\) & c \(42^{\circ}\) & d \(85^{\circ}\) \\
\end{tabular}
\end{center}

\(\begin{array}{lll}\mathbf{a} \frac{1}{\sqrt{2}} & \mathbf{b}-\frac{1}{\sqrt{2}}\end{array}\)
3 a \(30^{\circ}, 150^{\circ}\)
b \(35^{\circ}, 145^{\circ}\)
c \(151^{\circ}\)

4

\begin{center}
\begin{tabular}{c|c|c|c|c|c}
\(\theta\) & \(30^{\circ}\) & \(120^{\circ}\) & \(150^{\circ}\) & \(90^{\circ}\) & \(135^{\circ}\) \\
\hline
\(\sin \theta\) & \(\frac{1}{2}\) & \(\frac{\sqrt{3}}{2}\) & \(\frac{1}{2}\) & 1 & \(\frac{1}{\sqrt{2}}\) \\
\hline
\(\cos \theta\) & \(\frac{\sqrt{3}}{2}\) & \(-\frac{1}{2}\) & \(-\frac{\sqrt{3}}{2}\) & 0 & \(-\frac{1}{\sqrt{2}}\) \\
\hline
\end{tabular}
\end{center}

\(\begin{array}{lll}5 & \text { a } 44.42 & \text { b } 6.47\end{array}\)

\(6 \begin{array}{ll}\text { a } 140^{\circ} & \text { b } 155^{\circ}\end{array}\)

7 a The two angles are supplementary, and the sines of two supplementary angles are the same.
b \(h=b \sin A\)
c \(h=a \sin B\)

\(859^{\circ}\)

\(9 \quad 25.1 \mathrm{~m}\)
10 a \(234^{\circ} \mathrm{T}\)
b \(1285 \mathrm{~m}\)
11 a \(341^{\circ} \mathrm{T}\)
b \(1282 \mathrm{~m}\)

\section*{Exercise 12F}
\begin{center}
\begin{tabular}{llrl}
1 & a \(8.65 \mathrm{~cm}\) & b \(3.82 \mathrm{~cm}\) &  \\
 & c \(16.34 \mathrm{~cm}\) & d \(6.93 \mathrm{~cm}\) &  \\
2 & \(17.91 \mathrm{~cm}\) & \(\mathbf{3}\) & \(6.2 \mathrm{~m}\) \\
\(\mathbf{4}\) & \(410 \mathrm{~km}\) & \(\mathbf{5}\) & \(103 \mathrm{~km}\) \\
\end{tabular}
\end{center}

6 a \(14.38 \mathrm{~m}\)
b \(7.01 \mathrm{~m}\)
c \(50.43 \mathrm{~m}^{2}\)

7 From \(\triangle B X A\)

\[
c^{2}=h^{2}+x^{2} \text { and } x=c \cos (180-A)=-\cos A
\]

From \(\triangle B X C\),

\[
a^{2}=(b+x)^{2}+h^{2}
\]

Hence,

\[
a^{2}=b^{2}+2 b x+x^{2}+c^{2}-x^{2}
\]

\[
=b^{2}+c^{2}-2 b c \cos A
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-060}
\end{center}

\section*{Exercise \(12 \mathrm{G}\)}
1 a \(x^{2}=y^{2}+z^{2}-2 y z \cos R\)

b \(b^{2}=a^{2}+c^{2}-2 a c \cos B\)

c \(p^{2}=q^{2}+r^{2}-2 q r \cos B\)
2 a \(52.6^{\circ}\)
b \(54.8^{\circ}\)

c \(78.5^{\circ}\)

3 a \(A=39.8^{\circ}, B=45.4^{\circ}, C=94.8^{\circ}\)

b \(A=B=29.0^{\circ}, C=122.1^{\circ}\)

c \(A=21.0^{\circ}, B=18.6^{\circ}, C=140.4^{\circ}\)

\(419.4^{\circ}\)

\(5108.4^{\circ}\)

\(616^{\circ}, 34^{\circ}, 130^{\circ}\)

\(\begin{array}{ll}7 & 92.1^{\circ}\end{array}\)

8 a i \(21.8^{\circ}\)

ii \(4.4 \mathrm{~cm}\)

b \(9.5 \mathrm{~cm}, 11.8 \mathrm{~cm}\)

\section*{Exercise \(12 \mathrm{H}\)}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
1 & a \(39.5 \mathrm{~cm}^{2}\) & b \(29.8 \mathrm{~cm}^{2}\) & c \(48.5 \mathrm{~cm}^{2}\) \\
\hline
2 & a \(10.08 \mathrm{~cm}^{2}\) & b \(46.98 \mathrm{~cm}^{2}\) & c \(43.30 \mathrm{~cm}^{2}\) \\
\hline
3 & a \(73.2^{\circ}\) & b \(56.8^{\circ}\) & c \(134 \mathrm{~cm}^{2}\) \\
\hline
4 & a \(36.3^{\circ}\) & b \(21 \mathrm{~cm}^{2}\) &  \\
\hline
5 & a \(57 \mathrm{~cm}^{2}\) & b \(30 \mathrm{~cm}^{2}\) & c \(199 \mathrm{~cm}^{2}\) \\
\hline
6 & a \(8.5 \mathrm{~cm}\) & b \(9.6 \mathrm{~cm}\) & c \(36.7^{\circ}\) \\
\hline
7 & \(22^{\circ}\) &  &  \\
\hline
8 & a \(152.6 \mathrm{~m}\) & b \(159.7^{\circ}\) & c \(6213.6 \mathrm{~m}^{2}\) \\
\hline
9 & a \(108^{\circ}\) & b \(47.55 \mathrm{~cm}^{2}\) & c \(16.18 \mathrm{~cm}\) \\
\hline
 & d \(36^{\circ}\) & e \(72^{\circ}\) & f \(36^{\circ}\) \\
\hline
 & g \(76.94 \mathrm{~cm}^{2}\) & h \(172.05 \mathrm{~cm}^{2}\) &  \\
\hline
10 & \(98 \mathrm{~cm}^{2}\) &  &  \\
\hline
11 & \multicolumn{3}{|c|}{\[
\begin{aligned} Area \(=\frac{1}{2} \times(k a) \times(k b) \sin C & =\frac{1}{2} k^{2} \times a b \sin C \\
& =k^{2} \times\) area of triangle \(A B C\end{aligned}
\])} \\
\hline
12 & \multicolumn{3}{|c|}{\[
\begin{aligned} Area & \(=\frac{1}{2} \times a \times h \\
& =\frac{1}{2} a b \sin (180-C) \\
& =\frac{1}{2} a b \sin C\end{aligned}
\])} \\
\hline
\end{tabular}
\end{center}

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-061}
13 a \(31.29^{\circ}\)
b \(42.85 \mathrm{~cm}^{2}\)
c \(34 \mathrm{~cm}\)
d \(42.85 \mathrm{~cm}^{2}\)

\section*{Review exercise}
1 a \(31.0^{\circ}\)
b \(33.6^{\circ}\)
c 10.58

\begin{center}
\begin{tabular}{llcl}
2 & a 10 & b \(18 \sqrt{2}\) & c \(25 \sqrt{3}\) \\
3 & \(A=26.4^{\circ}, B=117.3^{\circ}, C=36.3^{\circ}\) &  &  \\
4 & a \(92^{\circ}\) & b \(1211 \mathrm{~m}^{2}\) &  \\
5 & 48 & 7 & \(24.47 \mathrm{~cm}^{2}\) \\
8 & a \(49.0^{\circ}\) & b \(71.0^{\circ}\) &  \\
 & d \(142.5 \mathrm{~cm}^{2}\) & \(\mathrm{~cm}\) &  \\
9 & \(141 \mathrm{~m}\) &  &  \\
\(\mathbf{1 0}\) & a \(\angle C=91^{\circ}, B C \approx 5.44 \mathrm{~cm}, A C \approx 4.31 \mathrm{~cm}\) &  &  \\
 & b \(B C \approx 4.46 \mathrm{~cm}, \angle C \approx 90.19^{\circ}, \angle B \approx 41.81^{\circ}\) &  &  \\
 & c \(A B \approx 11.17 \mathrm{~cm}, \angle B \approx 61.88^{\circ}, \angle A \approx 38.12^{\circ}\) &  &  \\
 & d \(A C \approx 19.08 \mathrm{~cm}, \angle A \approx 33.00^{\circ}, \angle C \approx 27.00^{\circ}\) &  &  \\
 & e \(\angle B \approx 105.05^{\circ}, \angle C \approx 43.95^{\circ}, A B \approx 10.78 \mathrm{~cm}\) &  &  \\
f \(\angle C \approx 149.00^{\circ}, \angle A \approx 13.00^{\circ}, B C \approx 10.92 \mathrm{~cm}\) &  &  &  \\
\end{tabular}
\end{center}

\section*{Challenge exercise}
\(1 \frac{1}{2} a b \sin C=\frac{1}{2} a c \sin B, \frac{b}{\sin B}=\frac{c}{\sin C}\)
2 a i \(\frac{h}{\tan 12^{\circ}}\)
ii \(\frac{h}{\tan 9^{\circ}}\)
b \(47.5 \mathrm{~m}\)
c \(318^{\circ} \mathrm{T}\)

4 b \(b=a \cos C+c \cos A, c=a \cos B+b \cos A\)

7 If given angle \(A\) and sides \(b\) and \(a\) :

0 triangles if \(a<b \sin A\) or \(A\) is obtuse and \(a<b\)

1 triangle if \(A\) is obtuse and \(a>b\) or \(A\) is acute and \(a=b \sin A\)

2 triangles if \(A\) is acute and \(b \sin A<a<b\)

\section*{Chapter 13 answers}
\section*{Exercise 13A}
1 a ii \(90^{\circ}\)

b ii The angle at the centre is twice any angle at the circumference standing on the same minor arc.

iii The reflex angle at the centre is twice any angle at the circumference standing on the same major arc.

2 a \(\alpha=90^{\circ}\) (Thales' theorem),

\(\beta=75^{\circ}\) (sum of angles in a triangle is \(180^{\circ}\) )

b \(\theta=90^{\circ}\) (Thales' theorem)

c \(\theta=10^{\circ}\left(\angle J L K=90^{\circ}\right.\), Thales' theorem, sum of angles in a triangle is \(180^{\circ}\) )

d \(\gamma=70^{\circ}(O S=O T\) radii of circle and base angles of an isosceles triangle are equal),

\(\alpha=40^{\circ}\) (sum of angles in a triangle is \(180^{\circ}\) ),

\(\theta=140^{\circ}\) (straight angle),

\(\beta=20^{\circ}\) (base angles of an isosceles triangle)

e \(\theta=55^{\circ}(O Z=O Y\) radii of circle, base angles of an isosceles triangle are equal and the sum of angles in a triangle is \(\left.180^{\circ}\right)\),

\(\beta=35^{\circ}\) (external angle of isosceles triangle is \(70^{\circ}\) )

f \(\theta=80^{\circ}\left(\angle A O C=20^{\circ}\right.\) and \(\triangle A O C\) is isosceles \()\)
circumference standing on the same arc)

b \(\gamma=44^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc)

c \(\alpha=190^{\circ}\) (angle at the centre is twice angle at the circumference standing on the same arc)

d \(\gamma=100^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc)

e \(\theta=129^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc)

f \(\alpha=40^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc)

g \(\beta=30^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc)

h \(\gamma=100^{\circ}\) (angle at the centre is twice angle at the circumference standing on the same arc)

i \(\theta=24^{\circ}\) (angle at the centre is twice angle at the circumference standing on the same arc)

a \(\alpha=120^{\circ}\) (sum of angles about a point is \(360^{\circ}\) ), \(\beta=60^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc)

b \(\theta=30^{\circ}\left(\angle A O B=60^{\circ}\right.\), sum of angles about a point is \(360^{\circ}\) and angle at the circumference is half angle at the centre standing on the same arc so \(\theta\) is half of \(\angle A O B\) )

c \(\theta=220^{\circ}\left(\angle S O R=140^{\circ}\right.\), angle at the centre is twice angle at the circumference standing on the same arc and \(\theta+\angle S O R=360^{\circ}\), sum of angles about a point is \(360^{\circ}\) )

d \(\alpha=\beta=40^{\circ}\) (any angle at the circumference is half angle at the centre standing on the same arc)

e \(\theta=320^{\circ}\left(\angle S O R=40^{\circ}\right.\), angle at the centre is twice angle at the circumference standing on the same arc and \(\theta+\angle S O R=360^{\circ}\), sum of angles about a point is \(360^{\circ}\) ) \(\alpha=20^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc, \(\left.\angle S O R=40^{\circ}\right)\)

f \(\alpha=100^{\circ}(O R=O Q\) radii so base angles of an isosceles triangle are both \(40^{\circ}\), sum of angles in a triangle is \(180^{\circ}\) ), \(\beta=140^{\circ}\) (sum of angles about a point is \(360^{\circ}\) ), \(\gamma=20^{\circ}(O R=O P\) radii, \(\gamma\) is a base angle of an isosceles triangle)

g \(\alpha=80^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc),

\(\beta=200^{\circ}\) (sum of angles about a point is \(360^{\circ}\) ),

\(\gamma=100^{\circ}\) (angles at the circumference is half angle at the centre standing on the same arc)

h \(\alpha=60^{\circ}\left(\angle D A B=90^{\circ}\right.\), Thales' theorem, so

\(\left.\alpha+30^{\circ}=90^{\circ}\right)\),

\(\beta=60^{\circ}(O A=O B\) radii, so \(\alpha=\beta\), base angles of isosceles triangle),

\(\gamma=30^{\circ}\left(\angle A B C=90^{\circ}\right.\), Thales' theorem, so \(\left.\gamma+\beta=90^{\circ}\right)\)

i \(\alpha=\beta=45^{\circ}\) (both are base angles in isosceles triangles with the third angle \(90^{\circ}\) )

a \(\alpha=90^{\circ}\) (Thales' theorem),

\(\beta=10^{\circ}\) (alternate angles, \(A B \| F G\) )

b \(\alpha=60^{\circ}(O P=O A=A P\) so the triangle is equilateral \()\), \(\beta=30^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc)

c \(\alpha=20^{\circ}\) (alternate angles, \(P O \| Q R\) ),

\(\gamma=40^{\circ}\) (angle at the centre is twice angle at the circumference standing on the same arc),

\(\beta=40^{\circ}\) (alternate angles, \(P O \| Q R\) ) d \(\alpha=220^{\circ}\) (sum of angles about a point),

\(\beta=110^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc),

\(\gamma=60^{\circ}\) (sum of angles in a quadrilateral is \(360^{\circ}\) )

e \(\alpha=200^{\circ}\) (sum of angles about a point),

\(\beta=100^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc),

\(\gamma=80^{\circ}\) (co-interior angles, \(P Q \| O R\) )

f \(\alpha=100^{\circ}\) (angle at the circumference is half angle at the centre standing on the same arc),

\(\beta=60^{\circ}\) (construct \(O A\), as \(O B=A B=O A\) the triangle

is equilateral),

\(\gamma=40^{\circ}\) (sum of angles in a quadrilateral is \(360^{\circ}\) )

6 a i \(O A=O P\) so \(\triangle A P O\) is isosceles with

\(\angle P A O=\angle A P O=\alpha\),

\(O B=O P\) so \(\triangle P B O\) is isosceles with

\(\angle P B O=\angle B P O=\beta\),

so \(\angle A P B=\alpha+\beta ; \angle X P B=\alpha+\beta\), external angle of \(\triangle A P B\)

ii \(\angle X P B+\angle A P B=180^{\circ}, \alpha+\beta+\alpha+\beta=180^{\circ}\), \(2(\alpha+\beta)=180^{\circ}, \alpha+\beta=90^{\circ}\)

b i \(O A=O P\) so \(\triangle A P O\) is isosceles with \(\angle P A O=\angle A P O=\alpha, O B=O P\) so \(\triangle P B O\) is isosceles with \(\angle P B O=\angle B P O=\beta, \angle A O M=2 \alpha\) and \(\angle B O M=2 \beta\), external angles of \(\triangle A P O\) and \(\triangle P B O\)

ii \(\angle A O M+\angle B O M=180^{\circ}\) so \(2 \alpha+2 \beta=180^{\circ}\)

iii Thus \(\alpha+\beta=90^{\circ}\)

7 a \(O B=O P\) so \(\triangle P B O\) is isosceles with \(\angle P B O=\angle B P O=\beta\), so \(\angle A P B=\beta\)

ii \(\angle A O B=2 \beta\), (exterior angle of triangle)

b i \(O A=O P\) so \(\triangle A P O\) is isosceles with \(\angle P A O=\angle A P O=\alpha\), \(O B=O P\) so \(\triangle P B O\) is isosceles with \(\angle P B O=\angle B P O=\beta\), so \(\angle A P B=\angle B P O-\angle A P O=\beta-\alpha\)

ii \(\angle X O B=2 \beta\) (exterior angle of triangle), \(\angle X O A=2 \alpha\) (exterior angle of triangle),

\(\angle A O B=2 \beta-2 \alpha=2(\beta-\alpha)=2 \angle A P B\)

8 a The diagonals bisect each other.

b A parallelogram with a right angle is a rectangle.

c The diagonals of a rectangle are equal and bisect each other, so \(O A=O B=O P\). A circle with diameter \(A B\) has centre \(O\) and passes through \(P\).

9 The angle at the centre is twice the angle at the circumference. As the horse moves from position 1 to position 2, angles are subtended both at the binoculars and the centre.

\section*{Exercise 13B}
1 b They are equal.

c They are equal.

2 a \(\alpha=50^{\circ}\) (angles at the circumference standing on the same arc)

b \(\alpha=\beta=20^{\circ}\) (angles at the circumference standing on the same arc)
c \(\alpha=20^{\circ}\) (angles at the circumference standing on the same arc)

\(\beta=40^{\circ}\) (angles at the circumference standing on the same arc)

d \(\alpha=40^{\circ}\) (angles at the circumference standing on the same arc)

\(\beta=30^{\circ}\) (angles at the circumference standing on the same arc)

e \(\theta=90^{\circ}\) (angles at the circumference standing on the same arc)

\(\alpha=40^{\circ}\left(\alpha+\theta=130^{\circ}\right.\), exterior angle of triangle \()\)

f \(\alpha=20^{\circ}\) (angles at the circumference standing on the same arc)

\(\theta=100^{\circ}\left(\angle Q P K=60^{\circ}\right.\), angles at the circumference standing on the same arc, sum of angles in triangle is \(180^{\circ}\) )

a \(\theta=80^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

b \(\alpha=100^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

\(\beta=95^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

c \(\alpha=40^{\circ}\) (angles at the circumference standing on the same arc)

\(\beta=45^{\circ}\) (angles at the circumference standing on the same arc)

\(\theta=35^{\circ}(80=\beta+\theta\) exterior angle \()\),

\(\gamma=60^{\circ}\) (sum of angles in triangle \(A C D\) is \(180^{\circ}\) )

d \(\alpha=50^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

\(\beta=90^{\circ}\) (sum of angles in triangle is \(180^{\circ}\) ),

\(\gamma=30^{\circ}\) (sum of angles in triangle is \(180^{\circ}\) )

e \(\alpha=110^{\circ}\) (straight angle),

\(\gamma=70^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

\(\beta=80^{\circ}\) (straight angle),

\(\theta=100^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

f \(\alpha=20^{\circ}\) (angles at the circumference standing on the same arc),

\(\beta=90^{\circ}\) (Thales' theorem),

\(\gamma=90^{\circ}\) (Thales' theorem),

\(\theta=70^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

a \(\alpha=70^{\circ}\) (co-interior angles, \(D C \| A B\) ),

\(\gamma=110^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary),

\(\beta=70^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

b \(\alpha=65^{\circ}\) (angles at the circumference standing on the same arc),

\(\gamma=65^{\circ}\) (alternate, \(T U \| S R\) )

\(\beta=65^{\circ}\) (alternate, \(T U \| S R\) )

c \(\alpha=30^{\circ}\left(\angle B Q M=30^{\circ}\right.\), alternate, \(\left.B Q \| A P\right)\),

\(\beta=30^{\circ}\) (angles at the circumference standing on the same arc),

\(\gamma=30^{\circ}\) (angles at the circumference standing on the same arc) d \(\beta=70^{\circ}\left(\angle J M L=90^{\circ}\right.\), Thales' theorem, sum of angles in triangle is \(180^{\circ}\) )

\(\alpha=20^{\circ}\left(\angle M J K=90^{\circ}\right.\), Thales' theorem, \(\left.\alpha+\beta=90^{\circ}\right)\)

e \(\alpha=70^{\circ}\) (alternate, \(A D \| B C\) ),

\(\beta=40^{\circ}(\triangle A O B\) is isosceles with base angles \(\alpha\), so \(\angle A O B=40^{\circ}\), alternate, \(\left.A D \| B C\right)\),

\(\gamma=40^{\circ}(\triangle B O C\) is isosceles with base angles \(\gamma\) and \(\beta)\),

\(\theta=70^{\circ}\left(\angle D O C=40^{\circ}\right.\), alternate, \(A D \| B C\) and \(\triangle D O C\) is isosceles with base angles \(\theta\) )

f \(\alpha=60^{\circ}\) (construct \(S O, \triangle R S O\) is equilateral),

\(\beta=120^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary),

\(\gamma=60^{\circ}\) (co-interior, \(S T \| R U\) )

a i \(\angle Q=\angle T=\theta\) (angles at the circumference standing on the same arc),

\(\angle P=\angle T=\theta(\) alternate, \(P Q \| S T), \angle P=\angle S=\theta\)

(angles at the circumference standing on the same arc),

so \(\angle P=\angle Q=\angle S=\angle T=\theta\)

ii \(\angle P=\angle Q=\angle S=\angle T=\theta, \triangle S M T\) and \(\triangle P M Q\)

are isosceles, so \(S M=T M\) and \(P M=Q M\), so

\(S M+M Q=T M+M P\) giving \(S Q=T P\)

b i \(\alpha=60^{\circ}(\triangle P Q R\) is equilateral),

\(\beta=120^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary),

\(\gamma=30^{\circ}\) (base angle of isosceles triangle \(P G Q\) )

ii \(\angle G R P=30^{\circ}, \angle G R Q=30^{\circ}\) (angles standing on the same arc),

\(\angle P M R=90^{\circ}\) so \(P Q \perp G R\)

a i \(\alpha=110^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary),

\(\beta=70^{\circ}\) (straight angle),

\(\gamma=110^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

ii As \(\angle T Q P\) and \(\angle S P Q\) are co-interior and supplementary then \(P S \| Q T\).

b i \(\alpha=130^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary),

\(\beta=50^{\circ}(\triangle A B M\) is isosceles, straight angle at \(A)\),

\(\gamma=50^{\circ}\left(\angle Q B A=130^{\circ}\right.\), straight angle at \(B\),

opposite angles in a cyclic quadrilateral are supplementary)

ii \(\angle Q P M\) and \(\angle B A P\) are co-interior and supplementary, so \(P Q \| A B\)

iii \(\angle Q P M=\angle P Q M, Q M P\) is isosceles with \(Q M=P M\), also \(B M=A M\), so \(Q B=P A\)

7 a i \(\angle A B S=90^{\circ}\) and \(\angle A B T=90^{\circ}\) (Thales' theorem)

ii \(\angle T B S=\angle A B S+\angle A B T=180^{\circ}\), so \(\angle T B S\) is a straight angle \(\therefore T, B\) and \(S\) are collinear

b i \(\angle A B C=90^{\circ}\); (opposite angles in a cyclic quadrilateral are supplementary), \(\angle A B D=90^{\circ}\) (Thales' theorem)

ii \(\angle C B D=\angle A B C+\angle A B D=180^{\circ}\), so \(\angle C B D\) is a straight angle \(\therefore C, B\) and \(D\) are collinear

iii \(\angle A N C=90^{\circ}\); by the converse of Thales' theorem \(\angle A N C\) is in a semicircle with \(A C\) the diameter
a \(\angle D B C=\angle D A C=\alpha\) (angles at the circumference standing on the same arc)

\(\angle B D C=\angle B A C=\beta\) (angles at the circumference standing on the same arc)

b \(\angle D B C+\angle B D C+\angle B C D=180^{\circ}\) (sum of angles in a triangle),

\(\alpha+\beta+\angle B C D=180^{\circ}\), so \(\angle B C D=180^{\circ}-\alpha-\beta\)

c \(\angle B A D=\alpha+\beta\); thus, \(\angle B A D+\angle B C D=180^{\circ}\)

a i \(\angle C=180^{\circ}-\theta\) (opposite angles in a cyclic quadrilateral are supplementary),

\(\angle C=\theta\) (opposite angles in a parallelogram are equal)

ii From a \(180^{\circ}-\theta=\theta\), so \(\theta=90^{\circ}\), so

\(\angle C=\angle A=90^{\circ}\), similarly \(\angle B=\angle D=90^{\circ}\), so

\(A B C D\) is a rectangle.

b A rhombus is a parallelogram with all sides equal and so if a cyclic parallelogram is a rectangle, a cyclic rhombus must be a square.

