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\section*{Review of prisms and cylinders}
A polyhedron is a solid bounded by polygons called faces.

Two adjacent faces meet at an edge and two adjacent edges meet at a vertex.

A prism is a polyhedron that has two congruent and parallel faces and all its remaining faces are parallelograms.

A right prism is a prism in which the top and bottom polygons are vertically above each other, and the vertical polygons connecting their faces are rectangles. A prism that is not a right prism is often called an oblique prism.

Some examples of prisms are shown below.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-02(1)}
\end{center}

Right rectangular prism

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-02}
\end{center}

Oblique rectangular prism

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-02(2)}
\end{center}

Right triangular prism

When we refer to a prism we generally mean a right prism.

A prism with a rectangular base is called a rectangular prism, while a triangular prism has a triangular base.

You will notice that if we slice a prism by a plane parallel to its base, then the cross-section is congruent to the base and so has the same area as the base.

In this chapter, we will use the pronumeral \(S\) for the surface

\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-02(3)}
area of a solid and the pronumeral \(V\) for the volume of a solid.

\section*{Surface areas}
Surface area of a prism

The surface area of a prism is the sum of the areas of its faces.

A rectangular prism with dimensions \(a, b\) and \(c\) has six faces.

These occur in opposite pairs; the faces with areas \(a b, b c\) and \(c a\) each occur twice.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-02(4)}
\end{center}

Thus:

Surface area of a rectangular prism \(=2(a b+b c+c a)\)

We do not need to learn this formula. We can simply find the area of each face and take the sum of the areas. The same idea applies to all other types of prisms.

In the diagrams in this chapter, assume that all quadrilaterals are rectangles unless the context indicates otherwise.

\section*{Surface area of a prism}
To find the surface area of a prism, find the area of each face and calculate the sum of the areas.

\section*{Example 1}
Find the surface area of each prism.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-03(1)}
\end{center}

\(12 \mathrm{~cm}\) b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-03}
\end{center}

\section*{Solution}
a The prism has six faces.

Area of top rectangle \(=12 \times 5\)

\[
=60 \mathrm{~cm}^{2}
\]

Area of front rectangle \(=12 \times 8\)

\[
=96 \mathrm{~cm}^{2}
\]

Area of side rectangle \(=8 \times 5\)

\[
=40 \mathrm{~cm}^{2}
\]

Thus, \(S=2 \times 60+2 \times 96+2 \times 40\)

\[
=392 \mathrm{~cm}^{2}
\]

b We need to find the length \(A B\) in the diagram. We can find this using Pythagoras' theorem.

\[
A B^{2}=6^{2}+8^{2}
\]

\[
=100
\]

so \(A B=10 \mathrm{~cm}\)

Area of the sloping rectangle \(=10 \times 20\)

\[
=200 \mathrm{~cm}^{2}
\]

Area of each triangle \(=\frac{1}{2} \times 6 \times 8\)

\[
=24 \mathrm{~cm}^{2}
\]

Area of the base rectangle \(=6 \times 20\)

\[
=120 \mathrm{~cm}^{2}
\]

Area of the back rectangle \(=8 \times 20\)

\[
=160 \mathrm{~cm}^{2}
\]

Thus, \(S=24+24+120+160+200\)

\[
=528 \mathrm{~cm}^{2}
\]

\section*{Surface area of a cylinder}
A cylinder is a solid that has parallel circular discs of equal radius at the top and the bottom. Each cross-section parallel to the base is a circle, and the centres of these circular cross-sections lie on a straight line. If that line is perpendicular to the base, the cylinder is called a right cylinder. When we use the word 'cylinder' in this book, we will generally mean a right cylinder.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-04(2)}
\end{center}

Right cylinder

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-04(4)}
\end{center}

Oblique cylinder

We will use a dot \((\bullet)\) to indicate the centre of the circular base or top.

As we did with prisms, we find the surface area of a cylinder by adding up the area of the curved section of the cylinder, and the area of the two circles.