\(\angle M D C=180^{\circ}-\theta\) (co-interior, \(A B \| D C\) ) and \(\angle B C D=180^{\circ}-\theta(\) opposite angles in a cyclic quadrilateral are supplementary) \(\therefore \triangle M D C\) is isosceles \(\angle M B A=180^{\circ}-\theta\) (corresponding angles, \(A B \| D C\) ) and \(\angle M A B=180^{\circ}-\theta(\) straight angle at \(A) \therefore \triangle M A B\) is isosceles

b As \(\triangle M D C\) is isosceles, \(M D=M C\) and as \(\triangle M A B\) is isosceles, \(M A=M B\), so \(A D=B C\)

\section*{Exercise \(13 \mathrm{C}\)}
2 a \(x=3, \theta=53.1^{\circ}\)

b \(x=2 \sqrt{13}, \theta=33.7^{\circ}\)

c \(x=4 \sqrt{3}, \theta=30^{\circ}\)

3
a \(\quad\) i \(17.0 \mathrm{~cm}\)
ii \(8.5 \mathrm{~cm}\)
b \(\quad\) i \(24 \mathrm{~cm}\)
ii \(73.7^{\circ}\)
c i \(6 \mathrm{~cm}\)
ii \(106.3^{\circ}\)

4 a \(\theta=50^{\circ}\) and \(\angle U O T=50^{\circ}\) (chords of equal length subtend equal angles at the centre of a circle)

\(\alpha=65^{\circ}\) since \(\triangle U O T\) is isosceles

b \(\alpha=60^{\circ}(\triangle A B O\) is equilateral \()\),

\(\beta=240^{\circ}(\triangle A B O\) and \(\triangle B C O\) are equilateral, sum of angles at a point)

c \(\theta=30^{\circ}\left(\triangle R Q O\right.\) is equilateral, so \(\angle Q R O=60^{\circ}\) and \(\angle R Q P=90^{\circ}\), Thales' theorem)

d \(\alpha=45^{\circ}\) (base angle of isosceles triangle with third angle \(90^{\circ}\) ),

\(\beta=25^{\circ}\) (base angle of isosceles triangle with third angle \(130^{\circ}\) ),

\(\gamma=20^{\circ}\) (base angle of isosceles triangle with third angle \(\left.140^{\circ}\right)\)

e \(\theta=50^{\circ}\left(\angle H O F=130^{\circ}\right.\) as \(\triangle H O F\) is isosceles, \(\angle G O F\) is a straight angle)

f \(\alpha=36^{\circ}\) (10 congruent triangles in a circle),

\(\beta=72^{\circ}\) (base angle of isosceles triangle with third angle \(36^{\circ}\) )

a \(\angle F P O=60^{\circ}(\triangle F P O\) is equilateral \(), \angle F G O=30^{\circ}\) (angle at the centre is twice angle at the circumference standing on the same arc) \(\angle F M O=90^{\circ}\) (angle sum of triangle)

b \(F G=\sqrt{3}\)
6 a i \(A O=B O\) (radii of the circle), \(A M=B M\) (as \(M\) is midpoint of \(A B\) ),

\(O M\) is common \(\therefore \triangle A O M \equiv \triangle B O M(\mathrm{SSS})\)

ii \(\angle A M O=\angle B M O\) (matching angles,

\(\triangle A O M \equiv \triangle B O M)\) and

\(\angle A M O+\angle B M O=180^{\circ}\) (straight angle at \(M\) ), so

\(\angle A M O=\angle B M O=90^{\circ}\)

\(\angle A O M=\angle B O M\) (matching angles,

\(\triangle A O M \equiv \triangle B O M)\)

b i \(\angle A M O=\angle B M O=90^{\circ}\) (given), \(A O=B O\) (radii of the circle),

\(O M\) is common \(\therefore \triangle A O M \equiv \triangle B O M(\mathrm{RHS})\)

ii \(\angle A O M=\angle B O M\) (matching angles, \(\triangle A O M \equiv \triangle B O M)\) and

\(A M=B M\) (matching sides, \(\triangle A O M \equiv \triangle B O M)\) so \(O M\) bisects \(\angle A O B\)

c i \(\angle A O M=\angle B O M=\alpha\) (given), \(A O=B O\) (radii of the circle),

\(O M\) is common \(\therefore \triangle A O M \equiv \triangle B O M\) (SAS)

ii \(\angle A M O=\angle B M O\) (matching angles, \(\triangle A O M \equiv \triangle B O M)\) and \(\angle A M O+\angle B M O=180^{\circ}(\) straight angle at \(M)\), so \(\angle A M O=\angle B M O=90^{\circ} A M=B M\) (matching sides, \(\triangle A O M \equiv \triangle B O M)\)

7 a i Join \(O A\) and \(O P . O A=O P=O Q\) (radii of the circle) so \(\triangle A O P\) and \(\triangle A O Q\) are isosceles with base angles \(\theta\) making the third angle in both triangles \(180^{\circ}-2 \theta\), so \(\triangle A O P \equiv \triangle A O Q\) (SAS)

ii \(A P=A Q\) (matching sides, \(\triangle A O P \equiv \triangle A O Q\) )

b i Join \(O S \angle T F S=\angle S F O=\theta\) (given) and \(\triangle F S O\) is isosceles \((O F=O S\), radii \()\)

so \(\angle S F O=\angle F S O \therefore \angle T F S=\angle F S O\)

ii Join \(O F\). \(\angle T F S\) and \(\angle F S O\) are alternate and equal, so \(F T \| O S\)

c i \(S Q=S P\) (given), \(Q T=P T\) (given), \(S T\) is common \(\therefore \triangle S Q T \equiv \triangle S P T(\mathrm{SSS})\)

ii \(\angle P=\angle Q\) (matching angles, \(\triangle S Q T \equiv \triangle S P T)\)

iii \(\angle P+\angle Q=180^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

iv As \(\angle P=\angle Q\) from ii and \(\angle P+\angle Q=180^{\circ}\) from iii then \(\angle P=\angle Q=90^{\circ}\), so \(S T\) is a diameter (converse of Thales' theorem)

d i Join \(O R, O S, O T\) and \(O U . \triangle O S T\) is isosceles with \(O T=O S\) (radii of circle centre \(O\) through \(T\) ), angles opposite equal sides are equal so \(\angle O S T=\angle O T S\)

ii \(\triangle O U R\) is isosceles with \(O R=O U\) (radii of circle centre \(O\) through \(U\) ), angles opposite equal sides are equal so \(\angle O U R=\angle O R U\)

iii \(\angle O S T=\angle O T S(\) from i), \(\angle O U T=\angle O R S(\) from ii) and \(O R=O U\) (radii), so \(\triangle O R T \equiv \triangle O U S\) (AAS)

iv \(R T=S U\) (matching sides, \(\triangle O R T \equiv \triangle O U S\) )

8 a \(G O=F O\) (radii of the circle centre \(O\) ), \(P G=P F\) (radii of the circle centre \(P\) ), \(O P\) is common \(\therefore \triangle G O P \equiv \triangle F O P(\mathrm{SSS})\)

b \(\angle F O M=\angle G O M\) (matching angles, \(\triangle G O P \equiv \triangle F O P\) ) \(G O=F O(\) radii of the circle centre \(O)\), \(\angle F O M=\angle G O M\) (from b), \(O M\) is common \(\therefore \triangle G O M \equiv \triangle F O M(\mathrm{SAS})\)
c \(\angle G M O=\angle F M O\) (matching angles, \(\triangle F O M \equiv \triangle G O M\) ) and \(\angle G M O+\angle F M O=180^{\circ}\) (straight angle at \(M\) ), so \(\angle G M O=\angle F M O=90^{\circ}\) and \(O P \perp F G, F M=G M\) (matching sides, \(\triangle G O M \equiv \triangle F O M\) )

a \(\triangle P A M \equiv \triangle P B M(\mathrm{SSS})\), so \(\angle A M P=\angle B M P=90^{\circ}\)

b \(\triangle P A M \equiv \triangle P B M\) (SAS), so \(A P=B P\)

c Take three points \(A, B\) and \(C\) on the circle. Construct the perpendicular bisectors of \(A B\) and \(B C\). By parts a and \(\mathrm{b}\), the centre lies on the perpendicular bisectors of \(A B\) and \(B C\), so the intersection is the only centre of the circle.

d i Imagine the vertical line through the centre of the circle, and let \(P\) be any point on this line. Then \(P\) is equidistant from all the points on the circle.

ii A sphere

iii An infinite cylinder

iv A cylinder with hemispherical ends

v A plane perpendicular to the interval through the midpoint of the interval

vi One method is to take three chords, not in a plane. Take the plane perpendicular to each chord through its midpoint, then the intersection of the three planes is the centre of the sphere.

\section*{Exercise 13D}
1 a \(\alpha=90^{\circ}\) (tangent perpendicular to radius at common point on circle),

\(\beta=40^{\circ}\) (sum of angles in a triangle)

b \(\alpha=75^{\circ}\left(\angle O F P=90^{\circ}\right.\), sum of angles in a triangle \()\),

\(\beta=37.5^{\circ}\) (base angle of isosceles triangle with external angle \(75^{\circ}\) )

c \(\alpha=70^{\circ}\) (angle at the centre is twice angle at the circumference standing on the same arc),

\(\beta=20^{\circ}\left(\angle O T U=90^{\circ}\right.\), sum of angles in a triangle \()\)

d \(\alpha=62^{\circ}\left(\angle M T O=90^{\circ}\right.\), sum of angles in a triangle \()\),

\(\beta=31^{\circ}\) (angles at the circumference is half angle at the centre standing on the same arc)

e \(\theta=50^{\circ}\left(\angle U O T=40^{\circ}\right.\), angle at the centre is twice angle at the circumference standing on the same arc and \(\angle O T U=90^{\circ}\), sum of angles in a triangle)

f \(\alpha=90^{\circ}\left(\angle O T U=90^{\circ}\right.\), alternate angles, \(\left.T U \| B A\right)\), \(\beta=45^{\circ}\) (base angle of isosceles triangle with third angle \(90^{\circ}\) )

g \(\beta=70^{\circ}\) (alternate angle, \(D T \| A C\) ),

\(\alpha=20^{\circ}\left(\alpha+\beta=90^{\circ}=\angle D T O\right)\)

h \(\alpha=25^{\circ}\) (alternate angles, \(T R \| S D\) ), \(\beta=90^{\circ}\)

(Thales' theorem),

\(\gamma=65^{\circ}\) (sum of angles in a triangle)

i \(\alpha=35^{\circ}\left(\angle B T S=90^{\circ}\right)\),

\(\beta=55^{\circ}\left(\angle S Q T=90^{\circ}\right.\), Thales' theorem, sum of angles in triangle)

\(\theta=70^{\circ}\left(\angle S P T=90^{\circ}\right.\), Thales' theorem, sum of angles in triangle)

j \(\alpha=30^{\circ}\left(\angle D T O=15^{\circ}\right.\), angle at circumference is half angle at centre standing on same arc),

\(\beta=60^{\circ}\left(\angle O T B=90^{\circ}\right.\), sum of angles in a triangle \()\)

k \(\theta=30^{\circ}\) ( \(\triangle T X O\) is equilateral, \(\angle X T O=60^{\circ}\),

\(\left.\angle L T O=90^{\circ}\right)\),

\(\beta=120^{\circ}\left(\angle Y Z O=60^{\circ}\right.\), sum of opposite angles in a

cyclic quadrilateral)
I \(\alpha=30^{\circ}\left(\triangle T Q O\right.\) is equilateral, \(\angle T O Q=60^{\circ}\), angle at the circumference is half angle at the centre standing on the same arc),

\(\theta=30^{\circ}\left(\right.\) complementary with \(\left.\angle Q T O=60^{\circ}\right)\)

a \(x=5\) (tangents to a circle from an external point have equal length),

\(\alpha=70^{\circ}\) (base angle of an isosceles triangle),

\(\beta=40^{\circ}\) (sum of angles in a triangle),

\(\gamma=20^{\circ}\left(\angle O S P=90^{\circ}\right)\)

b \(x=8\) (tangents to a circle from an external point have equal length),

\(\alpha=70^{\circ}\) (base angle of an isosceles triangle),

\(\theta=140^{\circ}\left(\angle T=\angle S=90^{\circ}\right.\), sum of angles in a quadrilateral \()\)

c \(x=2\) (tangents to a circle from an external point have equal length),

\(y=3, z=3\) (tangents to a circle from an external point have equal length)

d \(x=7(S Q=4\), tangents to a circle from an external point have equal length, so \(R S=7\) )

e \(x=9(S B=4\) (equal tangents), \(S P=14\) and \(T P=14\) (equal tangents), \(T A=5\) (equal tangents), \(x=14-5\)

f \(\alpha=100^{\circ}\) (reflex \(\angle S O T=260^{\circ}\), angle at the centre is twice angle at the circumference standing on the same arc, sum of angles at point \(O\) ),

\(\beta=70^{\circ}\) (sum of angles in a quadrilateral),

\(\theta=20^{\circ}\left(\angle P S O=90^{\circ}\right)\)

3 a \(x=5\) (radius), \(y=8(O B=13\) by Pythagoras' theorem), \(\theta \approx 22.62^{\circ}\)

b \(x=\sqrt{133} \approx 11.53\)

c \(x \approx 19.23, y \approx 13.47\)

\(\angle P A B=90^{\circ}\) (tangent perpendicular to radius) and \(\angle V B A=90^{\circ}\) (tangent perpendicular to radius), \(\angle P A B\) and \(\angle V B A\) are alternate and equal therefore \(P Q \| U V\)

\(A B+C D=A P+P B+C R+D R\)

\(=A S+B Q+Q C+S D\) (tangents to a circle from an external point

\[
=A D+B C
\]

have equal length)

b Join \(O S\) and \(O T\). \(\angle O S P\) and \(\angle O T P\) are angles in a semi-circle, centre \(M\), so by Thales' theorem \(\angle O S P=\angle O T P=90^{\circ}\). As \(O S\) and \(O T\) are radii of circle, centre \(O, P S \perp O S\) and \(P T \perp O T\), so \(P S\) and \(P T\) are tangents to circle centre \(O\).

c \(P S=P T\) (tangents to a circle from an external point have equal length)

a i \(\angle P S O=\angle P T O\) (tangent perpendicular to radius), \(S O=T O\) (radii of circle), \(P O\) is common. Thus \(\triangle P S O \equiv \triangle P T O\) (RHS)

ii \(P S=P T\) (matching sides, \(\triangle P S O \equiv \triangle P T O\) ), \(\angle S P O=\angle T P O\) (matching angles, \(\triangle P S O \equiv \triangle P T O\) ), \(\angle S O P=\angle T O P(\) matching angles, \(\triangle P S O \equiv \triangle P T O)\)

b i \(\angle S P O=\angle T P O\) (from aii), \(S P=T P\) (from aii), \(P M\) is common

Thus \(\triangle P S M \equiv \triangle P T M\) (SAS)

ii \(S M=T M\) (matching sides, \(\triangle P S M \equiv \triangle P T M\) ), \(\angle S M P=\angle T M P(\) matching angles, \(\triangle P S M \equiv \triangle P T M)\) and

\(\angle S M P+\angle T M P=180^{\circ}(\) straight angle at \(M)\), so

\(\angle S M P=\angle T M P=90^{\circ}\).

Thus \(O P\) is the perpendicular bisector of \(S T\)
\(R O \perp A B, Q O \perp A C, P O \perp B C\) (tangent perpendicular to radius)

Area of \(\triangle A B C=\) area of \(\triangle A O C+\) area of \(\triangle B O C+\) area of \(\triangle A O B\)

\[
\begin{aligned}
& =\frac{1}{2} \times A C \times O Q+\frac{1}{2} \times B C \times O P+\frac{1}{2} \times A B \times O R \\
& =\frac{1}{2} \times \text { radius } \times(A C+B C+A B) \\
& =\frac{1}{2} \times(\text { radius of circle }) \times(\text { perimeter of triangle })
\end{aligned}
\]

a i \(\angle B T A=\alpha\) (base angle of isosceles triangle),

\(\angle B T O=\beta\) (base angle of isosceles triangle),

\(\alpha+\beta=90^{\circ}\) (tangent perpendicular to radius),

\(\beta=2 \alpha\) (exterior angle of \(\triangle A B T\) ).