\section*{Surface area of the curved surface}
Suppose we have a cylinder with base radius \(r\) and height \(h\). If we roll the cylinder along a flat surface through one revolution, as shown in the diagram, the curved surface traces out a rectangle. The width of the rectangle is the height of the cylinder, while the length of the rectangle is the circumference of the circle, which is \(2 \pi r\), so the area of the curved part is \(2 \pi r h\). Thus:

Curved surface area of cylinder \(=2 \pi r h\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-04}
\end{center}

\section*{Example 2}
Calculate the surface area of each solid, correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-04(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-04(1)}
\end{center}

\section*{Solution}
a This is a cylinder with radius \(3 \mathrm{~cm}\) and height \(8 \mathrm{~cm}\).

Area of curved surface \(=2 \pi r h\)

\[
\begin{aligned}
& =2 \times \pi \times 3 \times 8 \\
& =48 \pi \mathrm{cm}^{2}
\end{aligned}
\]

Area of a circular end \(=\pi r^{2}\)

\[
\begin{aligned}
& =\pi \times 3^{2} \\
& =9 \pi \mathrm{cm}^{2}
\end{aligned}
\]

Hence, \(S=48 \pi+9 \pi+9 \pi\)

\[
=66 \pi \mathrm{cm}^{2}
\]

\[
=207.35 \mathrm{~cm}^{2} \quad \text { (Correct to two decimal places.) }
\]

b Area of curved section \(=\frac{1}{2} \times 2 \pi r h\)

\[
\begin{aligned}
& =\pi \times 10 \times 15 \\
& =150 \pi \mathrm{cm}^{2}
\end{aligned}
\]

Area of a semicircle \(=\frac{1}{2} \times \pi r^{2}\)

\[
\begin{aligned}
& =\frac{1}{2} \pi \times 10^{2} \\
& =50 \pi \mathrm{cm}^{2}
\end{aligned}
\]

Area of top rectangle \(=20 \times 15\)

\[
=300 \mathrm{~cm}^{2}
\]

\[
\begin{aligned}
S & =150 \pi+50 \pi+50 \pi+300 \\
& =(250 \pi+300) \mathrm{cm}^{2} \\
& \approx 1085.40 \mathrm{~cm}^{2} \quad \text { (Correct to two decimal places.) }
\end{aligned}
\]

\section*{Volume}
\section*{Volume of a rectangular prism}
We have seen in earlier work that the volume of a right rectangular prism is given by:

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-05}
\end{center}

area of the base \(=A\)

Volume of a right rectangular prism \(=\) area of base \(\times\) height

\[
=A \times h
\]

That is, \(V=A \times h\)

Suppose that we have two solids of the same height. If the cross-sections of the two solids, taken at the same distance above the base, have the same area, it can be shown that the solids have the same volume. This is known as Cavalieri's principle.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-06(1)}
\end{center}

Cross-sectional areas are the same.

Cavalieri's principle allows us to say that the volume \(V\) of any rectangular prism, right or oblique, is given by \(V=A \times h\).

\section*{Volume of a prism}
The volume \(V\) of a prism, right or oblique, is given by the formula:

\[
V=A h
\]

where \(A\) is the area of the base and \(h\) is the height, as discussed previously.

The proof is in five steps. In all of the following, the height is \(h\).

\section*{Step 1: The base of the prism is a right-angled triangle}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-06(2)}
\end{center}

The right rectangular prism has volume \(V=A h\). If it is cut in half, we obtain two prisms of the same volume \(\frac{A h}{2}\) base area \(\frac{A}{2}\) and height \(h\).

Step 2: The base of the prism is a parallelogram of area \(A\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-06}
\end{center}

\(A=R-2 T\), where \(R\) is the area of the rectangle and \(T\) is the area of each right angled triangle.

The volume of the prism:

\[
\begin{aligned}
V & =h R-2 h T \\
& =h(R-2 T) \\
& =h A
\end{aligned}
\]

So the formula holds if the base of the prism is a parallelogram.

\section*{Step 3: The base is a triangle \(A B C\)}
The area of the triangle \(A B C\) is half the area of the parallelogram \(A B C D\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-07(1)}
\end{center}

The prism with triangular base \(A B C\) is half the volume of the prism with the parallelogram \(A B C D\) as the base. So the formula holds if the base of the prism is a triangle.