So \(3 \alpha=90^{\circ}, \alpha=30^{\circ}\).

ii \(\beta=60^{\circ}\)

b i \(O T=O B\) (radii) and \(O B=B A\) (given), so

\(O A=2 O T, \angle O T A=90^{\circ}, \sin \alpha=\frac{O T}{O A}=\frac{O T}{2 O T}=\frac{1}{2}\)

ii \(\sin \alpha=\frac{1}{2}, \alpha=30^{\circ}\),

so \(\angle T O A=60^{\circ}\) (sum of angles in a triangle), \(\triangle O B T\) is equilateral and \(\beta=60^{\circ}\)

a i \(\angle A T O=90^{\circ}\) (tangent perpendicular to radius),

\(\angle A T P=90^{\circ}\) (tangent perpendicular to radius)

ii \(\angle O T P=\angle A T O+\angle A T P=90^{\circ}+90^{\circ}=180^{\circ}\), so

\(\angle O T P\) is a straight angle so \(O, T\) and \(P\) are collinear

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-066(1)}
\end{center}

\(\angle A T O=90^{\circ}\) (tangent perpendicular to radius),

\(\angle A T P=90^{\circ}\) (tangent perpendicular to radius),

\(\angle A T O=\angle A T P=90^{\circ}\) so \(O, T\) and \(P\) are collinear

11 a i \(M F=M G\) (tangents to a circle from an external point have equal length)

\(M R=M S\) (tangents to a circle from an external point have equal length)

ii \(F R=M F-M R=M G-M S=G S\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-066}
\end{center}

\(M F=M G\) (tangents to a circle from an external point have equal length)

\(M R=M S\) (tangents to a circle from an external point have equal length)

\[
F S=F M+M S=G M+M R=G R
\]

12 a \(M A=M T\) (tangents to a circle from an external point have equal length)

\(M B=M T\) (tangents to a circle from an external point have equal length)

\(\therefore M A=M T=M B\)

b Circle centre \(M\) and passing through \(A, B\) and \(T\) has a diameter \(A B\), so \(\angle A T B\) is in a semicircle, hence by Thales' theorem \(\angle A T B=90^{\circ} \therefore A T \perp B T\)

\section*{Exercise 13E}
2 a \(\alpha=35^{\circ}\) (alternate segment theorem)

b \(\theta=40^{\circ}\) (alternate segment theorem)

c \(\beta=110^{\circ}\) (alternate segment theorem)

d \(\gamma=150^{\circ}\) (alternate segment theorem)

e \(\alpha=70^{\circ}\) (alternate segment theorem),

\(\beta=110^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

f \(\beta=125^{\circ}\) (alternate segment theorem),

\(\alpha=55^{\circ}\) (opposite angles in a cyclic quadrilateral are supplementary)

3 a \(\alpha=50^{\circ}\) (alternate segment theorem),

\(\beta=50^{\circ}\) (alternate angles, \(F G \| L M\) alternate segment theorem),

\(\gamma=80^{\circ}\) (straight angle)

b \(\beta=70^{\circ}\) (base angles of isosceles triangle are equal),

\(\alpha=70^{\circ}\) (alternate segment theorem),

\(\gamma=40^{\circ}\) (angles in a triangle)

c \(\beta=80^{\circ}\) (alternate segment theorem)

d \(\beta=35^{\circ}(\Delta T S F\) is isosceles \()\),

\(\alpha=35^{\circ}\) (alternate segment theorem)

e \(\alpha=130^{\circ}\) (alternate segment theorem),

\(\beta=80^{\circ}\left(\angle B T S=50^{\circ}\right.\), straight angle at \(T\) and \(\triangle B T S\) is isosceles, tangents to a circle from an external point have equal length)

f \(\alpha=50^{\circ}\) (alternate segment theorem),

\(\beta=80^{\circ}(\triangle U A T\) is isosceles, tangents to a circle from an external point have equal length),

\(\gamma=55^{\circ}(\triangle B U S\) is isosceles, tangents to a circle from an external point have equal length)

4 a \(\angle B A O=90^{\circ}-\theta\left(\angle O A T=90^{\circ}\right.\), tangent perpendicular to radius)

b \(\angle A O B=180^{\circ}-2\left(90^{\circ}-\theta\right),(\triangle A O B\) is isosceles with base angles \(\angle B A O\) and \(\angle O B A\), sum of angles in a triangle), so \(\angle A O B=2 \theta\)

c \(\angle A P B=\theta\) (angle at the circumference is half angle at the centre standing on the same arc)

5 a \(\angle A N B=180-\alpha\) (opposite angles in a cyclic quadrilateral are supplementary)

b \(\angle A B N=90^{\circ}\) (Thales' theorem)

\(\angle N A B=180-(\angle A B N+\angle A N B)\)

\(=180-(180-\alpha+90)\)

\(=\alpha-90\)

c \(\angle S A B=90+(\alpha-90)\)

\(=\alpha\)
d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-067}
\end{center}

\(\angle A O B=2 \alpha\) on major arc \(A B\) ( \(\angle A P B\) on same arc)

\(\angle A O B=360-2 \alpha\) on minor arc, \(A B\)

\(\angle O A B=\frac{180-\angle A O B}{2} \quad(\triangle A O B\) is isosceles \()\)

\(=\frac{180-(360-2 \alpha)}{2}\)

\(=\alpha-90\)

\(\angle S A B=\angle S A O+\angle O A B\)

\(=90+(\alpha-90)\)

\(=\alpha\)

6 a \(\angle L B A=180^{\circ}-\theta(\) opposite angles in a cyclic quadrilateral are supplementary)

\(\angle T B A=\theta(\) straight angle at \(B), \angle G T A=\theta\) (alternate segment theorem)

b \(\angle G T A\) and \(\angle L K T\) are alternate and equal so \(L K \| F G\)

7 a \(\angle G T B=\theta\) (alternate segment theorem), \(\angle Q T A=\theta\) (vertically opposite to \(\angle G T B\) )

b \(\angle Q P T=\theta\) (alternate segment theorem), \(\angle Q P T\) and \(\angle G F T\) are alternate and equal so \(F G \| Q P\)

8 a \(\angle P=\theta(\) alternate, \(F G \| Q P)\)

b \(\angle G T B=\theta\) (alternate segment theorem), \(\angle A T Q=\theta\) (alternate segment theorem), so \(\angle G T B=\angle A T Q\), since \(A T B\) is a tangent then \(G T Q\) must be collinear for \(\angle G T B\) and \(\angle A T Q\) to be vertically opposite.

\section*{Exercise 13F}
1 a \(x=8\) (products of the intervals on intersecting chords are equal)

b \(x=6\) (products of the intervals on intersecting chords are equal)

c \(x=\frac{28}{3}\) (products of the intervals on intersecting chords are equal)

d \(x=6\) (tangent and secant from an external point)

e \(x=12\) (tangent and secant from an external point)

f \(x=\frac{9}{2}\) (tangent and secant from an external point)

g \(x=22\) (secants from an external point)

h \(x=6\) (secants from an external point)

i \(x=3\) (secants from an external point)

j \(x(x+5)=24 ; x=3\) (products of the intervals on intersecting chords are equal)

k \(6^{2}=4(x+4) ; x=5\) (tangent and secant from an external point)

l \(x(x+8)=48 ; x=4\) (secants from an external point)

\(2 \triangle M A Q\) is similar to \(\triangle M P B\) (AAA)

\(\frac{A M}{P M}=\frac{Q M}{B M}\) (matching sides in similar triangles)

Therefore, \(A M \times B M=Q M \times P M\)

\(3 \quad A M \times B M=T M^{2}\) and \(P M \times Q M=T M^{2}\)

Therefore, \(A M \times B M=P M \times Q M\)
4 a \(\triangle O G M \equiv \triangle O H M\) (RHS as \(O G=O H\), radii, \(\angle O M G=\angle O M H\), given, and \(O M\) is common), so \(M G=M H\) (corresponding sides)

b \(G M \times H M=A M \times B M\), so \(g \times g=a \times b, g^{2}=a b\)

c diameter is \(a+b\), so the radius is \(\frac{a+b}{2}\)

d \(0 \leq G M \leq\) radius, so \(0 \leq g \leq\) radius and \(g=\sqrt{a b}\), so

\(0 \leq \sqrt{a b} \leq \operatorname{radius}\left(=\frac{a+b}{2}\right)\)

\(5 \quad P T^{2}=P A \times P B\) (tangent and secant from an external point) and \(P S^{2}=P A \times P B\) (tangent and secant from an external point), so \(P T=P S\)

6 a \(\angle M S A=\angle M B S\) (alternate segment theorem), \(\angle S M A=\angle B M S\) (common), so \(\triangle M S A\) is similar to \(\triangle M B S(\mathrm{AA})\)

b \(\frac{B S}{S A}=\frac{S M}{A M}\) (ratio of matching sides in similar triangles are equal), so \(\frac{a}{x}=\frac{t}{m}\)

c \(\angle M T A=\angle M B T\) (alternate segment theorem), \(\angle T M A=\angle B M T\) (common), so \(\triangle M T A\) is similar to \(\triangle M B T\) (AA) and \(\frac{B T}{T A}=\frac{T M}{A M}\) (ratio of matching sides in similar triangles are equal),

so \(\frac{y}{b}=\frac{t}{m}\)

d From \(\mathbf{b}\) and \(\mathbf{c} \frac{a}{x}=\frac{t}{m}=\frac{y}{b}\), so \(a b=x y\)

\section*{Review exercise}
\begin{center}
\begin{tabular}{lll}
a \(\alpha=110^{\circ}\) & b \(\alpha=17^{\circ}\) & c \(\alpha=40^{\circ}\) \\
d \(\alpha=260^{\circ}\) & e \(\alpha=110^{\circ}\) & f \(\alpha=46^{\circ}\) \\
\end{tabular}
\end{center}

2 a \(\beta=42^{\circ}, \alpha=84^{\circ}\)

b \(\alpha=62^{\circ}, \beta=124^{\circ}\)

c \(\alpha=53^{\circ}, \beta=37^{\circ}\)

\(3 \begin{array}{lll}\mathbf{a} \alpha=32^{\circ} & \text { b } \alpha=52^{\circ}, \beta=38^{\circ} \quad \text { c } \alpha=45^{\circ}, \beta=45^{\circ}\end{array}\)

4

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-067(1)}
\end{center}

\(\angle D B C=\angle C A D(\) angles subtended by the same arc) \(\angle B C A=\angle B D A\) (angles subtended by the same arc) \(\triangle A P D\) is similar to \(\triangle B P C\) (AAA)

5

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-067(2)}
\(\angle B D A=\angle B C A=90^{\circ}\) (Thales' theorem)

\(A B=B A\) (common)

\(\triangle A B C \equiv \triangle B A D(\mathrm{RHS})\)

Thus, \(A D=B C\) (matching sides of congruent triangles)

Alternatively, use Pythagoras' theorem

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-068(5)}
\end{center}

Let \(\angle B C A=\alpha\)

\(\angle B D A=\alpha(\) angles subtended by the same arc)

\(\angle D A C=\alpha(\) alternate angles \(B C \| A D)\)

\(\angle B P A=2 \angle A C B\) (exterior angle of \(\triangle A P D)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-068(4)}
\end{center}

\(\angle A B C=180-\angle C D A\) (cyclic quadrilateral)

\(\angle E B C=180-\angle A B C\) (straight line)

\(\angle E B C=\angle C D A\)

8

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-068(1)}
\end{center}

\(\triangle P Q R \equiv \triangle P S R(\mathrm{AAS})\)

Let \(\angle Q P R=\angle R P S=\beta\) and \(\angle Q R P=\angle S R P=\alpha\)

\(2 \alpha+2 \beta=180^{\circ}\)

\(\alpha+\beta=90^{\circ}\)

Thus \(\angle P Q R=\angle R S P=90^{\circ}\)

9

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-068(3)}
\end{center}

Join \(B\) to \(E\). Let \(\angle E D C=\alpha\).

\(\angle C B E=180^{\circ}-\alpha\) (opposite angles of cyclic quadrilateral)

\(\angle A B E=\alpha\) (supplementary angles)

\(\angle A F E=180^{\circ}-\alpha\) (opposite angles of a cyclic quadrilateral)

\(\angle A F E\) and \(\angle C D E\) are co-interior angles and \(\angle C D E+\angle A F D=180^{\circ}\)

Thus, \(A F \| C D\)
10

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-068(2)}
\end{center}

Let \(\angle R V P=\alpha\)

\(\angle R T P=\angle R T P=\alpha\) (angles subtended by the same arc)

\(\angle S T Q=\alpha\) (vertically opposite)

\(\angle Q T S=\angle S V Q\) (angles subtended by the same arc)

Thus, \(\angle Q V S=\angle P V R\)

11 a \(\angle A D C=\frac{1}{2} \angle A Q C=\angle A Q X\) (angle subtended at the centre is twice the angle subtended at the circumference in the small circle)

\(\angle A B C=\frac{1}{2} \angle A D C\) (angle subtended at the centre is twice the angle subtended at the circumference in the large circle)

\(\therefore \angle A Q X=2 \angle A B C\)

b \(\angle A C B=90^{\circ}\) (Thales' theorem)

\(A B^{2}=B C^{2}+A C^{2}\) (Pythagoras' theorem in \(\triangle A B C\) )

\(=B C^{2}+4\left(A Q^{2}-X Q^{2}\right)(\) Pythagoras' theorem in \(\triangle A Q X)\)

\section*{Challenge exercise}
1 a \(\angle B C P=90^{\circ}\) (Thales' theorem)

\(\angle P=\angle A\) (angles on the circumference standing on the same arc)

In \(\triangle B C P, \sin P=\frac{a}{B P}\), so \(B P=\frac{a}{\sin A}\)

So \(2 R=\frac{a}{\sin A}\)

b \(\angle P=180^{\circ}-\angle A\) (opposite angles in a cyclic quadrilateral are supplementary)

In \(\triangle B C P, \sin \left(180^{\circ}-A\right)=\frac{a}{B P}\), so \(2 R=\frac{a}{\sin A}\),

since \(\sin \left(180^{\circ}-\alpha\right)=\sin \alpha\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-068}
\end{center}

2 a \(\triangle B R I \equiv \triangle B P I\) (AAS) as \(\angle R B I=\angle P B I(\angle B\) is bisected), \(\angle B R I=\angle B P I\) (right angles), and \(B I\) is common \(\therefore\)

\(I R=I P\) (corresponding sides of congruent triangles) \(\triangle C P I \equiv \triangle C Q I\) (AAS) as \(\angle P C I=\angle Q C I\) ( \(\angle C\) is bisected), \(\angle C P I=\angle C Q I\) (right angles), and \(C I\) is common \(\therefore I P=I Q\) (corresponding sides of congruent triangles)

b From a \(I P=I Q=I R, \angle A R I=\angle A Q I\) (right angles), and \(A I\) is common

\(\therefore \triangle A R I \equiv \triangle A Q I(\mathrm{RHS})\). So \(\angle R A I=\angle Q A I\)

(corresponding angles of congruent triangles)

\(\therefore I A\) bisects \(\angle A\)

c \(I P=I Q=I R=\) radius, \(I R \perp A B, I P \perp B C, I Q \perp A C\), so \(A B, B C\) and \(C A\) are tangents to the circle

3 a \(\angle C L H=90^{\circ}\) so \(C H\) is a diameter of a circle that passes through \(L\) (converse of Thales' theorem) and \(\angle C K H=90^{\circ}\) so \(C H\) is a diameter of a circle that passes through \(K\) (converse of Thales' theorem), so \(C, K, H\) and \(L\) are concyclic.

b \(C, K, H\) and \(L\) are concyclic, so \(\angle H C L=\angle H K L\) (angles on the circumference standing on the same arc) and \(\angle A K L=\angle H K L=\theta\)

c \(\angle A K B=90^{\circ}\) so \(A B\) is a diameter of a circle that passes through \(K\) (converse of Thales' theorem) and \(\angle B L A=90^{\circ}\) so \(A B\) is a diameter of a circle that passes through \(L\) (converse of Thales' theorem), so \(B, K, L\) and \(A\) are concyclic. As \(B, K, L\) and \(A\) are concyclic \(\angle A B L=\angle A K L\) (angles on the circumference standing on the same arc) and \(\angle A B L=\angle A K L=\theta\)

d \(A M C\) is similar to \(A L B\) (AA) as \(\angle A B L=\angle A C M\) (both \(\theta\) ), \(\angle B A L=\angle C A M\) (common) \(\therefore \angle B L A=\angle C M A=90^{\circ}\) (corresponding angles in similar triangles) meaning \(C M\) is an altitude of \(\triangle A B C\)

a \(M G: G O=2: 1\) (given), \(A G: G F=2: 1\) (centroid property),

\(\angle A G M=\angle F G O\) (vertically opposite) \(\therefore G M A\) is similar to \(\triangle G O F\) (SAS)

b \(\angle O F G=\angle M A G\) (corresponding angles in similar triangles), so \(M A \| O F\) (alternate angles equal) \(\therefore\) line \(A M \perp C B\) since \(F O \perp C B\), so \(M\) lies on the altitude from \(A\)

c By the same argument, \(B M \perp A C\) and \(C M \perp A B\). Hence, \(M=H\) from question 3 .

a \(\angle A B C+\angle A P C=180^{\circ}\) (as \(A P C B\) is a cyclic quadrilateral), \(\angle A B C+\angle A D C=180^{\circ}\) (given), so \(\angle A P C=\angle A D C\)

b \(D\) is on \(A P\) and \(\angle A P C=\angle A D C\), so \(D\) and \(P\) coincide.

a \(P M \times C M=A M \times B M\) (The product of the intervals on intersecting chords are equal)

b \(D M \times C M=A M \times B M\) (given) and \(P M \times C M=A M \times D M\) (from a), then \(D M=P M . D\) is on \(M P\) and \(D M=P M\), so \(P\) and \(D\) coincide.

7

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-069(3)}
\end{center}

a Let the intersection of \(A A^{\prime}\) and \(B B^{\prime}\) be \(M\).

\(A M=B M\) and \(A^{\prime} M=B^{\prime} M\) (tangents to a circle from an external point have equal length)

As \(A A^{\prime}=A M+A^{\prime} M\) and \(B B^{\prime}=B M+B^{\prime} M\), then \(A A^{\prime}=B B^{\prime}\)

b \(Y C^{\prime}=Y B^{\prime}, X C=X A, X A^{\prime}=X C^{\prime}, Y C=Y B\), (tangents to a circle from an external point have equal length),

\(X Y=X A+A A^{\prime}-Y C^{\prime}(\) eqn 1) and

\(X Y=Y B^{\prime}+B B^{\prime}-C X(\) eqn 2)

eqn \(1+\mathbf{e q n} 2\) gives

\(2 X Y=X A+A A^{\prime}-Y C^{\prime}+Y B^{\prime}+B B^{\prime}-C X\)

\(2 X Y=X A-C X+A A^{\prime}+B B^{\prime}-Y C^{\prime}+Y B^{\prime}\)

\(2 X Y=A A^{\prime}+B B^{\prime}\)

\(2 X Y=2 A A^{\prime}\) so \(X Y=A A^{\prime}\) c The indirect common tangents \(\left(A A^{\prime}\right.\) and \(\left.B B^{\prime}\right)\) become the same common tangent at the point of contact and \(A A^{\prime}=B B^{\prime}=X Y\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-069(2)}
\end{center}

\(\angle D\) is constant as angles on the circumference standing on the same arc are equal and \(\angle C\) is constant as angles on the circumference standing on the same arc are equal. So any triangles drawn as described are similar (AAA).

b Area \(=\frac{1}{2} B C \times B D \sin B\)

\(B\) is a constant. \(\therefore\) largest area when \(B C\) and \(B D\) are diameters. (If \(B C\) is a diameter, \(\angle C A B=90^{\circ}\). Therefore \(\angle B A D=90^{\circ}\). Thus \(B D\) is a diameter by the converse of Thales' theorem.)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-069(4)}
\end{center}

\(C A=C P\) and \(C B=C P\) (tangents to a circle from an external point have equal length), so \(C A=C P=C B\) \(\therefore C\) is the midpoint of \(A B\).

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-069(1)}
\end{center}

Let the centres of the circles be \(O\) and \(Q\).

As the circles touch at \(P\), the radii \(O P\) and \(Q P\) are \(\perp\) to the common tangent at \(P\), so \(O P Q\) is a line. \(\triangle D O P\) is isosceles as \(O D=O P\) (radii), so \(\angle O D P=\angle O P D\) and \(\triangle E Q P\) is isosceles as \(Q E=Q P\) (radii), so

\(\angle Q P E=\angle Q E P\).

\(\angle O P D=\angle Q P E\) (vertically opposite)

\(\therefore \angle O D P=\angle Q E P\), so \(O D \| Q E\) (alternate and equal angles). Finally the tangent at \(D\) is perpendicular to \(O D\) and the tangent at \(E\) is perpendicular to \(E Q\), so the tangents must also be parallel.

10

Area of \(\triangle A B C=\frac{1}{2} a b \sin C\)

\[
\begin{aligned}
& =\frac{1}{2} b c \sin A \\
& =\frac{1}{2} a c \sin B
\end{aligned}
\]

\[
\begin{aligned}
\therefore \frac{2 \times \text { Area of } \triangle A B C}{a b c} & =\frac{a b \sin C}{a b c}=\frac{b c \sin A}{a b c}=\frac{a c \sin B}{a b c} \\
& =\frac{\sin C}{c}=\frac{\sin A}{a}=\frac{\sin B}{b}
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-069}
\end{center}

\section*{Chapter 14 answers}
\section*{Exercise 14A}
\begin{center}
\begin{tabular}{llll}
1 & a 3 & b 3 & c 11 \\
d 0 & e 4 & f 3 &  \\
g 4 & h 6 &  &  \\
\(\mathbf{2}\) & a -4 & b -3 & c -1 \\
 & d -2 & e -3 & f -2 \\
g -10 & h -4 &  &  \\
3 & a \(\frac{3}{2}\) & b \(\frac{5}{2}\) & c \(\frac{5}{2}\) \\
 & d \(\frac{5}{2}\) & e \(\frac{7}{2}\) & f \(-\frac{5}{2}\) \\
 & g \(\frac{7}{3}\) & h \(\frac{1}{6}\) &  \\
4 & a 32 & b 729 & c 1000 \\
 & d \(\frac{1}{1000}\) & e \(\frac{1}{10} 000\) & f 625 \\
 & g 5 & h 60 & i 13 \\
5 & a 9 & b \(\sqrt{2}\) & c 4 \\
d 2 & e 3 & f 10 &  \\
\end{tabular}
\end{center}

\(6 \begin{array}{lll}\text { a } \log _{\sqrt{2}} 2=2 & \text { b } \log _{10} 0.001=-3 \quad \text { c } \log _{\frac{1}{2}} 2=-1\end{array}\) d \(\log _{32} 1024=2 \quad\) e \(\log _{10} N=x \quad\) f \(\log _{5} 5 \sqrt{2}=\frac{3}{2}\) g \(\log _{5} 1=0 \quad\) h \(\log _{13} 13=1\)

\begin{center}
\begin{tabular}{|c|c|c|}
\hline
a \(2^{5}=32\) & b \(3^{4}=81\) & c \(10^{-3}=0.001\) \\
\hline
d \(3^{\frac{1}{2}}=27 \sqrt{3}\) & e \(b^{x}=y\) & f \(a^{x}=N\) \\
\hline
\end{tabular}
\end{center}

9 a \(\log _{3} 10\)

b \(\log _{7} 2\)

e \(\log _{5} 10\)

c 1

d -1

10 a \(\log _{2} 105\)

b \(\log _{3} 5\)

c 0 d 0

11 a \(2 \alpha+\beta\)

b \(\beta+2 \gamma\)

c \(\alpha+\beta+\gamma+\delta\)

d \(7 \alpha+\beta+6 \gamma\)

e \(\beta+4 \gamma\)

g \(a \alpha+b \beta+c \gamma+d \delta\)

h 1

f \(\alpha+\beta+2 \gamma+\delta\)

12 a \(x y=x+y\)

b \(x^{2}=\frac{1}{10} y^{3}\)

c \(y=125 x^{2}\)

d \(\frac{1+y}{1-y}=7^{x}\)

\(13 \log _{2} V=\log _{2} \frac{4 \pi}{3}+3 \log _{2} r\),

or \(\log _{2} V=2+\log _{2} \pi-\log _{2} 3+3 \log _{2} r\)

\(14 x=\frac{1}{b} \log _{10} \frac{y}{a}\)

\(15 A=P \times 10^{b t}\)

Exercise 14B

1 a \(\frac{5}{2} \quad\) b \(\frac{5}{3}\)

\begin{center}
\begin{tabular}{llll}
2 & a 1.1292 & b 0.6826 & c 1.4650 \\
 & d 2.3347 & e 0.9622 & f -0.7124 \\
3 & a 2.3219 & b 2.6309 & c 0.4307 \\
 & d 1.7604 & e -2.8074 & f -1.4650 \\
4 & a -0.4225 & b 0.1587 & c -5.3847 \\
\end{tabular}
\end{center}

d 1

\(\begin{array}{lll}5 & \text { a } 1 & \text { b } 1\end{array}\)
6 a \(\frac{3}{2}\)
b \(\frac{7}{2}\)
c \(\frac{5}{3}\)
d \(\frac{3}{11}\)
e 12
f \(\frac{133}{60}\)

\section*{Exercise \(14 \mathrm{C}\)}
1 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-070(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-070(2)}
\end{center}

2

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(x\) & 0.01 & 0.1 & 1 & 10 & 100 \\
\hline
\(\log _{10} x\) & -2 & -1 & 0 & 1 & 2 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-070}
\end{center}

b

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(x\) & \(\frac{1}{36}\) & \(\frac{1}{6}\) & 1 & 6 & 36 \\
\hline
\(\log _{6} x\) & -2 & -1 & 0 & 1 & 2 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-070(3)}
\end{center}

3 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-071(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-071(6)}
\end{center}

c \(y=2^{-x}\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-071(2)}

4

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-071(7)}
\end{center}

b They have the same \(x\)-intercept, no \(y\)-intercept and the same asymptote, the \(y\)-axis.

c \(\log _{2} x\) grows faster than \(\log _{3} x\) as \(x\) grows

5

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-071(4)}
\end{center}

b \(y\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-071(3)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-071}
\end{center}

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-071(5)}

6

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-071(8)}
\end{center}

\section*{Exercise 14D}
14 hours 19 minutes

2 a \(M=0.72 \times 5^{t} \mathrm{~g}\)

b \(18 \mathrm{~g}\)

c 10 hours 20 minutes

d 39.94 days

e This is unrealistic, as there will not be enough food to keep the bacteria alive, so they will not multiply at the same rate for long.

3 a \(4 \times 1.02^{t-1976}\) billion
b 28.98 billion
c 2022

4 a \(750 \times 1.04^{t-1970}\) million
b 1995
c 2014

d 2050 (this is clearly impossible, because if the population of China increases at the same rate, the world population will increase at a rate getting closer to \(4 \%\), not the assumed \(2 \%\) )

5 a \(M_{0}=10, k=8.864 \times 10^{-4}\)

b 339.59 years
6 a 5
b 14

\(\begin{array}{ll}7 & 6\end{array}\)
8 a \(\$ 62985.60\)
b \(\$ 65816.20\)
c 55.5
d 38.3

\section*{Review exercise}
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
1 & a 4 & b 3 & c 9 \\
\hline
 & d 0 & e -3 & f -6 \\
\hline
 & g 4 & \(\mathbf{h}-3\) &  \\
\hline
2 & a 4 & b 2 & c 2 \\
\hline
 & d 8 & e 5 & f 5 \\
\hline
3 & \begin{tabular}{l}
a \(\log _{2} 1024=10\) \\
c \(\log _{6} 1=0\) \\
e \(\log _{3} b=x\) \\
\end{tabular} &  & \begin{tabular}{l}
\(=x\) \\
\(=1\) \\
\(5=4\) \\
\end{tabular} \\
\hline
4 & \begin{tabular}{l}
a \(3^{4}=81\) \\
d \(b^{a}=c\) \\
\end{tabular} & \begin{tabular}{l}
b \(2^{6}=64\) \\
e \(a^{c}=b\) \\
\end{tabular} & c \(10^{-2}=0.01\) \\
\hline
5 & \begin{tabular}{l}
a \(\log _{2} 55\) \\
d \(-\log _{3} 4\) \\
\end{tabular} & \begin{tabular}{l}
b \(\log _{2} 35\) \\
e 1 \\
\end{tabular} & \begin{tabular}{l}
c \(\log _{6} 77\) \\
f 1 \\
\end{tabular} \\
\hline
6 & \begin{tabular}{l}
a \(\log _{2} 140\) \\
d -2 \\
\end{tabular} & b 0 & c -2 \\
\hline
7 & \begin{tabular}{l}
a 1.2323 \\
d 3.2362 \\
\end{tabular} & \begin{tabular}{l}
b 1.2091 \\
e 0.75 \\
\end{tabular} & \begin{tabular}{l}
c 2.8928 \\
f -1.2619 \\
\end{tabular} \\
\hline
8 & \begin{tabular}{l}
a 2.8074 \\
d 3.0339 \\
\end{tabular} & \begin{tabular}{l}
b 3.9656 \\
e -2.3219 \\
\end{tabular} & \begin{tabular}{l}
c 2.0704 \\
f -2.4650 \\
\end{tabular} \\
\hline
 & a \(\frac{19}{2}\) & b 27 & c -1 \\
\hline
 & d 10000 & e \(-\frac{59}{2}\) & f 10 \\
\hline
\end{tabular}
\end{center}

10
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-072(1)}

c
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-072}

\(11 \begin{array}{ll}\text { a } y=10 x & \text { b } y+1=100 x\end{array}\)

\(12 \log _{2} 3-1\)

13 1.1

\begin{center}
\begin{tabular}{ll}
14 a \(\log _{2} 40\) & b \(\log _{2}\left(\frac{32}{5}\right)\) \\
15 a 4 years & b 7 years \\
\end{tabular}
\end{center}

\section*{Challenge exercise}
1 From Pythagoras' theorem, \(a^{2}=c^{2}-b^{2}=(c-b)(c+b)\); now take \(\log\) of both sides

\(2 \quad 1\)
4 a \(\frac{9}{7}\)
b 10 or \(10^{\frac{-5}{2}}\)

c 4 or \(2^{-\frac{5}{4}}\)

d 9 or 27

5 a \(x=3, y=5\)

b \(x=5, y=2\)

c \(x=-\frac{63}{8}, y=-\frac{9}{4}\)

d \(x=2, y=3\)

6 Change base: \(\frac{\log _{a} x \log _{a} x}{\log _{a} b}=\log _{a} b\), so \(\log _{a} x= \pm \log _{a} b\),

\(x=b\) or \(\frac{1}{b}\)

\(7 \quad a=\frac{2 b}{3}\)

8 b Suppose \(\log _{10} n=\frac{p}{q}\), then \(10^{\frac{p}{q}}=n, 10^{p}=n^{q}\), then for \(n\) to be an integer, it must contain an equal power of 2 and 5, and no other prime factors; that is, it is a power of 10 .

10 Let \(\alpha=\frac{\log _{a} x}{y-z}=\frac{\log _{a} y}{z-x}=\frac{\log _{a} z}{x-y}\)

\(\log _{a} x=\alpha y-\alpha z\)

\(\log _{a} y=\alpha z-\alpha x\)

\(\log _{a} z=\alpha x-\alpha y\)

Add together

\(\log _{a} x+\log _{a} y+\log _{a} z=0\)

\[
x y z=1
\]

Now, \(x=a^{\alpha y-\alpha z}, y=a^{\alpha z-\alpha x}\) and \(z=a^{\alpha x-\alpha y}\)

\[
\begin{aligned}
x^{x} y^{y} z^{z} & =a^{\alpha y x-\alpha z x+\alpha z y-\alpha x y+\alpha x z-\alpha y z} \\
& =a^{0} \\
& =1
\end{aligned}
\]

\(11 x=2 a\) or \(x=5 a\)

\section*{Chapter 15 answers}
\section*{Exercise 15A}
\(1 \frac{3}{13}\)
\(2 \quad \frac{7}{12}\)
\(3 \frac{3}{5}\)
\(4 \frac{4}{5}\)
\(5 \frac{8}{11}\)
6 a \(\xi=\{5,6,7,8,9\}\)
b \(P(5)=\frac{1}{10}, P(6)=\frac{1}{10}, P(7)=\frac{1}{4}, P(8)=\frac{7}{20}, P(9)=\frac{1}{5}\)
c \(\frac{4}{5}\)
\(7 \quad \mathbf{a} \xi=\left\{\begin{array}{l}(1,2),(1,4),(1,7),(2,2),(2,4),(2,7),(3,2), \\ (3,4),(3,7)\end{array}\right\}\) b equally likely, \(\frac{1}{9} \quad\) c \(\frac{2}{9}\)

8 a \(\xi=\{2,3,4,5,6,7\}\)

b \(P(2)=\frac{1}{12}, P(3)=\frac{1}{6}, P(4)=\frac{1}{4}, P(5)=\frac{1}{4}\),

\[
P(6)=\frac{1}{6}, P(7)=\frac{1}{12}
\]

c \(P(\) sum is less than 5\()=\frac{1}{2}\)
9 a \(\frac{1}{4}\)
b \(\frac{5}{18}\)
c \(\frac{1}{6}\)
10
a \(\xi=\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}\)
b \(\frac{5}{12}\)
c \(\frac{1}{6}\)
d \(\frac{5}{36}\)
e \(\frac{1}{3}\)

11 a \(\xi=\left\{\left(R_{1}, R_{1}\right),\left(R_{1}, R_{2}\right),\left(R_{1}, R_{3}\right),\left(R_{1}, W_{1}\right),\left(R_{1}, W_{2}\right)\right.\),

\(\left(R_{1}, Y\right),\left(R_{2}, R_{1}\right),\left(R_{2}, R_{2}\right),\left(R_{2}, R_{3}\right),\left(R_{2}, W_{1}\right),\left(R_{2}, W_{2}\right)\),

\(\left(R_{2}, Y\right),\left(R_{3}, R_{1}\right),\left(R_{3}, R_{2}\right),\left(R_{3}, R_{3}\right),\left(R_{3}, W_{1}\right)\), \(\left(R_{3}, W_{2}\right),\left(R_{3}, Y\right),\left(W_{1}, R_{1}\right),\left(W_{1}, R_{2}\right),\left(W_{1}, R_{3}\right)\), \(\left(W_{1}, W_{1}\right),\left(W_{1}, W_{2}\right),\left(W_{1}, Y\right),\left(W_{2}, R_{1}\right),\left(W_{2}, R_{2}\right)\), \(\left(W_{2}, R_{3}\right),\left(W_{2}, W_{1}\right),\left(W_{2}, W_{2}\right),\left(W_{2}, Y\right),\left(Y, R_{1}\right)\), \(\left.\left(Y, R_{2}\right),\left(Y, R_{3}\right),\left(Y, W_{1}\right),\left(Y, W_{2}\right),(Y, Y)\right\}\)
b i \(\frac{7}{18}\)
ii \(\frac{11}{18}\)
b \(\frac{33}{400}\)
c \(\frac{29}{200}\)

12 a \(\frac{1}{8}\)

\section*{Exercise 15B}
\(1 \frac{3}{5}\)

\[
2 \frac{12}{13}
\]

\(3 \frac{13}{15}\)
\(4 \quad \frac{19}{20}\)
5 a \(\frac{1}{10}\)
b \(\frac{1}{5}\)
c \(\frac{2}{5}\)
d \(\frac{4}{5}\)
6 a \(\frac{1}{4}\)
b \(\frac{4}{13}\)
c \(\frac{1}{13}\)
d \(\frac{25}{52}\)
e 0
f \(\frac{8}{13}\)
\(\begin{array}{rr}7 & \text { a } \frac{1}{6} \\ & \text { d } \frac{2}{3}\end{array}\)
b \(\frac{1}{2}\)
c 0
e \(\frac{1}{3}\)
f \(\frac{2}{3}\)
8 a \(\frac{29}{100}\)
b \(\frac{23}{100}\)
c \(\frac{49}{100}\)
d \(\frac{13}{20}\)
e \(\frac{11}{100}\)
f \(\frac{13}{20}\)
g \(\frac{77}{100}\)
h \(\frac{1}{8}\)
i \(\frac{7}{8}\)

9

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-073(1)}
\end{center}

b i \(\frac{3}{10} \quad\) ii \(\frac{2}{5}\)

iii 0
10

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-073(2)}
a \(\frac{1}{8}\)
b \(\frac{21}{40}\)
c \(\frac{7}{8}\)
d \(\frac{7}{20}\)

11

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-073(3)}
a \(\frac{4}{5}\)
b \(\frac{1}{5}\)
c \(\frac{22}{65}\)
d \(\frac{32}{65}\)
12 a \(\frac{1}{5}\)
b \(\frac{111}{1000}\)
c \(\frac{9}{100}\)
d \(\frac{11}{500}\)
e \(\frac{9}{500}\)
f \(\frac{1}{100}\)
g \(\frac{1}{500}\)
13 a \(\frac{6}{85}\)
b \(\frac{61}{85}\)
c \(\frac{16}{85}\)
14 a \(\frac{7}{26}\)
b \(\frac{4}{13}\)
c \(\frac{1}{2}\)
\(15 \frac{7}{15}\)

Exercise 15C
1 a \(\frac{17}{48}\)
b \(\frac{11}{24}\)
c \(\frac{13}{48}\)
d \(\frac{13}{31}\)
e \(\frac{4}{13}\)
f \(\frac{9}{22}\)
2 a \(\frac{1}{3}\)
b \(\frac{1}{2}\)
\(3 \frac{1}{3}\)

4 a \(\frac{4}{13}\)

b \(\frac{1}{13}\)

5 a \(\frac{4}{11}\)

b \(\frac{7}{11}\)

c \(\frac{2}{5}\)

d \(\frac{1}{3}\)

6 a \(\frac{7}{20}\)

b \(\frac{53}{100}\)

c \(\frac{20}{77}\)

d \(\frac{1}{2}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-073}
\end{center}

8

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
 & \(\mathbf{1}\) & \(\mathbf{2}\) & \(\mathbf{3}\) & \(\mathbf{4}\) & \(\mathbf{5}\) & \(\mathbf{6}\) \\
\hline
\(\mathbf{1}\) & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
\(\mathbf{2}\) & 1 & 0 & 1 & 2 & 3 & 4 \\
\hline
\(\mathbf{3}\) & 2 & 1 & 0 & 1 & 2 & 3 \\
\hline
\(\mathbf{4}\) & 3 & 2 & 1 & 0 & 1 & 2 \\
\hline
\(\mathbf{5}\) & 4 & 3 & 2 & 1 & 0 & 1 \\
\hline
\(\mathbf{6}\) & 5 & 4 & 3 & 2 & 1 & 0 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|}
\hline
Outcome & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
Probability & \(\frac{1}{6}\) & \(\frac{5}{18}\) & \(\frac{2}{9}\) & \(\frac{1}{6}\) & \(\frac{1}{9}\) & \(\frac{1}{18}\) \\
\hline
\end{tabular}
\end{center}

c \(\frac{2}{3}\)
d \(\frac{1}{3}\)
e \(\frac{4}{5}\)
f \(\frac{1}{5}\)
9 a \(\frac{1}{13}\)
b \(\frac{9}{19}\)
c \(\frac{3}{4}\)
d \(\frac{9}{19}\)
e \(\frac{2}{5}\)
10 a \(\frac{3}{14}\)
b \(\frac{9}{28}\)
c \(\frac{1}{3}\)
11 a \(\frac{1}{6}\)
b \(\frac{1}{2}\)
120.3

\section*{Exercise 15D}
1 No. Example: \(P(M \cap Y e s) \neq P(M) \times P(Y e s)[0.6 \times 0.55\)

\[
=0.33 \neq 0.25]
\]

2 No. Example: \(P(<30 \cap Y e s) \neq P(<30) \times P(Y e s)[0.5625 \times\) \(0.6875 \approx 0.387 \neq 0.4375\) ]

3 a 0.27

c \(P(M) \neq P(M \mid>10 \mathrm{~km} / \mathrm{h})\)

b \(\frac{21}{38} \approx 0.55\)

d \(P(\) Minor \(\cap\) Not speeding \() \neq P(\) Minor \() \times P(\) Not speeding \()\)

\[
\left[\frac{41}{200} \neq \frac{146}{200} \times \frac{43}{200}\right]
\]

4 a \(\frac{3}{35}\)
b \(\frac{18}{35}\)
c \(\frac{12}{35}\)
d \(\frac{23}{35}\)
\(5 \quad P(A)=P(A \mid B)=\frac{3}{8}\) or \(P(A \cap B)=P(A) \times P(B)=\frac{3}{20}\) or \(P(B)=P(B \mid A)=\frac{2}{5}\)

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
 & \multicolumn{7}{|c|}{Die 2} \\
\hline
\multirow{7}{*}{Die 1} &  & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
 & 1 & \((1,1)\) & \((1,2)\) & \((1,3)\) & \((1,4)\) & \((1,5)\) & \((1,6)\) \\
\hline
 & 2 & \((2,1)\) & \((2,2)\) & \((2,3)\) & \((2,4)\) & \((2,5)\) & \((2,6)\) \\
\hline
 & 3 & \((3,1)\) & \((3,2)\) & \((3,3)\) & \((3,4)\) & \((3,5)\) & \((3,6)\) \\
\hline
 & 4 & \((4,1)\) & \((4,2)\) & \((4,3)\) & \((4,4)\) & \((4,5)\) & \((4,6)\) \\
\hline
 & 5 & \((5,1)\) & \((5,2)\) & \((5,3)\) & \((5,4)\) & \((5,5)\) & \((5,6)\) \\
\hline
 & 6 & \((6,1)\) & \((6,2)\) & \((6,3)\) & \((6,4)\) & \((6,5)\) & \((6,6)\) \\
\hline
\end{tabular}
\end{center}

6 a

b \(P(A)=\frac{6}{36}=\frac{1}{6}, P(B)=\frac{6}{36}=\frac{1}{6}, P(A \cap B)=\frac{1}{36}\).

Events independent since \(P(A \cap B)=P(A) \times P(B)\).

c \(P(A \cap B) \neq 0\). Therefore, events \(A\) and \(B\) are not mutually exclusive.

\section*{Exercise 15E}
1 a \(\frac{1}{16}\)

b \(\frac{9}{16}\)

c \(\frac{3}{8}\)

d \(\frac{3}{16}\)

e \(\frac{3}{16}\)

f \(\frac{7}{16}\)

2 a \(\frac{1}{17}\)

b \(\frac{19}{34}\)

c \(\frac{13}{34}\)

d \(\frac{13}{68}\)

e \(\frac{13}{68}\)

f \(\frac{15}{34}\)

3 a \(\frac{10}{39}\)

b \(\frac{20}{39}\)

c \(\frac{14}{39}\)

d \(\frac{5}{39}\)

4 a \(\frac{10}{91}\)

b \(\frac{15}{91}\)

c \(\frac{30}{91}\)

d \(\frac{3}{91}\)

5 a \(\frac{5}{36}\)

b \(\frac{1}{18}\)

c \(\frac{5}{18}\)

d \(\frac{5}{18}\)

e \(\frac{17}{18}\)
6 a \(\frac{351}{1891}\)

b \(\frac{595}{1891}\)

c \(\frac{945}{1891}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-074}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-074(1)}
\end{center}

9 a \(\frac{1}{216}\)

b \(\frac{125}{216}\)

c \(\frac{1}{8}\)

d \(\frac{1}{8}\)

e \(\frac{5}{216}\)

f \(\frac{5}{216}\)

\(10 P(G)=P(G \mid C)=\frac{4}{9}\)

Review exercise
1 a i \(\frac{9}{25}\)
ii \(\frac{4}{25}\)
iii \(\frac{12}{25}\)
iv \(\frac{16}{25}\)
b i \(\frac{1}{10}\)
ii \(\frac{3}{5}\)
iii \(\frac{2}{5}\)
2 a \(\frac{1}{10}\)
b \(\frac{1}{2}\)
c \(\frac{1}{20}\)

3 a

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
 & \(\mathbf{1}\) & \(\mathbf{2}\) & \(\mathbf{3}\) & \(\mathbf{4}\) & \(\mathbf{5}\) & \(\mathbf{6}\) \\
\hline
\(\mathbf{2}\) & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
\(\mathbf{4}\) & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
\(\mathbf{6}\) & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\(\mathbf{8}\) & 9 & 10 & 11 & 12 & 13 & 14 \\
\hline
\end{tabular}
\end{center}

b i \(P(A)=\frac{1}{2}\)
ii \(P(B)=\frac{1}{8}\)
c \(A \cap C=\{9,10,11,12\} ; P(A \cap C)=\frac{5}{12}\)
d No. \(P(A) \times P(C)=\frac{1}{2} \times \frac{11}{12}=\frac{11}{24} \neq \frac{10}{24}=P(A \cap C)\)

\(4 \frac{1}{3}\)

\(5 \frac{2}{5}\)
6 a \(\frac{13}{20}\)
b \(\frac{13}{20}\)
c \(\frac{2}{5}\)
\(7 \frac{1}{2}\)
\(\begin{array}{lllll}\text { a } \frac{33}{100} & \text { b } \frac{7}{50} & \text { c } \frac{1}{25} & \text { d } \frac{29}{100} & \text { e } \frac{1}{10}\end{array}\)

\section*{Challenge exercise}
1 a i 330
ii 150
iii 215
iv 215

b 2

c i 0.28

ii 0.288

iii 0.49

2 a \(\frac{22}{119}\)

b \(\frac{11}{30}\)

c \(\frac{6}{11}\)

3 a i 1296

ii 360

iii 810

b \(\frac{671}{1296}\)

\(4 \frac{2}{3} \quad 5 \quad \frac{10(n-3)}{n(n-1)}\) for \(n>4\)

7 a 210

b \(\frac{5}{21}\)

\section*{Chapter 16 answers}
\section*{Exercise 16A}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-075(1)}
\end{center}

\(\frac{q}{\sqrt{p}}=4\), for each pair \((p, q)\)

c \(q=4 \sqrt{p}\)

3 a \(m=3 n^{2}\)
i 75
ii 3

b \(a=10 \sqrt{b}\)
i 40
ii 6.25

4 a \(R=4 s\)

b \(P=0.12 T\)

c \(a=4 \sqrt{b}\)

d \(V=8 r^{3}\)

5 a \(y=\frac{x}{4}\)

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(\boldsymbol{x}\) & 2 & 8 & 12 & 18 \\
\hline
\(\boldsymbol{y}\) & \(\frac{1}{2}\) & 2 & 3 & \(\frac{9}{2}\) \\
\hline
\end{tabular}
\end{center}

b \(y=8 x\)

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(x\) & 2 & 3 & 6 & 15 \\
\hline
\(y\) & 16 & 24 & 48 & 120 \\
\hline
\end{tabular}
\end{center}

\(6 \quad 130 \mathrm{~km}\)

\(\begin{array}{ll}7 & 210.5 \mathrm{~m}^{2}\end{array}\)

\(8 \quad 1852 \mathrm{~kW}\)

\(9 \quad 86.4 \mathrm{~m}\)

10 a The surface area is multiplied by 9.

b The radius is multiplied by \(\sqrt{3}\).