\section*{Step 4: The base of the prism is a polygon}
As an example, the convex hexagon can be cut up into four triangles.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-07(2)}
\end{center}

Clearly the volume of the prism is:

\[
\begin{aligned}
V & =A_{1} h+A_{2} h+A_{3} h+A_{4} h \\
& =\left(A_{1}+A_{2}+A_{3}+A_{4}\right) h \\
& =A h,
\end{aligned}
\]

where \(A\) is the area of the hexagon.

This argument applies to any right prism.

\section*{Step 5: Oblique prisms}
Use Cavalieri's principle exactly as above.

\section*{Volume of a cylinder}
The area of a regular polygon inscribed in a circle approximates the area of that circle. The greater the number of sides, the better the approximation. A cylinder has a circular base. Since \(V=A h\) holds for any polygon-based prism, it seems reasonable that the volume of the cylinder should be the area of its circular base multiplied by the height.
\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-07}

Thus, the volume of a cylinder with radius \(r\) and height \(h\) is equal to the area of the circular crosssection, \(\pi r^{2}\), multiplied by the height, \(h\).

\[
\text { Volume of a cylinder }=\pi r^{2} h
\]

Cavalieri's principle shows that this also applies to an oblique cylinder.

\section*{Example 3}
Calculate the volume of each solid.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-08(2)}
\end{center}

\(6 \mathrm{~cm}\)

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-08}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-08(1)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-08(3)}
\end{center}

\section*{Solution}
a \(V=A h\)

\(=6 \times 4 \times 5\)

\(=120 \mathrm{~cm}^{3}\)

c This is a rectangular prism of height \(10 \mathrm{~cm}\).

\[
\begin{aligned}
V & =A h \\
& =8 \times 6 \times 10 \\
& =480 \mathrm{~cm}^{3}
\end{aligned}
\]

b The solid is a triangular prism.

\[
\begin{aligned}
V & =A h \\
& =(\text { area of triangular base }) \times(\text { height of prism }) \\
& =\left(\frac{1}{2} \times 10 \times 3\right) \times 15 \\
& =225 \mathrm{~cm}^{3}
\end{aligned}
\]

d This is a cylinder of radius \(2.5 \mathrm{~cm}\) and height \(14 \mathrm{~cm}\).

\[
\begin{aligned}
V & =\pi r^{2} h \\
& =\pi \times 2.5^{2} \times 14 \\
& =87.5 \pi \mathrm{cm}^{3}
\end{aligned}
\]

\section*{Surface area and volume}
\begin{itemize}
  \item A prism is a polyhedron with two parallel congruent polygonal faces and all other faces parallelograms.
  \item The surface area of a prism is the sum of the areas of its faces.
  \item The surface area of the curved part of a cylinder with radius \(r\) and height \(h\) is given by: Curved surface area \(=2 \pi r h\)
  \item The volume of a prism is given by the product of the area of the base \(A\) and the height \(h\) : \(V=A h\)
  \item The volume of a cylinder with radius \(r\) and height \(h\) is given by: \(V=\pi r^{2} h\)
\end{itemize}

\section*{Exercise 6A}
1 Calculate the surface area of each solid.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-09(1)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-09(7)}
\end{center}

\(12 \mathrm{~cm}\)

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-09(2)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-09(3)}
\end{center}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-09}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-09(5)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-09(8)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-09(4)}
\end{center}

h
\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-09(6)}

2 The base of a right prism of height \(12 \mathrm{~cm}\) is an equilateral triangle.

a If the side length of the triangle is \(6 \mathrm{~cm}\), use Pythagoras' theorem to calculate the height of the triangle.

b Calculate the surface area of the prism.

3 A cube has surface area \(486 \mathrm{~m}^{2}\). Find its side length.

4 A cylinder of base radius \(8 \mathrm{~cm}\) has curved surface area \(72 \pi \mathrm{cm}^{2}\). Find its height.

5 A rubbish bin is in the shape of an open cylinder.

a If the bin has radius \(15 \mathrm{~cm}\) and height \(40 \mathrm{~cm}\), find its total surface area (do not include the lid).

b If the bin has radius \(15 \mathrm{~cm}\) and curved surface area \(3500 \mathrm{~cm}^{2}\), find the height of the bin, correct to one decimal place.

6 Calculate the volume of each solid.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(6)}
\end{center}

\(10 \mathrm{~cm}\)

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(9)}
\end{center}

g

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10}
\end{center}

j

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(2)}
\end{center}

e

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(5)}
\end{center}

h

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(4)}
\end{center}

\(\mathbf{k}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(1)}
\end{center}

c
\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-10(8)}

i

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(10)}
\end{center}

I

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(7)}
\end{center}

7 In the diagram below, \(A B C F E D\) is a right triangular prism, with \(A B=A C\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-10(11)}
\end{center}

a If \(A B=8 \mathrm{~cm}\) and \(B C=12 \mathrm{~cm}\), find the height of \(\triangle A B C\), correct to two decimal places.

b If \(A D=24 \mathrm{~cm}\), find the volume of the prism, correct to two decimal places.

8 A manufacturer of drink cans produces cylindrical cans with a volume of \(1000 \mathrm{~cm}^{3}\).

a If the radius of the can is \(4 \mathrm{~cm}\), find the height of the can, correct to one decimal place.

b If the height of the can is \(8 \mathrm{~cm}\), find the radius of the can, correct to two decimal places.

9 a If a cube has volume \(64 \mathrm{~cm}^{3}\), find its side length.

b If a cube has volume \(400 \mathrm{~cm}^{3}\), find its side length, correct to one decimal place.

10 A modern sculpture consists of three cubes, with side lengths \(0.5 \mathrm{~m}, 1 \mathrm{~m}\) and \(1.5 \mathrm{~m}\), respectively, placed on top of each other as shown in the diagram opposite. Calculate the surface area of the sculpture. Do not include the base of the large cube in your calculations.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-11(3)}
\end{center}

11 The figure shows a trapezoidal prism. Its ends \(A B C D\) and \(E F G H\) are congruent trapezia, with \(D C=4 \mathrm{~cm}\) and \(A B=6 \mathrm{~cm}\). If \(A E=15 \mathrm{~cm}\) and the volume is \(300 \mathrm{~cm}^{3}\), find the height of the trapezium \(A B C D\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-11(2)}
\end{center}

12 A farmer is making a trough that needs to be filled with water once each day. The trough is in the shape of a prism with pentagonal ends, as shown opposite. The farmer has 60 horses that each drink about \(10 \mathrm{~L}\) of water per day. Using the fact that \(1 \mathrm{~cm}^{3}=1 \mathrm{~mL}\), what is the smallest length the trough needs to be to water the horses each day?

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-11}
\end{center}

13 A cylindrical vase of radius \(5 \mathrm{~cm}\) and height \(14 \mathrm{~cm}\) just fits into a box. Find:

a the volume of the cylinder, correct to two decimal places

b the volume of the box

c the volume of unused space inside the box, correct to two decimal places

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-11(1)}
\end{center}

\section*{Pyramids}
A pyramid is a polyhedron with a polygonal base and triangular sides that meet at a point called the vertex. The pyramid is named according to the shape of its base.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-12(3)}
\end{center}

Square pyramid

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-12(1)}
\end{center}

Triangular pyramid

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-12(2)}
\end{center}

Hexagonal pyramid

If we drop a perpendicular from the vertex of the pyramid to the base, then the length of the perpendicular is called the height of the pyramid.

A right pyramid is one whose vertex is directly above the centre of its base.

\section*{Volume of a pyramid}
Here is a method for determining the formula for the volume of a right, square-based pyramid.

Consider a cube of side length \(2 x\). If we draw the long diagonals as shown, then we obtain 6 square pyramids, one of which is shaded in the diagram. Each of these pyramids has base area \(2 x \times 2 x\) and height \(x\). Let \(V\) be the volume of each pyramid.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-12}
\end{center}

\(2 x\)