11 a \(m\) is multiplied by 32

b \(m\) is multiplied by \(\frac{1}{32}\)

c \(n\) is tripled

d \(n\) is divided by 4

12 a \(a\) is increased by \(11.8 \%\)

b \(a\) is decreased by approximately \(4.08 \%\)

13 a \(p\) is increased by \(6.27 \%\)

b \(p\) is decreased by \(1.70 \%\)

c \(q\) is increased by \(33.1 \%\)

d \(q\) is decreased by \(27.1 \%\)
Exercise 16B

1 a

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(\frac{1}{\boldsymbol{a}}\) & 1 & \(\frac{1}{2}\) & \(\frac{1}{3}\) & \(\frac{1}{4}\) & \(\frac{1}{5}\) \\
\hline
\(\boldsymbol{b}\) & 15 & 7.5 & 5 & 3.75 & 3 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-075}
\end{center}

b \(b=\frac{15}{a}\)

2 a

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(\boldsymbol{x}\) & 1 & 2 & 5 & 10 \\
\hline
\(\boldsymbol{y}\) & 100 & 25 & 4 & 1 \\
\hline
\(\boldsymbol{x}^{2} \boldsymbol{y}\) & 100 & 100 & 100 & 100 \\
\hline
\end{tabular}
\end{center}

b \(y=\frac{100}{x^{2}}\)
3 a \(v \propto \frac{1}{t}\)
b \(m \propto \frac{1}{\sqrt{n}}\)
c \(s \propto \frac{1}{t^{3}}\)

\(4 x y=6\) or \(y=\frac{6}{x}\)
a 4
b 9

\(5 a b^{2}=24\)
a \(\frac{8}{3}\)
b \(2 \sqrt{2}\)

\(6 p^{2} q=100\)
a \(\frac{10}{3}\)
b \(\frac{25}{4}\)

\(7 \quad k=12\) so \(y=\frac{12}{x}\)

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(x\) & 1 & 2 & 3 & 4 & \(\frac{1}{2}\) \\
\hline
\(y\) & 12 & 6 & 4 & 3 & 24 \\
\hline
\end{tabular}
\end{center}

\(8 \quad k=32\) so \(y=\frac{32}{x^{2}}\)

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(\boldsymbol{x}\) & 2 & 4 & 8 & 10 & 16 \\
\hline
\(\boldsymbol{y}\) & 8 & 2 & 0.5 & 0.32 & 0.125 \\
\hline
\end{tabular}
\end{center}

\(955 \mathrm{~km} / \mathrm{h}\)

\(10 \quad 54\)

\(\begin{array}{lll}11 \text { a } 1 \frac{1}{3} \text { units } & \text { b } 2 \mathrm{~m} \\ 12 & \text { a } m \text { is divided by } 4 & \text { b } m \text { is multiplied by } 4 \\ \text { c } n \text { is divided by } 4 & \text { d } n \text { is tripled }\end{array}\)

13 a \(a\) is decreased by approximately \(13.04 \%\) b \(a\) is increased by approximately \(13.64 \%\)

14 a \(p\) is decreased by approximately \(24.87 \%\)

b \(p\) is increased by approximately \(37.17 \%\)

c \(q\) is decreased by approximately \(5.90 \%\)

d \(q\) is increased by approximately \(7.72 \%\)
a i The height is divided by 4 .

ii The radius is divided by 3 .

b \(k_{1}=36\)

\section*{Exercise 16C}
\(1 a=12 b c\)

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(\boldsymbol{a}\) & 12 & 24 & 24 & 48 & 72 \\
\hline
\(\boldsymbol{b}\) & 1 & 1 & 2 & 2 & 2 \\
\hline
\(\boldsymbol{c}\) & 1 & 2 & 1 & 2 & 3 \\
\hline
\end{tabular}
\end{center}

\(2 r=\frac{24 s}{t}\)

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\(\boldsymbol{r}\) & 24 & 12 & 24 & 48 & 4 \\
\hline
\(\boldsymbol{s}\) & 1 & 1 & 2 & 2 & 2 \\
\hline
\(\boldsymbol{t}\) & 1 & 2 & 2 & 1 & 12 \\
\hline
\end{tabular}
\end{center}

\(3 y=5\)

4 a \(y=\frac{k x^{2}}{\sqrt{z}}\)

b \(y=\frac{27}{4}\)

5 a \(a=k b c^{3}\)

b \(b=32\)

6 a \(H=k i^{2} R t\)

b 2916 units

\(7 \quad 90 \mathrm{~N}\)

\(86 \sqrt{30} \mathrm{~cm} \approx 32.86 \mathrm{~cm}\)

9 a \(r=12\)

b \(n=24\)

10 a \(y\) is doubled

b \(y\) is multiplied by 13.5

11 a \(y\) is multiplied by approximately 1.154

b \(y\) is multiplied by approximately 1.562
12 a \(F\) is divided by 4
b \(F\) is multiplied by 12
\(131.65 \mathrm{~m} / \mathrm{s}^{2}\)

\section*{Review exercise}
1 a \(x\) is directly proportional to \(y\)

b \(p\) is directly proportional to the square of \(n\)

c \(a\) is directly proportional to the square root of \(b\)

d \(p\) is directly proportional to the cube of \(q\)

2 a i \(p=48\)

ii \(q=\frac{81}{8}\)

b \(a=\frac{5 b^{2}}{4}\)

\[
\text { i } a=\frac{125}{4} \quad \text { ii } a=180
\]

3 a \(k=12, y=12 x\)

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(x\) & 0 & 1 & 2 & 3 \\
\hline
\(y\) & 0 & 12 & 24 & 36 \\
\hline
\end{tabular}
\end{center}

b \(k=1.5, y=1.5 x\)

\begin{center}
\begin{tabular}{|l|l|c|c|c|}
\hline
\(\boldsymbol{x}\) & 2 & 8 & 12 & 18 \\
\hline
\(\boldsymbol{y}\) & 3 & 12 & 18 & 27 \\
\hline
\end{tabular}
\end{center}

4 a \(y\) is multiplied by 8

b \(y\) is multiplied by 27

c \(y\) is divided by 64

5 a \(m\) is multiplied by \(\sqrt{2}\) (increase by \(41 \%\) since \(\sqrt{2} \approx 1.41\) )

b \(m\) is halved

6 a \(a\) is increased by \(10.25 \%\)

b \(a\) is decreased by \(15.36 \%\)
7 a \(y=\frac{80}{3}\)
b \(x=60\)
8 a \(a=2\)
b \(b=\frac{4 \sqrt{2}}{3}\)
\(9 \quad z=27 \sqrt{2}\)
\(10 x=\frac{128}{3}\)

\section*{Challenge exercise}
1 a \(R=\frac{0.00224 L}{D^{2}}\)

b \(5.736 \approx 5.4 \mathrm{ohm}\)

c resistance is halved

d resistance is increased by \(21.9 \%\)

\(2 \quad a=k c\) and \(b=K c\)

\(a+b=(k+K) c\) so \((a+b) \infty c\)

\(a-b=(k-K) c\) so \((a-b) \infty c\)

\(\sqrt{a b}=\sqrt{k c \times K c}=\sqrt{k K} c\) so \(\sqrt{a b} \infty c\)

\(3 y=\frac{150 x}{19}+\frac{1080}{19 x^{2}} \quad 4 \quad y=\frac{12}{5} \quad 5 \quad \frac{2 a b C}{(a+b)^{2}}\)

\(6 \quad a+b=k(a-b)\)

\[
a^{2}+2 a b+b^{2}=k^{2}\left(a^{2}-2 a b+b^{2}\right)
\]

\[
\begin{aligned}
2 a b k^{2}+2 a b & =k^{2} a^{2}-a^{2}+k^{2} b^{2}-b^{2} \\
2 a b\left(k^{2}+1\right) & =\left(k^{2}-1\right)\left(a^{2}+b^{2}\right) \\
a^{2}+b^{2} & =\frac{k^{2}+1}{k^{2}-1} \times 2 a b \\
a^{2}+b^{2} & \infty a b
\end{aligned}
\]

\section*{Chapter 17 answers}
\section*{Exercise 17A}
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-076}
a \(0,60,-90900,0\)

b \(39,0,11712,32\)

c \(-1,-55,295001,1\)

d \(-27,0,0,-20\)

6 a \(x^{2}-10 x+25, x^{2}, 2,25\)

b \(x^{2}+5 x-50, x^{2}, 2,-50\)

c \(-3 x^{7}+x^{3},-3 x^{7}, 7,0\)

d \(x^{3}+12 x^{2}+36 x, x^{3}, 3,0\)

e \(3 x^{7}+6 x^{5}+3 x^{3}, 3 x^{7}, 7,0\)

f \(-6 x,-6 x, 1,0\)

g \(x^{3}+9 x^{2}+26 x+24, x^{3}, 3,24\)

h \(3 x^{2}+12 x+14,3 x^{2}, 2,14\)

7 a \(a=6\)

b \(b=-1\)

c \(a=-15, d=8\)
a \(P(x)=x^{2}-5 x+1\)
b \(Q(x)=2 x^{4}+10 x\)
c \(R(x)=7 x^{4}+7 x^{3}+7 x^{2}+7 x+7\)

\section*{Exercise 17B}
1 a \(3 x^{3}-3 x^{2}-x+5,-x^{3}+3 x^{2}+7 x+5\)

b \(x+3,4 x^{3}-6 x^{2}-9 x+7\)

c \(8 x^{2}-6 x, 12\)

d \(x^{4}+x^{3}-2 x^{2}+2 x-2, x^{4}-x^{3}\)

e \(0,10 x^{3}+4 x^{2}-2 x-10\)

2 a \(12 x^{5}+4 x^{3}-12 x^{2}+20\)

b \(-16 x^{5}+10 x^{3}-7 x^{2}+27\)

c \(-5 x^{5}+5 x^{3}-5 x^{2}+15\)

d 0

3 a \(5 x^{6}-2 x^{5}+7 x^{4}\)

b \(x^{6}-1\)

c \(x^{6}-1\)

d \(x^{5}+3 x^{4}-x^{3}-3 x^{2}\)

e \(x^{4}+x^{2}+1\)

4 a equals the sum of the degrees of \(P(x)\) and \(Q(x)\)

b the product of the constant terms of \(P(x)\) and \(Q(x)\)

5
a \(x^{2}-14 x+49\)
b \(x^{4}-6 x^{2}+9\)
c \(x^{6}-14 x^{4}+49 x^{2}\)
d \(9 x^{10}+30 x^{8}+25 x^{6}\)
e \(x^{4}+2 x^{3}+3 x^{2}+2 x+1\)
f \(x^{8}+2 x^{6}+3 x^{4}+2 x^{2}+1\)

6 a is twice the degree of \(P(x)\)

b equal to the square of the constant term of \(P(x)\)

c the leading coefficient of \(P(x)\) is 1 or -1

7 a \(-5 x^{4}+5 x^{3}-7 x-3\)

b \(6 x^{4}-x^{3}-15 x^{2}-2 x+3\)

c \(-4 x^{4}-3 x^{3}+26 x^{2}+28 x+2\)

8 a \(x^{4}-5 x^{2}+4\)

b \(x^{4}+10 x^{3}+35 x^{2}+50 x+24\)

\section*{Exercise 17C}
a \(68=11 \times 6+\)

b \(1454=12 \times 121+2\)

c \(2765=21 \times 131+14\)
2 a \(x^{2}+6 x+1=(x+2)(x+4)-7\)

b \(x^{3}-5 x^{2}-12 x+30=(x+5)\left(x^{2}-10 x+38\right)-160\)

c \(5 x^{3}-7 x^{2}-6=(x-3)\left(5 x^{2}+8 x+24\right)+66\)

d \(x^{4}+3 x^{2}-3 x=(x+2)\left(x^{3}-2 x^{2}+7 x-17\right)+34\)

e \(4 x^{3}-4 x^{2}+1=(2 x+1)\left(2 x^{2}-3 x+\frac{3}{2}\right)-\frac{1}{2}\)

f \(x^{4}+3 x^{3}-3 x^{2}-4 x+1=(x+1)\left(x^{3}+2 x^{2}-5 x+1\right)\)

3 a \(R(x)=-5 x+1, Q(x)=x^{2}-x+3\)

b \(Q(x)=x^{2}-2 x+5, R(x)=15-8 x\)

4 a \(x^{3}+5 x^{2}-x+2=\left(x^{2}+x+1\right)(x+4)-6 x-2\)

b \(x^{3}-4 x^{2}-3 x+7=\left(x^{2}-2 x+3\right)(x-2)-10 x+13\)

c \(x^{4}+5 x^{2}+3=\left(x^{2}-3 x-3\right)\left(x^{2}+3 x+17\right)+60 x+54\)

d \(x^{5}-3 x^{4}-9 x^{2}+9=\left(x^{3}-x^{2}+x-1\right)\left(x^{2}-2 x-3\right)-\)

\[
9 x^{2}+x+6
\]

5 a 0

b 0 or 1

c 3 or higher

d 4

6 a \(P(x)=(x+5)(x+3)(x-7)\)

b \(P(x)=(x+2)^{2}(x+3)^{2}\)

7

a i \(x^{4}-3 x^{3}-5 x^{2}+x-7\)

\[
=(x+5)\left(x^{3}-8 x^{2}+35 x-174\right)+863
\]

ii -870

b i \(x^{4}-3 x^{3}-5 x^{2}+x-7\)

\[
=\left(x^{2}+5\right)\left(x^{2}-3 x-10\right)+16 x+43
\]

ii \(a=-15, b=-50\)

\section*{Exercise 17D}
1 a 38 , not a factor
b 0 , is a factor
c -274 , not a factor
d 0 , is a factor
2 a 0 , is a factor
b -6 , not a factor
c 38 , not a factor
d 20 , not a factor
3 a 0 , is a factor
b -14 , not a factor
c 1 , not a factor
d -4 , not a factor
e 6 , not a factor
f 0 , is a factor
4 a \(x+1, x+2, x-2\)
b \(x+1, x-1, x+4\)
\(5 \quad \mathbf{a} k=4\)
b \(m=4\)
6 a \(p=28\)
b \(b=-23\)
7 a \(a=-14, b=-22\)
b \(a=9, b=5\)
c \(a=17, b=21, c=0\)

\section*{Exercise 17E}
\(1 \quad\) a \(P(x)=(x-12)(x+9) \quad\) b \(x^{2}-3 x-108\)

\(2 \quad P(x)=(x-1)(x+1)(x-2)(x+2)\)

3 b \(P(x)=(x-1)\left(x^{2}-5 x+6\right)\)

c \(P(x)=(x-1)(x-2)(x-3)\)

\(4 \quad \mathbf{a}(x+1)(x+2)(x+3)\)

b \((x+1)(x-1)(x-7)\)

c \((x+1)(x+5)(x-3)\)

d \((x-5)(x+3)^{2}\)

e \((x-1)^{2}(x+3)\)

f \((x-1)(x+2)^{2}\)

5

a \((x+1)(x-1)(x-2)(x-3) \quad\) b \((x+1)^{2}(x+5)^{2}\)

6 a \(3(x-2)(x-1)(x+5)\)

b \(5(x-2)(x-1)(x+2)\)

c \(x(x-2)(x+1)(x+2)\)

d \(x(x-1)^{2}(x+3)^{2}\)

\(7 \quad \mathbf{a}(x-1)\left(x^{2}+3 x+5\right)\)

b \((x+3)\left(x^{2}+x+1\right)\)

c \(x^{2}(x-2)\left(x^{2}+6 x-3\right)\)

d \((x-2)(x+3)\left(x^{2}+3 x+1\right)\)

8 If \(P(\alpha)=0\), then, \(a_{n} \alpha^{n}+a_{n-1} \alpha^{n-1}+a_{n-2} \alpha^{n-2}+\ldots+\) \(a_{1} \alpha+a_{0}=0\)

Therefore, \(a_{0}=-\alpha\left(a_{n} \alpha^{n-1}+a_{n-1} \alpha^{n-2}+a_{n-2} \alpha^{n-3}+\ldots+a_{1}\right)\) and \(\alpha\) divides \(a_{0}\)

\section*{Exercise 17F}
1 a \(-7,5,-6\)
b \(3,-1\)
c \(2,4,6,8\)
d \(0,7,-8\)

2 a \(3,-3+\sqrt{17},-3-\sqrt{17}\)

b \(-5, \frac{1}{3}(1+\sqrt{7}), \frac{1}{3}(1-\sqrt{7})\)

c \(0,7,-6\)

d \(2,5, \sqrt{10},-\sqrt{10}\)
3
a \(-2,-1,5\)
b \(-2,2,3\)
c \(-7,0,1,3\)
d \(-2,-1,1,7\)

4
a 1,5
b \(-3,2\)
c 1,5
d \(-3,2\)
e \(-6,-1,0\)
f \(-1,1,2\)

5 a \((x-3)\left(x^{2}-4 x-1\right) ; 3,2-\sqrt{5}, 2+\sqrt{5}\)

b \((x+1)\left(x^{2}+3 x+7\right) ;-1\)

c \(x(x-2)(x-3)\left(x^{2}+3 x-1\right) ; 0,2,3\),

\[
\begin{aligned}
& \quad \frac{1}{2}(-3-\sqrt{13}), \frac{1}{2}(-3+\sqrt{13}) \\
& \mathbf{d}(x+1)(x-1)(x-2)\left(x^{2}+2 x+2\right) ;-1,1,2
\end{aligned}
\]

\section*{Exercise 17C}
1

a \(x=2, x=4\)

\begin{center}
\begin{tabular}{l|lll}
Sign of \(y\) & + & 0 & 0 \\
\hline
\(x\) values & 2 & 4 &  \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-078(6)}
\end{center}

b \(x=0, x=2, x=4\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-078(2)}

c \(x=-3, x=1, x=3\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-078(1)}

d \(x=-2, x=-1, x=3\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-078(3)}

2 a \(x=1\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-078(4)}
\end{center}

b \(x=1\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-078(5)}

c \(x=1\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-078}

3 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(7)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(6)}
\end{center}

\(4 \quad \mathbf{a} \quad y=(x+2)(x-1)(x+4)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(10)}
\end{center}

b \(x>1,-4<x<-2\)

c \(x<-4,-2<x<1\)

5

a \(y=3 x(x-4)(x+4)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(11)}
\end{center}

b \(y=(x-6)(x+6)(x-2)(x+2)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(3)}
\end{center}

6 a \(P(x)=(x-2)(x-4)(x-6)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(5)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(12)}
\end{center}

b \(P(x)=(x+3)^{3}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(1)}
\end{center}

c \(P(x)=(x-2)^{3}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(4)}
\end{center}

\(7 \mathbf{a}-\mathbf{b}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(8)}
\end{center}

\(8 a-b\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(9)}
\end{center}

\(9 \mathbf{a}-\mathbf{b}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(2)}
\end{center}

\(P(x)=-(x+3)^{4}\)

Exercise 17H

1 a \(x=0, x=2\)

\begin{center}
\begin{tabular}{c|cc}
Sign of \(y\) & -0 & + \\
\hline
\(x\) values & 0 & 0 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(13)}
\end{center}

b \(x=2, x=4\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-079(14)}
\end{center}

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-079}
c \(x=0, x=-3\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-080(3)}

d \(x=-2, x=-1\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(9)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(6)}
\end{center}

2

a \(x=4, x=-4\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(7)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080}
\end{center}

b \(x=-1, x=4\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(5)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(13)}
\end{center}

c \(x=0, x=4\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-080(8)}

d \(x=0, x=-2\)
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-080(4)}

3 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(11)}
\end{center}

4 a \(y=(x+3)^{3}(x-1)^{2} \overbrace{}^{y}\)
b \(x>1,-3<x<1\)
c \(x<-3\)

5 a \(y=3(x-1)(x+1)(x+3)^{4}(x-3)^{4}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(12)}
\end{center}

b \(y=5 x^{2}\left(4-x^{2}\right)=-5 x^{2}(x-2)(x+2)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(2)}
\end{center}

6 a \(P(x)=(x-2)^{3}(x-4)^{3}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-080(10)}
\end{center}