The volume of the cube is \((2 x)^{3}=8 x^{3}\). Hence:

\[
\begin{aligned}
6 \times V & =8 x^{3} \\
V & =\frac{4}{3} \times x^{3}
\end{aligned}
\]

so

Now the area of the base of each pyramid is \((2 x)^{2}=4 x^{2}\) and the height of each pyramid is \(x\), so in this case we can write:

\[
\begin{aligned}
V & =\frac{4}{3} \times x^{3} \\
& =\frac{1}{3} \times 4 x^{2} \times x
\end{aligned}
\]

or \(\quad V=\frac{1}{3} \times\) area of base \(\times\) height

It is possible to extend this result to any pyramid by using geometric arguments and Cavalieri's principle. We then have the following important result.

\[
\begin{aligned}
\text { Volume of a pyramid } & =\frac{1}{3} \times \text { base area } \times \text { height } \\
\text { That is, } V & =\frac{1}{3} A h
\end{aligned}
\]

\section*{Example 4}
Calculate the volume of a square pyramid with base of side length \(200 \mathrm{~m}\) and height \(300 \mathrm{~m}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-13(1)}
\end{center}

\section*{Solution}
\[
\begin{aligned}
V & =\frac{1}{3} \times \text { base area } \times \text { height } \\
& =\frac{1}{3} \times 200 \times 200 \times 300 \\
& =4000000 \mathrm{~m}^{3}
\end{aligned}
\]

The volume of the pyramid is \(4000000 \mathrm{~m}^{3}\).

\section*{Surface area of a pyramid}
To find the surface area of a pyramid, we need to find the areas of the surfaces bounding the pyramid, which will consist of a number of triangles together with the base. You will often have to use Pythagoras' theorem in calculating the surface area of a pyramid.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-13(2)}
\end{center}

\section*{Example 5}
\(V A B C D\) is a right, square-based pyramid with vertex \(V\) and base \(A B C D\), with \(V\) vertically above the centre of the square base. The height of the pyramid is \(4 \mathrm{~cm}\) and the side length of the base is \(6 \mathrm{~cm}\). Find the surface area of the pyramid.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-13}
\end{center}

\section*{Solution}
We need to find the height \(V E\) of triangle \(V B C\), using Pythagoras' theorem.

\[
\begin{aligned}
V E^{2} & =V O^{2}+O E^{2} \\
& =4^{2}+3^{2} \\
& =25
\end{aligned}
\]

Hence, \(V E=5 \mathrm{~cm}\).
\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-14(1)}

\[
\text { Area of } \begin{aligned}
\triangle V C B & =\frac{1}{2} \times C B \times V E \\
& =\frac{1}{2} \times 6 \times 5 \\
& =15 \mathrm{~cm}^{2}
\end{aligned}
\]

Area of base \(=6 \times 6\)

\(=36 \mathrm{~cm}^{2}\)

Surface area \(=4 \times 15+36\)

\[
=96 \mathrm{~cm}^{2}
\]

Hence, the surface area of the pyramid is \(96 \mathrm{~cm}^{2}\).

\section*{Pyramids}
\begin{itemize}
  \item A pyramid is a polyhedron with a polygonal base and triangular faces that meet at a point called the vertex.
  \item The pyramid is named according to the shape of its base.
  \item The volume of a pyramid with base area \(A\) and height \(h\) is given by:
\end{itemize}

\(V=\frac{1}{3} A h\)

\section*{Exercise 6B}
In this exercise, assume all pyramids are right pyramids unless otherwise stated.

1 In the diagram opposite, \(V A B C\) is a triangular pyramid. The base \(\triangle A B C\) is right-angled, with \(A C=6 \mathrm{~cm}\) and \(B C=8 \mathrm{~cm}\). The vertex \(V\) of the pyramid is vertically above \(A\), and \(V A=7.5 \mathrm{~cm}\). Calculate the volume of the pyramid.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-14}
\end{center}

2 Calculate the volume of each of the following pyramids. The area, \(A \mathrm{~cm}^{2}\), of the shaded face and the height, \(h \mathrm{~cm}\), are given in each case.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15}
\end{center}

\(A=24, h=10\)

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15(7)}
\end{center}

\(A=72, h=9\) b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15(3)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15(2)}
\end{center}

3 Calculate the volume of each pyramid.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15(8)}
\end{center}

height \(=12 \mathrm{~cm}\)

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15(5)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15(9)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15(6)}
\end{center}

4 When it was built, the Great Pyramid of Cheops in Egypt had a height of \(145.75 \mathrm{~m}\) and its base was a square of side length \(229 \mathrm{~m}\). Find its volume in cubic metres, correct to one decimal place.

5 Calculate the volume of each solid, giving the answer correct to two decimal places.

\(\mathbf{a}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-15(4)}
\end{center}

6 In the diagram opposite, \(V A B C D\) is a square pyramid and \(O\) is the centre of square \(A B C D\). If the height of the pyramid \(V O=12 \mathrm{~cm}\) and \(A B=10 \mathrm{~cm}\), find:

a the length \(O M\), where \(M\) is the midpoint of \(B C\)

b the length \(V M\)

c the area of \(\triangle V C B\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-16(3)}
\end{center}

d the surface area of the pyramid

7 Use a technique similar to that in question 6 to find the surface areas of these square pyramids, assuming the vertex is above the centre of the square.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-16(1)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-16}
\end{center}

8 In the diagram opposite, \(V A B C D\) is a rectangular pyramid. If \(V O=6 \mathrm{~cm}\) is the height of the pyramid, \(C D=10 \mathrm{~cm}\) and \(B C=8 \mathrm{~cm}\), find:

a the height of \(\triangle V B C\)

b the height of \(\triangle V D C\)

c the surface area of the pyramid

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-16(2)}
\end{center}

9 In the diagram opposite, \(V A B C D\) is a square pyramid, with vertex \(V\) directly above \(D\). If \(A B=8 \mathrm{~cm}\) and \(V D=6 \mathrm{~cm}\) :

a name each triangle in the diagram and state what type of triangle it is

b find the length of:

i \(V C\)

ii \(V B\)

iii \(V A\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-16(4)}
\end{center}

c find the surface area of the pyramid

\section*{Cones}
A cone is a solid that is formed by taking a circle and a point, called the vertex, which lies above or below the circle. We then join the vertex to each point on the circle. Of course, the cone can be in any orientation.

If the vertex is directly above or below the centre of the circular base, we call the cone a right cone. In this section, the only cones we consider are right cones, which we will simply call cones.

The distance from the vertex of the cone to the centre of the circular base is called the height, \(h\), of the cone.

The distance from the vertex of the cone to the circumference of the circular base is called the slant height, \(\ell\), of the cone.

We will use a dot \((\bullet)\) to indicate the centre of the base.

\section*{Surface area of a cone}
\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-17(1)}
\end{center}

To calculate the surface area of a cone, we need to find the area of each surface. To find the area of the curved surface, we cut and open up the curved surface to form a sector, as shown below.
\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-17(3)}

\begin{itemize}
  \item The arc length of the sector \(=\) circumference of the circular base of the cone \(=2 \pi r\)
  \item The proportion of a whole circle \(=\frac{\text { arc length }}{\text { whole circumference }}=\frac{2 \pi r}{2 \pi \ell}=\frac{r}{\ell}\)
  \item Area of sector
\end{itemize}

\[
A=\left(\frac{r}{\ell}\right) \times \pi \ell^{2}=\pi r \ell
\]

Thus, the surface area of the curved part of a cone is given by:

\section*{Curved surface area \(=\pi r \ell\),}
where \(r\) is the base radius and \(\ell\) is the slant height.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-17}
\end{center}

\section*{Example 6}
Find the surface area of each cone. Give your answer correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-17(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-17(4)}
\end{center}

\section*{Solution}
a We have \(r=4\) and \(\ell=6\).

Area of curved surface \(=\pi r \ell\)

\[
\begin{aligned}
& =\pi \times 4 \times 6 \\
& =24 \pi \mathrm{cm}^{2}
\end{aligned}
\]

Area of base \(=\pi r^{2}\)

\[
=16 \pi \mathrm{cm}^{2}
\]

Thus, \(S=24 \pi+16 \pi\)

\[
=40 \pi \mathrm{cm}^{2}
\]

\[
\approx 125.66 \mathrm{~cm}^{2} \quad(\text { Correct to two decimal places. })
\]

Hence, the surface area of the cone is approximately \(125.66 \mathrm{~cm}^{2}\).

b We first find the slant height \(A B\).