ICE-EM MATHEMATICS YEAR 10 ICE-EM Mathematics 10 3ed ISBN 978-1-108-40 Photocopying is restricted under law and this material must not be transferred to another party Updated June 2022
b \(P(x)=(x+3)^{3}(x-1)^{2}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-081(4)}
\end{center}

\(7 \mathbf{a}-\mathbf{b}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-081}
\end{center}

\(\mathbf{a}-\mathbf{b}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-081(2)}
\end{center}

\section*{Review exercise}
12 a \(2 w^{3}\)

b \(a=-2, b=4\)

13 a \(x\)-intercepts: \(-1,0,2 y\)-intercept: 0

b \(x\)-intercepts: \(-2,1,3 y\)-intercept: 6

c \(x\)-intercepts: \(-2,-1,1 \quad y\)-intercept: -2

d \(x\)-intercepts: \(-1,-\frac{2}{3}, 3 y\)-intercept: -6

e \(x\)-intercepts:, \(-4,-\frac{2}{5}, 2 y\)-intercept: -16

f \(x\)-intercepts: \(-\frac{1}{2}, \frac{1}{3}, 1 y\)-intercept: 1

14 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-081(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-081(3)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-081(1)}
\end{center}

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-081(5)}
f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-082(1)}
\end{center}

Challenge exercise

\(1 a=-3\)

\(2\left(x^{2}-2 x+2\right)\left(x^{2}+2 x+2\right)\)

\(3 b=14, a=-11\)

4 b \(x^{3}-6 x^{2}+11 x-6\)

\(5 \quad P(x)=6 x^{5}-15 x^{4}+10 x^{3}\)

\(6 \quad a=1\) and \(b=2\)

\(74 x\)

8 c \(\left(x^{4}+2 x^{2}+2 x\right)^{2}+\left(x^{3}-2 x^{2}-4\right)^{2}\)

\(9 \frac{(A-B)}{a-b} x-\frac{(B a-A b)}{a-b}\)

\(10 x=\frac{1+\sqrt{5}}{2}, y=\frac{1-\sqrt{5}}{2} ; x=\frac{1-\sqrt{5}}{2}, y=\frac{1+\sqrt{5}}{2}\);

\(x=1, y=1 ; x=-2, y=-2\)

\section*{Chapter 18 answers}
\section*{Exercise 18A}
1 a range \(=17, \mathrm{IQR}=9\)

b range \(=9, \mathrm{IQR}=3\)

c range \(=9, \mathrm{IQR}=2.5\)

d range \(=17, \mathrm{IQR}=9\)

2 a lower quartile \(=27\), median \(=32\), upper quartile \(=38.5\), \(\mathrm{IQR}=11.5\)

b lower quartile \(=64\), median \(=76\), upper quartile \(=82\), \(\mathrm{IQR}=18\)

3 mean \(=5.7\), mode \(=10\), median \(=5.5\), interquartile range \(=6\)

4

a

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\begin{tabular}{c}
number of \\
data items \\
\end{tabular} & \begin{tabular}{c}
lower quartile \\
position \\
\end{tabular} & \begin{tabular}{c}
median \\
position \\
\end{tabular} & \begin{tabular}{c}
upper quartile \\
position \\
\end{tabular} \\
\hline
100 & 25.5 & 50.5 & 75.5 \\
\hline
101 & 25.5 & 51 & 76.5 \\
\hline
\end{tabular}
\end{center}

5 a \(38 \mathrm{~cm}\)

b \(161.5 \mathrm{~cm}\)

c \(22.5 \mathrm{~cm}\)

\(6 \quad \mathrm{IQR}=1.4 \mathrm{~cm}\)

7 a median \(=\$ 149\), lower quartile \(=\$ 143\),

upper quartile \(=\$ 153.50\)

b \(\mathrm{IQR}=\$ 10.50\)

8

35710121213 and 35610111213 (others are possible)
9 No. 11122318 , mean \(=4\), lower quartile \(=1\)

and upper quartile \(=3\)

11313141416 , mean \(=12\), lower quartile \(=13\)

and upper quartile \(=14\)

10 a 35

b 13.7

\section*{Exercise 18B}
\(1 \$ 49\) and \(\$ 10.50\), respectively

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-082(3)}
\end{center}

5 a median \(=75\), upper quartile \(=78.5\), lower quartile \(=69.5, \mathrm{IQR}=9\)

b

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-082}
6 a \(50 \%\)
b \(25 \%\)
c \(50 \%\)
d \(25 \%\)

7 no

8 It depends on the spread of the values between the lower quartile and the median compared with those between the median and the upper quartile.
9 a \(\mathrm{B}\)
b B
c \(\mathrm{B}\)
d B
10 a \(\mathrm{A}\)
b B
c \(\mathrm{B}\)
d B

e Class A. Lowest mark is higher, and lower quartile, median and upper quartile are all higher. Only the maximum mark is lower.

11 a

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
channel & minimum & \begin{tabular}{c}
lower \\
quartile \\
\end{tabular} & median & \begin{tabular}{c}
upper \\
quartile \\
\end{tabular} & maximum \\
\hline
A & 7 & 10.5 & 15 & 16.3 & 23 \\
\hline
B & 7.8 & 11 & 14.6 & 16.7 & 25 \\
\hline
C & 8 & 10 & 13 & 14.5 & 16 \\
\hline
\end{tabular}
\end{center}

b channel A c channel B, channel A, channel C

d channel A, channel B, channel C

e If the criterion is the highest rating for the lowest rating show, then Channel \(\mathrm{C}\) is the winner.

\section*{Exercise \(18 \mathrm{C}\)}
1 a \(\mathrm{Q}_{1}=151.5, \mathrm{Q}_{3}=171.5\), median \(=164\)

b \(\mathrm{IQR}=20\)

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-082(2)}
\end{center}

2 a \(\mathrm{Q}_{1}=67000, \mathrm{Q}_{3}=100000\), median \(=81500\)

b \(\mathrm{IQR}=33000\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-083}
\end{center}

3 a iii, symmetric

b i, negative skew

c ii, positive skew

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-083(6)}
\end{center}

d symmetric

\(5 \mathbf{a}\) and \(\mathbf{b}\) are outliers; \(\mathbf{c}\) is not an outlier

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-083(3)}
\end{center}

b median \(=26.5, \mathrm{Q}_{1}=18.5, \mathrm{Q}_{3}=34, \mathrm{IQR}=15.5\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-083(5)}
\end{center}

c The distribution is symmetric. There are no outliers.

7 a median \(=38.5, \mathrm{Q}_{1}=32.5, \mathrm{Q}_{3}=44, \mathrm{IQR}=11.5\)

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-083(1)}
\end{center}

c 62 is an outlier

8 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-083(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-083(4)}
\end{center}

\[
\begin{array}{cc}
Q_{1}=5 \underset{2}{\text { median }} & Q_{3}=7.5 \\
=6
\end{array}
\]

c The distribution has a slight positive skew.

minimum \(=4, \mathrm{Q}_{1}=5\), median \(=6, \mathrm{Q}_{3}=7.5\),

maximum \(=13,12 \mathrm{~kg}\) and \(13 \mathrm{~kg}\) are outliers

\section*{Exercise 18D}
\(1 \quad 15.54\)

2 City A : 30.43, City B : 31.86; City B has the greater mean daily maximum temperature.

\(360 \mathrm{~kg}\)

\(4 \quad 41.2\)

5 a i \(\bar{x}=5, \sigma=3.16\)

ii \(\bar{x}=5, \sigma=1.34\)

b Both data sets have the same mean but data set \(\mathbf{i}\) is more spread out.

6

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(\boldsymbol{x}_{\boldsymbol{i}}\) & \(\boldsymbol{f}_{\boldsymbol{i}}\) & \(\boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}}\) & \(\left(\boldsymbol{x}_{\boldsymbol{i}}-\overline{\boldsymbol{x}}\right)\) & \(\boldsymbol{f}_{\boldsymbol{i}}\left(\boldsymbol{x}_{\boldsymbol{i}}-\overline{\boldsymbol{x}}\right)^{\mathbf{2}}\) \\
\hline
\(\mathbf{1}\) & 2 & 2 & -2 & 8 \\
\hline
\(\mathbf{2}\) & 7 & 14 & -1 & 7 \\
\hline
\(\mathbf{3}\) & 6 & 18 & 0 & 0 \\
\hline
\(\mathbf{4}\) & 1 & 4 & 1 & 1 \\
\hline
\(\mathbf{5}\) & 2 & 10 & 2 & 8 \\
\hline
\(\mathbf{6}\) & \(\mathbf{2}\) & 12 & 3 & 18 \\
\hline
 & Total \(=60\) & Total \(=60\) &  & Total \(=60\) \\
\hline
\end{tabular}
\end{center}

\(\sigma=1.45, \bar{x}=3\)

7 a \(\bar{x}=9.13, \sigma=3.61\)

b \(\bar{x}=14, \sigma=3.46\)

8 a \(\bar{x}=30.55\)

b 10

c 10.22

9 a 6.15

b 10

c 2.46

10 a \(\bar{x}=\frac{a+b+c}{3}\) and the

sum of the deviations \(=a-\bar{x}+b-\bar{x}+c-\bar{x}\)

\[
\begin{aligned}
& =(a+b+c)-3 \bar{x} \\
& =(a+b+c)-3 \times \frac{a+b+c}{3} \\
& =0
\end{aligned}
\]

b \(\bar{x}=\frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}\) and

the sum of the deviations \(=\left(x_{1}-\bar{x}\right)+\left(x_{2}-\bar{x}\right)+\)

\[
\begin{aligned}
& \left(x_{1}-\bar{x}\right)+\ldots+\left(x_{n}-\bar{x}\right) \\
= & \left(x_{1}+x_{2}+x_{3}+\ldots+x_{n}\right)-n \bar{x} \\
= & \left(x_{1}+x_{2}+x_{3}+\ldots+x_{n}\right)- \\
& n \times \frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n} \\
= & 0
\end{aligned}
\]

\section*{Exercise 18E}
1 a \(\bar{x}=9, \sigma=5.45\)
b \(\bar{x}=9, \sigma=1.31\)
c \(\bar{x}=9, \sigma=5.45\)

All three data sets have the same mean. Data set \(\mathbf{b}\) is less spread out than the other two, with a standard deviation of 1.31. Data sets \(\mathbf{a}\) and \(\mathbf{c}\) have the same mean and standard deviation. (The size of \(\sigma\) for a would be much smaller if the outlier 22 is omitted. It drops from 5.45 to 1.34 . This does not happen when any one value is omitted from c.)

2 a i between 32.5 and 37.5

ii between 30 and 40

b i between 35 and 45

ii between 30 and 50

c i between 27 and \(43 \quad\) ii between 19 and 51

3 a Mathematics : \(\bar{x}=16.13 \sigma=2.63\);

English \(: \bar{x}=15.87 \sigma=2.83\)

b The mathematics result is about 0.05 standard deviations below the mean, whereas the English result is 0.66 standard deviations below the mean. The mathematics mark is better.

4 a David's mark for English is one standard deviation below the mean.

David's mathematics mark is more than one standard deviation below the mean.

His English mark is better.

b Akira's mark for English is two standard deviations above the mean.

Akira's mathematics mark is less than two standard deviations above the mean.

Her English mark is better.

c Katherine's mark for English is one standard deviation below the mean.

Katherine's mathematics mark is half a standard deviation below the mean.

Her mathematics mark is better.

d Daniel's mark for English is more than one standard deviation above the mean.

Daniel's mathematics mark is one standard deviation above the mean.

His English mark is better.
a \(\mathbf{i} \bar{x}=6.5, \sigma=1.71\)
ii \(\bar{x}=6.5, \sigma=1.38\)
iii \(\bar{x}=6.5, \sigma=1.98\)
b i \(\bar{x}=11.5, \sigma=1.71\)
ii \(\bar{x}=11.5, \sigma=1.38\)
iii \(\bar{x}=11.5, \sigma=1.98\)
ii \(\bar{x}=13, \sigma=2.77\)
c i \(\bar{x}=13, \sigma=3.42\)
iii \(\bar{x}=13.0, \sigma=3.96\)

6 a 1111155555

b 1111199999

c 1111559999

7 a 1111111111

b 1111111111

c 5555555555

8 a average annual salary increases by about \(\$ 5200\)

b the standard deviation remains unchanged

c an increase of \(\$ 2900\)

\section*{Exercise \(18 \mathrm{~F}\)}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-084(2)}
\end{center}

b The rainfall is high in January and February and decreases to reach a minimum in September. The rainfall from September to December increases significantly.
2

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-084(4)}
\end{center}

3 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-084}
\end{center}

b The number of births fell over the first half of the year. From July until the end of the year, the birth rate was roughly constant. The maximum occurred in January and the minimum in October.

4 a 12

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-084(3)}
\end{center}

b The team improved its position quite markedly during the first half of the season and maintained a position in the top five teams between rounds 7 and 18 (except for round 12, when it was sixth). However, during the last five rounds, the team's position deteriorated again to eighth at the end of the season.

5 a fourth quarter

b third quarter

c yes

1st quarter sales: \(45,51,55\)

2 nd quarter sales: \(63,69,71\)

3rd quarter sales: \(67,75,79\)

4th quarter sales: \(43,39,49\)

6 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-084(1)}
\end{center}

b Car sales per quarter have shown a general upward trend, with major fluctuations.

c It seems that the car dealer is able to sell more cars in the third and fourth quarters each year than in the first and second quarters.

\section*{Exercise 18G}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-085(5)}
\end{center}

2

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-085(4)}
\end{center}

b The number of rainy days per month generally increases as the average monthly rainfall increases.

3 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-085(8)}
\end{center}

b There is no apparent relationship between carbohydrate content and fat content.

4 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-085(3)}
\end{center}

b The time taken to complete a puzzle tends to decrease as IQ increases.

5 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-085(7)}
\end{center}

b The scatter plot does not support the claim 'the greater the kicks, the greater the handballs'. This would only be supported by an upwards trend from left to right.

6

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-085(2)}
\end{center}

b i The best team is F, with easily the greatest 'goals for' and very nearly the least 'goals against'.

ii The worst team is E, with the least 'goals for' and the greatest 'goals against'.

iii \(\mathrm{J}\) is better than \(\mathrm{H}\) - while the 'goals for' are about the same, the 'goals against' clearly favour \(\mathbf{J}\).

7 a \(B=\mathrm{v}\)

b \(C=\) ii

c \(D=\) iv

d \(E=\) viii

e \(F=\) vi

f \(G=\) iii

g \(H=\mathrm{i}\)

h \(I=\) vii

8
a \(A=\) iii
b \(B=\) iv
c \(C=\) viii
d \(D=\) vii
e \(E=\) vi
f \(F=\mathrm{v}\)
g \(G=\mathrm{i}\)
h \(H=\) ii

9 a i 54

ii 55

iii 36

iv 84

v 67 or 68

b i 52

ii 53

iv 84

v 66

iii 32

\section*{Exercise 18H}
1 i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-085}
\end{center}

ii

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-085(1)}
\end{center}

iii

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-085(6)}
iv

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-086}
\end{center}

a i increasing

ii decreasing

iii increasing

iv decreasing

b i weak

ii strong

iii moderate

iv weak

3 a \(161 \mathrm{~cm}\), interpolation

b \(185 \mathrm{~cm}\), interpolation

c \(195 \mathrm{~cm}\), extrapolation

4 a number of seeds \(=40 \times(\) weight \(\mathrm{kg})+260\)

b 468 seeds

c \(8.5 \mathrm{~kg}\)

5 a Using points \((4,69)\) and \((8,59)\) we get the equation,

time \(=-\frac{5}{2} \times(\) fitness level \()+79\)

b 71.5 seconds

c 7 (rounded up from 6.8)

d Interpolation. A fitness level of 3 and time of 62 seconds is within the data range already obtained.

6 a Too few points to judge if there is a relationship between variables. If there is, it appears weak at best, so using the line of best fit to make predictions would be unreliable.

b The form of the scatterplot is not linear. Hence, using a linear equation to make predictions would be inappropriate.

\(7 \$ 4300300\), extrapolation

\section*{Review exercise}
1 a 6.7 seconds

b 14.3 seconds

c 2.4 seconds

d no
2 a Grade A; 29 hours, Grade B; 26 hours

b Grade A; 4 hours, Grade B; 8 hours

c Grade A; 2 hours, Grade B; 4 hours

d Grade A; \(20(50 \%\) of 40\()\), Grade B; \(10(25 \%\) of 40\()\)

e Grade A; symmetrical, Grade B; positively skewed

f In samples of 40 batteries, a Grade B battery lasted longer than any other (32 hours)

3 a See below on this page

b See below on this page

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-086(2)}
\end{center}

d The mean would increase by one (from 2.6 to 3.6), but the standard deviation would remain the same (1.93).

5 Kathryn performed better in history, relative to the rest of the class, since she achieved a score that was a full standard deviation above the mean, whereas in mathematics her score was \(\frac{2}{3}\) of a standard deviation above the mean.

6 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-086(1)}
\end{center}

b In the first quarter (Jan-March)

c Sales figures are seasonally changing, with greatest sales in and around the summer period, and least sales in the winter to early spring period (July-September). Nevertheless, it can be observed that there is a general upwards trend in sales.

7 a height \((\mathrm{cm})=2.5 \times\) tibia \((\mathrm{cm})+72.5\) (Determined using points \((45,185)\) and \((31,150)\).)

b \(182.5 \mathrm{~cm}\)

c \(29 \mathrm{~cm}\)

d The answer to (b) is interpolation since a tibia length of \(44 \mathrm{~cm}\) is within the data range already collected. The answer to (c) is extrapolation since a height of \(145 \mathrm{~cm}\) is outside the data range already collected.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-086(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-086(3)}
\end{center}

\[
Q_{1}=100 \text { median }=106 \quad Q_{3}=117.5
\]

\section*{Chapter 19 answers}
\section*{Exercise 19A}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-087(6)}
\end{center}

\section*{5}
a \(1,0,0\)

c \(-1,0,0\)

e \(1,0,0\)

6 a \(\frac{1}{\sqrt{2}}\)

d -1

7 a \(\frac{1}{4}\)

e \(3 \quad \mathbf{3} \frac{3}{4}\)

e second

b 0,1

d \(0,-1\)

f \(\frac{1}{0}\) is not defined

b \(-\frac{1}{2}\)

c 0

e \(-\frac{1}{\sqrt{2}}\)

f \(-\frac{3 \sqrt{3}}{2}-1\)
b \(\frac{3}{4}\)
c \(\frac{1}{3}\)
d \(\frac{3}{4}\)

g 1

h 1

8 a \(\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}=\frac{a^{2}+b^{2}}{c^{2}}=1\), by Pythagoras' theorem

b \(\cos ^{2} \theta=\left(\frac{-b}{c}\right)^{2}=\frac{b^{2}}{c^{2}}\), the rest remains the same

c it is still true
9 a \(\frac{2}{\sqrt{3}}\)
b 1
c \(\frac{2}{\sqrt{3}}\)
d -2
e -2
f \(\frac{1}{\sqrt{3}}\)
g \(-\frac{1}{\sqrt{3}}\)
h \(\frac{2}{\sqrt{3}}\)

10 a From question \(8, \sin ^{2} \theta+\cos ^{2} \theta=1\), now divide both sides of the equation by \(\cos ^{2} \theta\).

b The identity holds for all \(\theta\) between \(0^{\circ}\) and \(360^{\circ}\) except when \(\cos \theta=0\). That is, when \(\theta=90^{\circ}\) and \(270^{\circ}\).