By Pythagoras' theorem:

\[
\begin{aligned}
A B^{2} & =5^{2}+12^{2} \\
& =169
\end{aligned}
\]

\(A B=169 \mathrm{~cm}\)

Hence, \(r=5\) and \(\ell=13\).

Area of curved surface \(=\pi r \ell\)

\[
\begin{aligned}
& =\pi \times 5 \times 13 \\
& =65 \pi \mathrm{cm}^{2}
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-18}
\end{center}

Area of base \(=\pi r^{2}\)

\[
=25 \pi \mathrm{cm}^{2}
\]

Then \(S=65 \pi+25 \pi\)

\[
\begin{aligned}
& =90 \pi \mathrm{cm}^{2} \\
& \approx 282.74 \mathrm{~cm}^{2} \quad \text { (Correct to two decimal places.) }
\end{aligned}
\]

Hence, the surface area of the cone is approximately \(282.74 \mathrm{~cm}^{2}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-18(1)}
\end{center}

\section*{Example 7}
The curved surface area of a cone is \(44 \mathrm{~cm}^{2}\) and the base radius is \(2 \mathrm{~cm}\). Find, correct to two decimal places:
a the slant height of the cone
b the height of the cone

\section*{Solution}
a Curved surface area of a cone \(=\pi r l\)

Here \(r=2\), so \(2 \pi \ell=44 \mathrm{~cm}^{2}\).

\[
\begin{aligned}
\ell & =\frac{22}{\pi} \mathrm{cm} \\
& \approx 7.00 \mathrm{~cm} \quad \text { (Correct to two decimal places.) }
\end{aligned}
\]

Hence, the slant height is approximately \(7.00 \mathrm{~cm}\).

(continued over page)
b Using Pythagoras' theorem:

\[
\begin{aligned}
h^{2}+2^{2} & =\left(\frac{22}{\pi}\right)^{2} \\
\text { so } \quad h & =\sqrt{\left(\frac{22}{\pi}\right)^{2}-2^{2}}
\end{aligned}
\]

\[
\approx 6.71 \mathrm{~cm} \quad(\text { Correct to two decimal places. })
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-19(2)}
\end{center}

Hence, the height of the cone is approximately \(6.71 \mathrm{~cm}\).

\section*{Volume of a cone}
The formula for the volume of a cone is the same as the formula for the volume of a pyramid, which is \(\frac{1}{3} \times\) area of the base \(\times\) height. For a cone, the base area is \(\pi r^{2}\), so:

\[
\text { Volume of a cone }=\frac{1}{3} \pi r^{2} h,
\]

where \(r\) is the radius of the base and \(h\) is the height.

To illustrate this informally, imagine constructing a polygon inside the circular base of the cone and joining the vertex of the cone to each of the vertices of the polygon.

This would give us a polygonal pyramid with volume equal to:

\[
\frac{1}{3} \times \text { area of the base } \times \text { height }
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-19}
\end{center}

The more sides we take in the polygon, the area of the base gets closer and closer to \(\pi r^{2}\), so the volume of the cone equals \(\frac{1}{3} \pi r^{2} h\).

\section*{Example 8}
Calculate the volume of a cone with base radius \(5 \mathrm{~m}\) and height \(6 \mathrm{~m}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-19(1)}
\end{center}

\section*{Solution}
\[
\begin{aligned}
V & =\frac{1}{3} \pi r^{2} h \\
& =\frac{1}{3} \times \pi \times 5^{2} \times 6 \\
& =50 \pi \mathrm{m}^{3}
\end{aligned}
\]

The volume of the cone is \(50 \pi \mathrm{m}^{3}\).

\section*{Right cones}
\begin{itemize}
  \item In a cone, the distance between the vertex and the centre of the base is called the height, \(h\), of the cone.
  \item The length of a straight line joining the vertex to a point on the circumference of the circle is called the slant height, \(\ell\), of the cone.
  \item The surface area of the curved part of a cone is given by \(\pi r \ell\), where \(r\) is the base radius and \(\ell\) is the slant height.
  \item The volume of a cone is given by:
\end{itemize}

\(V=\frac{1}{3} \pi r^{2} h\), where \(r\) is the base radius and \(h\) is the height.