\section*{Exercise 19B}
1 a \(30^{\circ}, 150^{\circ}\)
b \(60^{\circ}, 240^{\circ}\)
c \(45^{\circ}, 315^{\circ}\)
d \(120^{\circ}, 240^{\circ}\)
e \(240^{\circ}, 300^{\circ}\)
f \(45^{\circ}, 225^{\circ}\)
2 a \(225^{\circ}, 315^{\circ}\)
b \(150^{\circ}, 330^{\circ}\)
c \(150^{\circ}, 210^{\circ}\)
d \(90^{\circ}\)
e \(90^{\circ}, 270^{\circ}\)
f \(0^{\circ}, 180^{\circ}, 360^{\circ}\)
3
a \(10^{\circ}, 170^{\circ}\)
b \(155^{\circ}, 205^{\circ}\)
c \(65^{\circ}, 245^{\circ}\)
d \(233^{\circ}, 307^{\circ}\)
e \(36^{\circ}, 324^{\circ}\)
f \(106^{\circ}, 286^{\circ}\)
g \(161^{\circ}, 199^{\circ}\)
h \(23^{\circ}, 203^{\circ}\)
i \(258^{\circ}, 282^{\circ}\)

\section*{Exercise 19C}
1 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-087(2)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-087}
\end{center}

2 a
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-087(1)}

3 a \(30^{\circ}\)

b \(0^{\circ}\)

b \(30^{\circ}\)

4 a \(30^{\circ}\)

b -1

5 a 0

e \(-\frac{1}{2}\)

f \(-\sqrt{3}\)

6 a \(-\frac{\sqrt{3}}{2}\)

d \(-\frac{1}{2}\)

7 a 0

b 1

e 1

f 0

b \(-\frac{1}{\sqrt{2}}\)

c \(0^{\circ}\)

d \(40^{\circ}\)

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-087(3)}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-087(5)}
\end{center}

\(-0^{\circ}=\)

\(\begin{array}{ll}\text { c } 0^{\circ} & \text { d } 0^{\circ}\end{array}\)

c \(0 \quad\) d \(\frac{1}{2}\)

g \(\frac{1}{2} \quad \mathbf{h} \frac{1}{\sqrt{2}}\)

\section*{Exercise 19D}
1

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-087(4)}
\end{center}

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\(\theta\) & \(0^{\circ}\) & \(10^{\circ}\) & \(20^{\circ}\) & \(30^{\circ}\) & \(40^{\circ}\) & \(50^{\circ}\) & \(60^{\circ}\) & \(70^{\circ}\) & \(80^{\circ}\) & \(90^{\circ}\) \\
\hline
\(\sin \theta\) & 0.00 & 0.17 & 0.34 & 0.50 & 0.64 & 0.77 & 0.87 & 0.94 & 0.98 & 1.00 \\
\hline
\end{tabular}
\end{center}

2

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-088(5)}
\end{center}

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\(\theta\) & \(0^{\circ}\) & \(10^{\circ}\) & \(20^{\circ}\) & \(30^{\circ}\) & \(40^{\circ}\) & \(50^{\circ}\) & \(60^{\circ}\) & \(70^{\circ}\) & \(80^{\circ}\) & \(90^{\circ}\) \\
\hline
\(\cos \theta\) & 1.00 & 0.98 & 0.94 & 0.87 & 0.77 & 0.64 & 0.50 & 0.34 & 0.17 & 0.00 \\
\hline
\end{tabular}
\end{center}

3
a i 0.5
ii -0.9
iii 0.9
iv 0.95
v -0.85
vi 0.40
vii -0.8
viii 0.55
b i \(30^{\circ}, 150^{\circ}\)
ii \(120^{\circ}, 240^{\circ}\)
iii \(65^{\circ}, 115^{\circ}\)
iv \(55^{\circ}, 305^{\circ}\)
v \(55^{\circ}, 125^{\circ}\)
vi \(145^{\circ}, 215^{\circ}\)
vii \(205^{\circ}, 335^{\circ}\)
viii \(105^{\circ}, 255^{\circ}\)
c \(45^{\circ}\) and \(225^{\circ}\)

4 a \(1,-1\)

b maximum: \(90^{\circ}\) (plus multiples of \(360^{\circ}\) ); minimum: \(270^{\circ}\) (plus multiples of \(360^{\circ}\) )

5 a \(1,-1\)

b maximum: \(0^{\circ}\) (plus multiples of \(360^{\circ}\) ); minimum: \(180^{\circ}\) (plus multiples of \(360^{\circ}\) )

6 a
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-088(4)}

b

\(\left(\cos \left(180^{\circ}+\theta\right), \sin \left(180^{\circ}+\theta\right)\right)\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-088}
\end{center}

\(\cos \left(180^{\circ}+\theta\right)=-\cos \theta\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-088(7)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-088(1)}
\end{center}

\(\cos \left(360^{\circ}-\theta\right)=\cos \theta\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-088(2)}
\end{center}

7 a,b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-088(3)}
\end{center}

\(\cos (-\theta)=\cos \theta\)

\(\sin (-\theta)=-\sin \theta\)

8

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\(\theta\) & \(0^{\circ}\) & \(15^{\circ}\) & \(30^{\circ}\) & \(45^{\circ}\) & \(60^{\circ}\) & \(75^{\circ}\) & \(90^{\circ}\) & \(105^{\circ}\) & \(120^{\circ}\) & \(135^{\circ}\) & \(150^{\circ}\) & \(165^{\circ}\) & \(180^{\circ}\) \\
\hline
\(3 \sin (2 \theta)\) & 0 & 1.5 & 2.6 & 3 & 2.6 & 1.5 & 0 & -1.5 & -2.6 & -3 & -2.6 & -1.5 & 0 \\
\hline
\(\theta\) &  & \(195^{\circ}\) & \(210^{\circ}\) & \(225^{\circ}\) & \(240^{\circ}\) & \(255^{\circ}\) & \(270^{\circ}\) & \(285^{\circ}\) & \(300^{\circ}\) & \(315^{\circ}\) & \(330^{\circ}\) & \(345^{\circ}\) & \(360^{\circ}\) \\
\hline
\(3 \sin (2 \theta)\) &  & 1.5 & 2.6 & 3 & 2.6 & 1.5 & 0 & -1.5 & -2.6 & -3 & -2.6 & -1.5 & 0 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-088(6)}
\end{center}

ICE-EM MATHEMATICS YEAR 10

ICE-EM Mathematics 10 3ed ISBN 978-1-108-40434-1 (C) The University of Melbourne / AMSI 2017 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party Updated June 2022

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\(\theta\) & \(0^{\circ}\) & \(15^{\circ}\) & \(30^{\circ}\) & \(45^{\circ}\) & \(60^{\circ}\) & \(75^{\circ}\) & \(90^{\circ}\) & \(105^{\circ}\) & \(120^{\circ}\) & \(135^{\circ}\) & \(150^{\circ}\) & \(165^{\circ}\) & \(180^{\circ}\) \\
\hline
\(4 \cos (2 \theta)\) & 4 & 3.5 & 2 & 0 & -2 & -3.5 & -4 & \(-3.5^{\circ}\) & -2 & 0 & 2 & 3.5 & 4 \\
\hline
\(\theta\) &  & \(195^{\circ}\) & \(210^{\circ}\) & \(225^{\circ}\) & \(240^{\circ}\) & \(255^{\circ}\) & \(270^{\circ}\) & \(285^{\circ}\) & \(300^{\circ}\) & \(315^{\circ}\) & \(330^{\circ}\) & \(345^{\circ}\) & \(360^{\circ}\) \\
\hline
\(4 \cos (2 \theta)\) &  & 3.5 & 2 & 0 & -2 & -3.5 & -4 & -3.5 & -2 & 0 & 2 & 3.5 & 4 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-089}
\end{center}

\section*{Exercise 19E}
1 a \(30^{\circ}, 150^{\circ}\)
b \(60^{\circ}, 120^{\circ}\)
c \(240^{\circ}, 300^{\circ}\)
d \(60^{\circ}, 300^{\circ}\)
e \(45^{\circ}, 225^{\circ}\)
f \(150^{\circ}, 330^{\circ}\)
g \(150^{\circ}, 210^{\circ}\)
h \(60^{\circ}, 240^{\circ}\)
2 a \(36^{\circ}, 144^{\circ}\)
b \(56^{\circ}, 304^{\circ}\)
c \(63^{\circ}, 117^{\circ}\)
d \(228^{\circ}, 312^{\circ}\)
e \(16^{\circ}, 344^{\circ}\)
f \(256^{\circ}, 284^{\circ}\)
3 a \(60^{\circ}, 120^{\circ}, 240^{\circ}, 300^{\circ}\)
b \(45^{\circ}, 135^{\circ}, 225^{\circ}, 315^{\circ}\)
c \(60^{\circ}, 120^{\circ}, 240^{\circ}, 300^{\circ}\)
d \(45^{\circ}, 135^{\circ}, 225^{\circ}, 315^{\circ}\)
e \(30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}\)
f \(30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}\)
4 a \(45^{\circ}, 225^{\circ}\)
b \(60^{\circ}, 240^{\circ}\)

\section*{Review exercise}
1 a \(35^{\circ}\)
b \(30^{\circ}\)
c \(50^{\circ}\)
d \(20^{\circ}\)
e \(70^{\circ}\)
f \(60^{\circ}\)
g \(60^{\circ}\)
h \(60^{\circ}\)
2 a \(\frac{1}{2},-\frac{\sqrt{3}}{2},-\frac{1}{\sqrt{3}}\)
b \(\frac{\sqrt{3}}{2},-\frac{1}{2},-\sqrt{3}\)
c \(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-1\)
d \(-\frac{\sqrt{3}}{2}, \frac{1}{2},-\sqrt{3}\)
e \(-\frac{1}{2},-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{3}}\)
f \(-\frac{1}{2}, \frac{\sqrt{3}}{2},-\frac{1}{\sqrt{3}}\)
g \(-\frac{\sqrt{3}}{2},-\frac{1}{2}, \sqrt{3}\)
h \(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},-1\)
3 a \(\frac{\sqrt{3}}{2}\)
b 1
d 1
e \(-\frac{1}{2}\)
c \(-\frac{1}{2}\)
f -1

4

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-089(4)}
\end{center}

5

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-089(1)}
\end{center}

6
a \(120^{\circ}, 240^{\circ}\)
b \(45^{\circ}, 135^{\circ}\)
c \(45^{\circ}, 225^{\circ}\)
d \(120^{\circ}, 300^{\circ}\)
e \(60^{\circ}, 120^{\circ}\)
f \(30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}\)
7 a \(-\frac{1}{\sqrt{2}}\)
b \(-\frac{1}{2}\)
c \(-\frac{1}{\sqrt{3}}\)
d \(\frac{1}{2}\)
e \(-\frac{1}{2}\)
f \(-\frac{1}{\sqrt{3}}\)
8 a \(35^{\circ}, 145^{\circ} \quad\) b \(126^{\circ}, 234^{\circ}\)
c \(65^{\circ}, 245^{\circ}\)
d \(235^{\circ}, 305^{\circ}\)
9 a \(60^{\circ}, 300^{\circ}\)
b \(120^{\circ}, 240^{\circ}\)
c \(60^{\circ}, 120^{\circ}\)
d \(240^{\circ}, 300^{\circ}\)
e \(270^{\circ}\)
f \(0^{\circ}, 360^{\circ}\)

\section*{Challenge exercise}
\(1 \frac{3}{4}\)

3

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-089(5)}
\end{center}

\(40^{\circ}, 180^{\circ}, 360^{\circ}\)

5

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-089(2)}
\end{center}

6

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-089(3)}
\end{center}

7 a \(60^{\circ}, 300^{\circ} \quad\) b \(30^{\circ}, 150^{\circ} \quad\) c \(30^{\circ}, 150^{\circ}, 270^{\circ}\)

8 b Area \(=\frac{1}{2} a b \sin (\alpha+\beta)=\frac{1}{2} y a \sin \alpha+\frac{1}{2} y b \sin \beta\)

\[
=\frac{1}{2}(a b \sin \alpha \cos \beta+a b \sin \beta \cos \alpha) \text {. }
\]

Thus \(\sin (\alpha+\beta)=\sin \alpha \cos \beta+\sin \beta \cos \alpha\)

c Use \(\alpha=45^{\circ}\) and \(\beta=30^{\circ}: \sin 75^{\circ}=\frac{\sqrt{2}+\sqrt{6}}{4}\)

\(9 \begin{array}{ll}\text { a } 2 \sin \theta \cos \theta & \text { b } \angle C G D=\theta+\theta=2 \theta\end{array}\)

c \(2 \sin \theta \cos \theta=\) area \(A B C=\) area \(B C D=\frac{1}{2} B C \times D E=\sin 2 \theta\)

\section*{Chapter 20 answers}
\section*{Exercise 20A}
1 a all real numbers

b all real numbers

c \(x \neq 0\)

d \(x \neq 2\)

e \(x \neq-4\)

f \(x \neq 2\)

g \(x \neq 2\) and \(x \neq-2\)

h \(x \neq 3\) and \(x \neq-3\)

2
a \(x \geq 0\)
b \(x \geq-7\)
c \(x \leq 7\)
d \(x \geq \frac{1}{7}\)
e \(x>0\)
f \(x>7\)

3 a all real numbers

b all real numbers

c \(x>0\)

d \(x>2\)

e \(x<0\)

f all real numbers

4 a function

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(7)}
\end{center}

b not a function

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(2)}
\end{center}

c function

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(10)}
\end{center}

d function

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(5)}
\end{center}

e not a function

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(4)}
\end{center}

f function

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(3)}
\end{center}

g function

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(8)}
\end{center}

\(\mathbf{h}\) not a function

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(9)}
\end{center}

i not a function

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(6)}
\end{center}

5 a \(y=2 x\) or \(y=-2 x\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090}
\end{center}

c no

d not a function

6 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-090(11)}
\end{center}

b
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-090(1)}

domain: \(x \geq 0\)

Exercise 20B

1 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-091(1)}
\end{center}

inverse: \(y=x-4\)

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-091(2)}
\end{center}

inverse: \(y=\frac{x-2}{2}\)

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-091(6)}
\end{center}

inverse: \(y=\frac{x+1}{2}\)

d, e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-091(7)}
\end{center}

\(y=\frac{x-2}{3}\) is the inverse of \(y=3 x+2\)

and \(y=3 x+2\) is the inverse of \(y=\frac{x-2}{3}\) f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-091(5)}
\end{center}

inverse: \(y=\frac{3 x+4}{2}\)

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-091(4)}
\end{center}

inverse: \(y=\frac{x}{5}\)

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-091(3)}
\end{center}

inverse: \(y=3 x\)

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-091}
\end{center}

ICE-EM Mathematics 10 3ed

ANSWERS TO EXERCISES Photocopying is restricted under law and this material must not be transferred to another party
j

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-092(5)}
\end{center}

k

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-092(1)}
\end{center}

I

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-092(4)}
\end{center}

2

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-092(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-092(7)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-092}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-092(2)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-092(3)}
\end{center}

ICE-EM MATHEMATICS YEAR 10 ICE-EM Mathematics 10 3ed ISBN 978-1-108-40434-1 C) The University of Melbourne / AMSI 2017
Photocopying is restricted under law and this material must not be transferred to another party
f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(3)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(8)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093}
\end{center}

Exercise 20C
1 a 3
b -17
c 0
d -9
e 204
f -53
2 a 6
b 2
c 11
d \(\frac{9}{4}\)
e 4
f 504
3 a 1
b 0
c -6
d \(\frac{3}{2}\)
e 3
f \(\frac{1}{3}\)
4 a \(\frac{1}{6}\)
b \(\frac{2}{5}\)
c -2
5 a \(x=0\) or \(x=4\)
b \(x=2\)
c \(x=-1\) or \(x=5\)
d no values of \(x\)
e \(x=2-\sqrt{5}\) or \(x=2+\sqrt{5}\)
f \(x=1\) or \(x=3\)

6 a \(a^{2}-4\)

b \(y^{2}+4 y\)

c \(4 b^{2}-4\)

d \(9 c^{2}+6 c-3\)

e \(x^{4}-4\)

f \(x^{6}-4\)

7 a true

b false

c false

d true

e true

f true

\begin{center}
\begin{tabular}{lll}
8 a false & b false & c true \\
d false & e false & f true \\
\end{tabular}
\end{center}

9 a all real numbers, \(y \leq 3\)

b \(x \neq 0, y \neq 0\)

c all real numbers, \(y>4\)

d \(-3 \leq x \leq 3,0 \leq y \leq 3\)

e all real numbers, \(y \leq 6\)

f all real numbers, \(y \geq 4\)

g all real numbers, \(y>-3\)

h all real numbers, \(y>7\)

i \(-5 \leq x \leq 5,-5 \leq y \leq 0\)

j all real numbers, all real numbers

k \(x \neq 0, y \neq 0\)

l \(x<7\), all real numbers

\(\mathbf{m}\) all real numbers, \(-1 \leq y \leq 1\)

n \(x \neq \ldots,-\frac{3 \pi}{2},-\frac{\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2}, \ldots\), all real numbers

\section*{Exercise 20D}
a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(10)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(2)}
\end{center}

2 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(5)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(7)}
\end{center}

Indeed in this case, \(-f(-x)=f(x)\) b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(4)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(9)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(6)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-093(1)}
\end{center}

\(y=-f(-x)\)

\[
c(x)
\]

a domain: all real numbers

range: \(y \geq 5\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(10)}
\end{center}

b domain: all real numbers range: \(y \geq 0\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(8)}
\end{center}

c domain: all real numbers range: \(y \geq 0\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(4)}
\end{center}

d domain: all real numbers

range: \(y>0\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(9)}
\end{center}

e domain: all real numbers range: \(y>1\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(3)}
\end{center}

f domain: all real numbers range: \(y>-4\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(2)}
\end{center}

g domain: \(x>0\) range: all real numbers

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(7)}
\end{center}

h domain: \(x>4\)

range: all real numbers

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(1)}
\end{center}

i domain: \(\mathrm{x}<0\)

range: all real numbers

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(11)}
\end{center}

4 a domain: all real numbers range: \(y \geq 2\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(6)}
\end{center}

b domain: all real numbers

range: \(y \geq 4\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094(5)}
\end{center}

c domain: \(x \geq 0\)

range: \(y \geq 0\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-094}
\end{center}

ICE-EM MATHEMATICS YEAR 10 ICE-EM MathemaTHEMAT Photocopying is restricted und
d domain: \(x \geq 0\)

range: \(y \geq 2\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095}
\end{center}

e domain: \(x \geq 2\)

range: \(y \leq 0\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095(7)}
\end{center}

f domain: \(x \geq-2\)

range: \(y \leq 2\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095(1)}
\end{center}

5

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095(5)}
\end{center}

6

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095(4)}
\end{center}

7

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095(10)}
\end{center}

\section*{Exercise 20E}
1 a 3

b 5

c 10

d \(a+3\)

e \(g(f(x))=x+3\)

The result is a translation of 3 units to the right.

2 a 0

b -6

c -4

d -2

e 3

f -4

g \(x^{2}-6\)

h \(x^{2}-4 x\)

i \(x-4\)

j \(x^{4}-8 x^{2}+12\)

k \(f(g(x)) \neq g(f(x))\)

3 a 2
b 2
c 4
e \(x\)
f \(x\)

d 4

g \(f(g(x))=x\) and \(g(f(x))=x, f\) and \(g\) are inverses of each other.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095(3)}
\end{center}

d 4

f \(x\)

g \(f(g(x))=x\) and \(g(f(x))=x, f\) and \(g\) are inverses of each other.

5 a \(g(x)=x-5\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095(6)}
\end{center}

b \(g(x)=\frac{x+2}{3}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095(2)}
\end{center}

c \(g(x)=\frac{x-2}{3}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-095(9)}
\end{center}

d \(g(x)=\frac{4-x}{3}\)

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-095(8)}
e \(g(x)=6-2 x\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-096(2)}
\end{center}

6 a \(g(x)=\sqrt[3]{x+2}\)

b \(g(x)=\sqrt[3]{2-x}\)

c \(g(x)=\sqrt[5]{\frac{x}{32}}\)

7 a domain: \(x \neq 0, g(x)=\frac{1}{x-1}, x \neq 1\)

b domain: \(x \neq-1, g(x)=\frac{1}{x}-1, x \neq 0\)

c domain: \(x \neq 2, g(x)=\frac{2(x+1)}{x-1}, x \neq 1\)

d domain: \(x \neq-2, g(x)=\frac{2 x}{3-x}, x \neq 3\)

8 a \(f(f(x))=f(5-x)=5-(5-x)=x\)

b \(f(f(x))=f(-x)=x\)

c \(f(f(x))=f\left(-\frac{1}{x}\right)=x\)

d \(f(f(x))=f\left(\frac{6}{x}\right)=x\)

e \(f(f(x))=f\left(\frac{2 x-2}{x-2}\right)=x\)

f \(f(f(x))=f\left(\frac{-3 x-5}{x+3}\right)=x\)

9 a domain: all real numbers; \(g(x)=\frac{x}{3}\); domain of \(g(x)\) : all real numbers b domain: all real numbers; \(g(x)=\frac{1}{3} \log _{2} x\); domain of
\(g(x): x>0\)

c domain: all real numbers; \(g(x)=\log _{7} \frac{x}{5}\); domain of \(g(x): x>0\)

d domain: \(x>0 ; g(x)=5^{x}\); domain of \(g(x)\) : all real numbers

e domain: \(x>0 ; g(x)=\frac{1}{3} \times 4^{\frac{x}{2}}\); domain of \(g(x)\) : all real numbers

f domain: \(x>3 ; g(x)=2^{x}+3\); domain of \(g(x)\) : all real numbers

g domain: all real numbers; \(g(x)=\log _{5}(x)+1\); domain of \(g(x): x>0\)

h domain: \(x>0 ; g(x)=4^{x-4}\); domain of \(g(x)\) : all real numbers i domain: all real numbers; \(g(x)=\frac{1}{3} \log _{5}(x-5)\); domain
of \(g(x): x>5\)

\(10 y=\sqrt{49-x^{2}},-7 \leq x \leq 0 ; y=\sqrt{49-x^{2}}, 0 \leq x \leq 7\);

\(y=-\sqrt{49-x^{2}},-7 \leq x \leq 0 ; y=-\sqrt{49-x^{2}}, 0 \leq x \leq 7\)

\section*{Review exercise}
1 a all real numbers

b \(x \neq 0\)

c \(x \neq 5\)

d \(x \neq-8\)

e \(x \geq 2\)

f all real numbers

g \(x \neq-5\)

h \(x \geq-6\)

2 a -4

b -3

c -3

d 12

e \(a^{2}-4\)

f \(a^{2}-4\)

g \(4\left(a^{2}-1\right)\)

h \(a^{2}-4 a\)

3 a 3

b 1 c 5
d 11
e \(3-2 a\)
f \(3+2 a\)
g \(3-4 a\)
h \(7-2 a\)

4 a domain: all real numbers; range: all real numbers

b domain: all real numbers; range: \(y \leq 4\)

c domain: \(x \neq-6\); range: \(y \neq 0\)

5 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-096}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-096(3)}
\end{center}

c

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-096(1)}
d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-097(5)}
\end{center}

6

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-097(1)}
\end{center}

7 a 4

b 11

c -20
d 3
e 31
f 4
g \(11-2 x^{2}\)
h \(-4 x^{2}-4 x+4\)
i \(4 x+3\)

j \(-x^{4}+10 x^{2}-20\)

k \(f(g(x)) \neq g(f(x))\)

8
a \(y=\frac{x+4}{3}\)
b \(y=\frac{2-x}{3}\)
c \(y=\sqrt[3]{x-2}\)
d \(y=\frac{1}{x}-2\)

\section*{Challenge exercise}
1 a \(f(a+b)=2(a+b)=2 a+2 b=f(a)+f(b)\)

\(f(k a)=2 k a=k f(a)\)

b Assume \(f(a+b)=f(a)+f(b)\)

Then \(a+b+2=a+2+b+2=a+b+4\), which is a contradiction.

Assume \(f(k a)=k f(a)\).

Then, \(k a+2=k a+2 k\).

Thus, \(k=1\).

2 a \(f(x+y)=2^{x+y}=2^{x} \times 2^{y}=f(x) f(y)\)

b \(x+y=x y \cdot x=2\) and \(y=2\), and \(x=y=0\)

3 a \(f(x)=x^{2}\) is even and \(f(x)=x^{3}\) is odd.

b Let \(f(x)\) and \(g(x)\) be even functions. The sum function \((f+g)(x)=f(x)+g(x)\) \((f+g)(-x)=f(-x)+g(-x)=f(x)+g(x)=(f+g)(x)\).

c Let \(f(x)\) and \(g(x)\) be even functions. The product function \((f g)(x)=f(x) g(x)\)

\((f g)(-x)=f(-x) \times g(-x)=f(x) g(x)=(f g)(x)\) d Let \(f(x)\) and \(g(x)\) be odd functions. The product function \((f g)(x)=f(x) g(x)\)

\((f g)(-x)=f(-x) \times g(-x)=-f(x) \times(-g(x))\)

\(=f(x) g(x)=(f g)(x)\)

e Let \(f(x)\) and \(g(x)\) be odd functions. Then \(f(g(-x))=f(-g(x))=-f(g(x))\)

\section*{Chapter 21 answers}
\section*{21A Review}
\section*{Chapter 11: Circles, hyperbolas and simultaneous equations}
\(1 \mathbf{a}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-097(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-097}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-097(2)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-097(3)}
\end{center}

2 a \((x-3)^{2}+y^{2}=16\)

b \((x+1)^{2}+(y-2)^{2}=3\)

3 a \((x-2)^{2}+(y+3)^{2}=4\), centre \((2,-3)\), radius 2

b \((x+1)^{2}+(y+4)^{2}=16\), centre \((-1,-4)\), radius 4

4

a asymptotes: \(x=0\) and \(y=0\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-098(5)}
\end{center}

b asymptotes: \(x=0, y=2 ; x\)-intercept \(x=\frac{1}{2}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-098(1)}
\end{center}

c asymptotes: \(y=0, x=2, x=2 ; y\)-intercept \(=-\frac{3}{2}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-098(9)}
\end{center}

d asymptotes: \(x=-3\) and \(y=-2\) \(x\)-intercept \(=-\frac{5}{2}, y\)-intercept \(=-\frac{5}{3}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-098(8)}
\end{center}

5
a \((3,12),(-2,-3)\)
b \((-2,-1),\left(\frac{3}{2}, 6\right)\)
c \(\left(-\frac{3}{2},-2\right),(1,3)\)
d \((-3,-2),\left(\frac{2}{3}, 9\right)\)

6 a \((\sqrt{5}, 2),(-\sqrt{5}, 2)\)

b \((1, \sqrt{3}),(1,-\sqrt{3})\)

c \((0,2),(-2,0)\)

d \((0,4),\left(\frac{8 \sqrt{2}}{3},-\frac{4}{3}\right)\)

\(7 \quad\left(-\frac{6}{5}, \frac{17}{5}\right),(2,-3)\)

\(8(-4,3),(6,8)\)
9 a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-098(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-098}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-098(2)}
\end{center}

d
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-098(6)}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-098(7)}
\end{center}

10 a

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-098(4)}
b
\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-099(3)}

\section*{Chapter 12: Further trigonometry}
1 a \(\theta \approx 43.2^{\circ}\)
b \(x \approx 2.9\)
c \(x \approx 4.9\)
2 a \(2.3 \mathrm{~m}\)
b \(1.9 \mathrm{~m}\)
\(363.4^{\circ}\)
\(4 \quad\) a \(13.5 \mathrm{~km}\)
b \(132.0^{\circ} \mathrm{T}\)
\(5 \quad 281.2^{\circ} \mathrm{T}\)
\(652.5 \mathrm{~m}\)
7 a \(4.619 \mathrm{~m}\)
b \(18.475 \mathrm{~m}\)
c \(4.619 \mathrm{~m}\)
8 a \(x=\sqrt{2}\)
b \(x=5(3+\sqrt{3})\)
\(9 x=50(\sqrt{3}-1)\)
\(10 \frac{20(3-\sqrt{3})}{3} \mathrm{~cm}\)
\(11 \quad 12 \mathrm{~km}\)
12 a \(B \approx 102.6^{\circ}, C \approx 27.4^{\circ}, A C \approx 6.4\)
b \(A \approx 71.2^{\circ}, B \approx 38.8^{\circ}, B C \approx 6.0\)
c \(C=35^{\circ}, A B \approx 1.3, A C \approx 2.17\)
\(13 \quad 14.8 \mathrm{~km}\)
\(14 \quad 34.03 \mathrm{~m}\)
\(15100(\sqrt{3}+1) \mathrm{m}\)
16 a \(24.8^{\circ}\)
b \(56.9^{\circ}\)
17 a \(36.37 \mathrm{~cm}\)
b \(50.8^{\circ}\)
c \(46.96 \mathrm{~cm}\)
\(1810.0^{\circ}\)
19 a \(45^{\circ}\)
b \(32.0^{\circ}\)
20 a \(P\) at \(B\) or \(C, \angle E P H \approx 35.26^{\circ}\)
b \(P\) is midpoint of \(B C, \angle E P H \approx 38.94^{\circ}\)

\section*{Chapter 13: Circle geometry}
a \(a=60, b=120\)
b \(c=20\)
c \(h=110, j=70\)
d \(d=50, e=40, f=10, g=40\)
e \(k=25, m=65, r=40, s=50, t=65\)

\[
\text { b } 6 \sqrt{3} \mathrm{~cm}
\]

\(3 C B=C A\) (equal tangents). Hence, \(\angle C A B=\angle C B A=45^{\circ}\) (isosceles triangle), \(\angle C A D=90^{\circ}\) (tangent to circle at \(A\) ). Therefore, \(B A\) bisects \(\angle C A D\).

iii \(30^{\circ} \quad\) iv \(30^{\circ}\)

\(B E=2 \mathrm{~cm} \quad 5 \quad T S=17 \mathrm{~cm}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-099}
\end{center}

\(6 E C=10, E D=6\)

7 If \(B P \times P Q=C P \times P R\)

then \(\frac{B P}{C P}=\frac{P R}{P Q}\)

\(\angle B P R=\angle C P Q\)

(vertically opposite)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-099(1)}
\end{center}

\(\triangle R P B\) is similar to \(\triangle Q P C\) (SAS)

\(\triangle A B Q\) is similar to \(\triangle A C R\) (AAA)

so \(\frac{A R}{A C}=\frac{A Q}{A B}\)

\(A R \times A B=A C \times A Q\)

8 a \(P Y=9 \mathrm{~cm}, X Y=5 \mathrm{~cm} \quad\) b \(P T=9 \mathrm{~cm}\)

c \(P X=4 \mathrm{~cm}\)

\(9 \angle B O C=150^{\circ}\)

10 a Let \(\angle Y A L=\beta\)

Then \(\angle L A B=\beta\) (given)

\(\angle A B L=\beta\) (alternate segment theorem)

So \(\triangle A L B\) is isosceles, \(A L=B L\)

b Let \(\angle X A K=\alpha\). Then \(\angle K A B=\alpha\) (given)

Therefore, \(2 \alpha+2 \beta=180^{\circ}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-099(2)}
\end{center}

\[
\alpha+\beta=90^{\circ}
\]

\(\angle K A L\) is a right angle. Therefore, \(K L\) is a diameter (converse of Thale's theorem)

Chapter 14: Indices, exponentials and

\section*{logarithms - part 2}
\begin{center}
\begin{tabular}{llll}
\(\mathbf{1}\) & a 4 & b 4 & c 10 \\
 & d 0 & e 5 &  \\
\(\mathbf{2}\) & a -5 & b -5 & c -4 \\
 & d -2 & e -4 & f -3 \\
 & g -11 & h -5 &  \\
3 & a \(\log _{2} 75\) & b \(\log _{2} 63\) & c \(\log _{2} 33\) \\
 & d \(\log _{3} 100\) & e \(\log _{7} 40\) & f 1 \\
g -1 & h 1 & i 1 &  \\
4 &  & b \(\log _{3} 20\) &  \\
 & c \(-\log _{5} 343=-3 \log _{5} 7\) & d \(\log _{11} \frac{3}{5}\) &  \\
5 125 & b 256 & c 620 &  \\
 & d \(\frac{1027}{6}\) & e -59 & f \(\frac{10001}{2}\) \\
\end{tabular}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-100(1)}
\end{center}

Chapter 15: Probability
1 a \(\frac{1}{3}\)
b \(\frac{1}{2}\)
2 a \(\frac{1}{4}\)
b \(\frac{1}{13}\)
c \(\frac{1}{52}\)
d \(\frac{3}{13}\)
\(3 \frac{1}{200}\)
\(4 \frac{5}{16}\)
5 a \(\frac{1}{12}\)
b \(\frac{1}{36}\)
c \(\frac{13}{18}\)
6 a \(\mathbf{i} \frac{4}{25}\)
ii \(\frac{2}{15}\)
b i \(\frac{12}{25}\)
ii \(\frac{8}{15}\)
7 a \(\frac{1}{2}\)
b \(\frac{77}{200}\)
c \(\frac{40}{77}\)
d \(\frac{2}{5}\)

\(8 \quad\) a \(p=\frac{4}{15} \quad\) b \(p=\frac{1}{3}\)

\(9 \quad\) a 0.75

b They are not independent as \(P(H \cap F) \neq P(H) P(F)\)
10 a 0.5
b 0.25

\section*{Chapter 16: Direct and inverse proportion}
1 a i \(\frac{1}{2}, y=\frac{1}{2} x\)

ii \begin{tabular}{|c|c|c|c|c|}
\hline
\(x\) & 1 & 4 & 8 & 20 \\
\hline
\(y\) & \(\frac{1}{2}\) & 2 & 4 & 10 \\
\hline
\end{tabular}

b i \(2, y=2 x^{2}\)

ii

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(x\) & 1 & 3 & 5 & \(\sqrt{7}\) \\
\hline
\(y\) & 2 & 18 & 50 & 14 \\
\hline
\end{tabular}
\end{center}

c i \(5, y=\frac{5}{x}\)

ii

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(x\) & 2 & 5 & 7 & 11 \\
\hline
\(y\) & \(\frac{1}{2}\) & 1 & \(\frac{5}{7}\) & \(\frac{5}{11}\) \\
\hline
\end{tabular}
\end{center}

d i \(1, y=\frac{1}{x^{2}}\)

ii

\includegraphics[max width=\textwidth, center]{2024_02_22_b89ee4b62067800f8970g-100}
\(2 y=9 \sqrt{x}\)
a \(y=18\)
b \(x=\frac{625}{9}\)
\(3 \quad 154 \mathrm{~cm}^{2}\)
\(4 y=\frac{40}{x^{2}}\)
a \(y=\frac{40}{81}\)
b \(x=\frac{2 \sqrt{10}}{3}\)
5 a \(2, c=2 a b^{2}\)

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
\(\boldsymbol{a}\) & 5 & 3 & 6 & 3 \\
\hline
\(\boldsymbol{b}\) & 1 & 2 & 2 & 3 \\
\hline
\(\boldsymbol{c}\) & 10 & 24 & 48 & 54 \\
\hline
\end{tabular}
\end{center}

\(6 \quad a=32 \quad 7 \quad z=243\)

8 a 2250 joules

b 8 times the original energy

\section*{Chapter 17: Polynomials}
1 a 3
b 5
c 8
d 0
e 4
f \(a^{3}-2 a+4\)

\(\begin{array}{ll}\text { a } a=7 & \mathbf{b} b=2\end{array}\)

3 a \(P(x)+Q(x)=-x^{3}+3 x^{2}+7\),

\(P(x)-Q(x)=3 x^{3}-3 x^{2}+8 x+7\)

b \(P(x)+Q(x)=x^{2}-3 x\),

\(P(x)-Q(x)=-6 x^{5}-x^{2}-3 x+14\)

c \(P(x)+Q(x)=-x+2\),

\(P(x)-Q(x)=8 x^{3}-10 x^{2}-11 x+10\)

4 a \(P(x)=(x+2)(x+6)-6\)

b \(P(x)=(x+6)\left(x^{2}-12 x+60\right)-330\)

c \(P(x)=(x-1)\left(5 x^{2}-2 x-2\right)-3\)
5 a 45
b \(\frac{25}{8}\)
c \(\frac{31}{8}\)

\(6 a=-1\)

\(7 \quad \mathbf{a}(x-1)(x+2)(2 x+3)\)

b \((x-2)(x+1)(2 x+3)\)

c \((x-2)(x-1)(x+1)(2 x+3)\)

8 a \(x=1\) or \(x=-2\) or \(x=-\frac{3}{2}\)

b \(x=2\) or \(x=1\) or \(x=-1\) or \(x=-\frac{3}{2}\)

\(9 \quad\) a \(P(1)=0\) for all \(x . \quad\) b \(k=7\)

c \(x=1\) or \(x=2\) or \(x=4\)
10 a \(2+\frac{5}{x-1}\)
b \(4+\frac{2-5 x}{x^{2}+2 x}\)

\section*{Chapter 18: Statistics}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-101(5)}
\end{center}

b There is some tendency for the heart mass to increase as the body mass increases. The relationship could be described as linear and of moderate strength.

3 As the scores on Test 1 increase, the scores on Test 2 increase.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-101(4)}
\end{center}

4 a 38

b 11.5

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-101(3)}
\end{center}

5 a 19

b approximately 150

c 4 hours

d approximately 50

6 a i \(a=38, c=61, e=88\)

ii 62

b i 17

ii 63

7 a public holiday, roadworks, accident
b i 340.8
ii 341

iii 53.5

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_b89ee4b62067800f8970g-101}
\end{center}

d George may have sampled over school holiday period. Community group may have sampled for a small number of days and obtained high counts, or may be exaggerating.

\section*{Chapter 19: Trigonometric functions}
\begin{center}
\begin{tabular}{llll}
1 a second & b third & c third &  \\
d fourth & e second & f fourth &  \\
g third & h fourth & b \(-\cos 30^{\circ}\) & c \(-\tan 20^{\circ}\) \\
a \(\sin 5^{\circ}-\sin 20^{\circ}\) & e \(-\cos 20^{\circ}\) & f \(\tan 5^{\circ}\) &  \\
g \(-\sin 5^{\circ}\) & h \(\cos 10^{\circ}\) & c \(\frac{\sqrt{3}}{2}\) &  \\
3 a \(-\frac{\sqrt{2}}{2}\) & b \(-\frac{\sqrt{2}}{2}\) & f \(\frac{\sqrt{2}}{2}\) &  \\
d \(-\sqrt{3}\) & e \(-\frac{1}{2}\) &  &  \\
a & h \(-\frac{\sqrt{3}}{2}\) &  &  \\
a \(\frac{1}{2}\) & b \(\frac{1}{4}\) & c &  \\
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{lll}
a \(60^{\circ}, 300^{\circ}\) & b \(120^{\circ}, 300^{\circ}\) & c \(45^{\circ}, 135^{\circ}\) \\
d \(210^{\circ}, 330^{\circ}\) & e \(150^{\circ}, 210^{\circ}\) & f \(135^{\circ}, 315^{\circ}\) \\
\end{tabular}
\end{center}

6 a \(15.93^{\circ}, 164.07^{\circ}\)
b \(156.42^{\circ}, 203.58^{\circ}\)
c \(66.00^{\circ}, 246.00^{\circ}\)
d \(243.84^{\circ}, 296.16^{\circ}\)
e \(45.49^{\circ}, 314.51^{\circ}\)
f \(111.12^{\circ}, 291.12^{\circ}\)
7 a \(\frac{1}{2}\)
b \(\frac{\sqrt{2}}{2}\)
c 1
d \(-\frac{\sqrt{3}}{2}\)
e \(\frac{\sqrt{3}}{2}\)
f \(-\frac{\sqrt{2}}{2}\)
8 a \(60^{\circ}, 300^{\circ}\)
b \(150^{\circ}, 210^{\circ}\)
c \(240^{\circ}, 300^{\circ}\)
d \(120^{\circ}, 240^{\circ}\)
e \(45^{\circ}, 225^{\circ}\)
f \(30^{\circ}, 210^{\circ}\)

\section*{Chapter 20: Functions and inverse functions}
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6 a domain: all real numbers, space and range: \(y \geq-3\)

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b domain: all real numbers, space and range: \(y \leq 6\)

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c domain: \(x>-3\)

range: all real numbers

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d domain: all real numbers

range: \(y>6\)

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e domain: all real numbers

range: \(y<6\)

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f domain: \(-4 \leq x \leq 4\)

range: \(0 \leq y \leq 4\)

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7

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8 a 1

b -1

c \(g(f(x))=2 x^{2}-3\) d \(f(g(x))=(2 x-3)^{2}\)
9 The domian for \(\mathbf{a}, \mathbf{b}, \mathbf{d}, \mathbf{e}\) and \(\mathbf{f}\) is all real numbers. The domain for \(\mathbf{c}\) is the real numbers \(>-3\).
a \(g(x)=\frac{x+3}{2}\)
b \(g(x)=2 x+1\)
c \(g(x)=\log _{2}(x+3)\)
d \(g(x)=3^{x}-1\)
e \(g(x)=\sqrt[3]{8-x}\)
f \(g(x)=\sqrt[3]{8+x}\)

\section*{21B Problem-solving}
\(1 \quad\) a \(10.2 \mathrm{~km}\)
b \(295.5^{\circ}\)
2 a i \(4 \sqrt{3} \mathrm{~m}\)
ii \(53^{\circ}\)

b \((5 \sqrt{3}+12) \mathrm{m}\)

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b

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c \(3.335 \mathrm{~km}\)
d \(3.205 \mathrm{~km}\)
4 a i \(60^{\circ}\) ii \(30^{\circ}\) iii \(30^{\circ}\) iv \(90^{\circ}\)
b i \(10 \mathrm{~km} \quad\) ii \(5 \mathrm{~km}\)
c \(\angle P A Y=\angle S X Y=90^{\circ}, \angle A P Y=\angle X S Y=60^{\circ}\), \(\angle A Y P=\angle S Y X=30^{\circ}\), so \(\triangle P A Y\) is similar to \(\triangle S X Y\) (AAA)

d \(2.5 \mathrm{~km}\)
e i \(\frac{15 \sqrt{3}}{2} \mathrm{~km}\)
ii \(\frac{30-15 \sqrt{3}}{2} \mathrm{~km}\)
f \(7.765 \mathrm{~km}\)

5 a \(3,4,5\) triangle \(\quad\) b \(\angle A P B\)

c \(A P=A Q\) (radii of smaller circle), \(B P=B Q\) (radii of larger circle), \(A B=A B\) (common), so \(\triangle A P E=\triangle A Q B(\mathrm{SSS})\).

d i \(53^{\circ}\) ii \(106^{\circ}\)

e \(53^{\circ}\) ( \(\angle P F Q\) is angle on circumference standing on the same arc as \(\angle P A Q\), the angle at the centre of circle)

6 a \(\angle C E P=\angle D E Q\) (common), \(\angle E C P=\angle E D Q=90^{\circ}\) (tangent \(\perp\) radius), so \(\triangle C P E\) is similar to \(\triangle D Q E\) (AAA).

b \(\angle G F P=\angle G H Q=90^{\circ}\) (tangent \(\perp\) radius), \(\angle F G P=\angle H G Q\) (vertically opposite angles at \(G\) ), so \(\triangle G F P\) is similar to \(\triangle G H Q\) (AAA).
c From \(\triangle C P E\) and \(\triangle D Q E, \frac{C E}{D E}=\frac{C P}{D Q}\)

From \(\triangle G F P\) and \(\triangle G H Q, \frac{F G}{G H}=\frac{P F}{H Q}=\frac{C P}{D Q}\) (radii), so

\[
\frac{C E}{D E}=\frac{F G}{G H} \text { so } \frac{C E}{F G}=\frac{D E}{G H}
\]

7 a deviation of \(4.30^{\circ}\)

b \(\sqrt{\left(275-220 \cos 10^{\circ}\right)^{2}+\left(220 \sin 10^{\circ}+75 \sqrt{3}\right)^{2}}\) \(\approx 177.94 \mathrm{~m}\)

\(8 \quad\) a \(x^{2}+y^{2}=16\)

b \(y=2 x-4\)

c \(P=\left(\frac{16}{5}, \frac{12}{5}\right)\), peg is \(3.2 \mathrm{~m}\) east and \(2.4 \mathrm{~m}\) north of the centre of the garden

9 a i \(h=r \tan \theta\)

b i \(\operatorname{Area}(A)=\frac{1}{2} \pi r^{2}\)

c \(\frac{1}{2} \pi r^{2}=r^{2} \tan \theta, \frac{\pi}{2}=\tan \theta\)

d \(57.5^{\circ}\) ii \(s=\frac{r}{\cos \theta}\)

ii Area \((B)=r^{2} \tan \theta\)

e \(4 \pi \approx 12.57 \mathrm{~mm}^{2}\)
10 a i \(60^{\circ}\)

ii \(30^{\circ}\)

b \(a+b=90\)

c i \(C E=(10-x) \mathrm{cm} \quad\) ii \(D E=(10-y) \mathrm{cm}\)

d \(x(10-x)=y(10-y) \Rightarrow 0=10(y-x)-\left(y^{2}-x^{2}\right)\) \(\Rightarrow 0=10(y-x)-(y-x)(y+x)\)

\(0=(y-x)(10-y-x)\)

e i \(\frac{x y}{2} \mathrm{~cm}^{2} \quad\) ii \(\frac{1}{2}(10-x)(10-y) \mathrm{cm}^{2}\)

f \(\quad \frac{x y}{2}=\frac{1}{2}(10-x)(10-y)\)

\(x y=100-10 y-10 x+x y\)

\(10(x+y)=100\)

\(x+y=10\)

g \(\frac{x(10-x)}{2} \mathrm{~cm}^{2}\)

h \(x=6\) or 4


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