\section*{Exercise 6C}
1 Calculate the total surface area of each cone, including the base. Give your answers correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-20(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-20(5)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-20(3)}
\end{center}

2 Calculate the total surface area of each cone.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-20(2)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-20}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-20(1)}
\end{center}

3 The curved surface of a cone of base radius \(4 \mathrm{~cm}\) has surface area \(40 \pi \mathrm{cm}^{2}\). Find:

a the slant height of the cone

b the height of the cone, correct to four decimal places

4 A cone has base radius \(10 \mathrm{~cm}\) and total surface area \(1000 \mathrm{~cm}^{2}\). Find, correct to two decimal places:

a the surface area of the curved part of the cone

b the slant height of the cone

c the height of the cone

5 a Calculate the area of the sector shown opposite. Give your answer correct to two decimal places.
b If the radii \(O A\) and \(O B\) are joined together to form the curved surface of a cone, find, correct to two decimal places:

i the slant height of the cone

ii the base radius of the cone

iii the height of the cone

6 A cone has radius \(r \mathrm{~cm}\) and height \(h \mathrm{~cm}\).

a If \(r=5\) and \(h=10\), find:

i the slant height of the cone

ii the surface area of the curved part of the cone

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-21(2)}
\end{center}

iii the angle of the sector we get if we cut the curved part of the cone

b If \(r=h\), find the angle of the sector that produces the curved part of the cone.

7 Calculate the volume of each solid. Give your answers correct to two decimal places.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-21(1)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-21(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-21(6)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-21(5)}
\end{center}

8 A cone with diameter \(6 \mathrm{~cm}\) has a volume of \(120 \mathrm{~cm}^{3}\). Find the height of the cone, correct to the nearest millimetre.

9 For the solid shown, find:

a the volume

b the total surface area

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-21(3)}
\end{center}

10 For the solid shown, find:

a the volume

b the total surface area

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-21}
\end{center}

11 A mound of earth is shaped like a cone. It is 6 metres high with a radius of 25 metres. Find the cost, to the nearest dollar, of moving the mound if it costs \(\$ 5\) to move one cubic metre.

\section*{Spheres}
The word sphere comes from the Greek word sphaira, meaning 'ball'. Every point on the surface of a sphere lies at a distance \(r\), called the radius of the sphere, from a fixed point \(O\), called the centre of the sphere. It is difficult to derive the formulas for the surface area and volume of a sphere. These formulas are discussed in the Challenge exercises.

We will use a dot \((\bullet)\) to indicate the centre of a sphere.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-22}
\end{center}

\section*{Surface area of a sphere}
The surface area of a sphere is given by:

\[
S=4 \pi r^{2},
\]

where \(r\) is the radius of the sphere.

\section*{Example 9}
Calculate, correct to two decimal places, the surface area of a sphere:
a with radius \(6 \mathrm{~cm}\)
b with diameter \(10 \mathrm{~cm}\)

\section*{Solution}
a We have

\[
\begin{aligned}
S & =4 \pi r^{2} \\
& =4 \times \pi \times 6^{2} \\
& =144 \pi \mathrm{cm}^{2} \\
& \approx 452.39 \mathrm{~cm}^{2} \quad \text { (Correct to two decimal places.) }
\end{aligned}
\]

The surface area is approximately \(452.39 \mathrm{~cm}^{2}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-22(1)}
\end{center}

b The diameter \(=10 \mathrm{~cm}\), so \(r=5 \mathrm{~cm}\).

\[
\text { Then } \begin{aligned}
S & =4 \pi r^{2} \\
& =4 \times \pi \times 5^{2} \\
& =100 \pi \mathrm{cm}^{2} \\
& \approx 314.16 \mathrm{~cm}^{2} \quad \text { (Correct to two decimal places.) }
\end{aligned}
\]

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-22(2)}
\end{center}

The surface area is approximately \(314.16 \mathrm{~cm}^{2}\).

\section*{Volume of a sphere}
The formula for the volume of a sphere is:

\[
V=\frac{4}{3} \pi r^{3}
\]

where \(r\) is the radius of the sphere.

\section*{Example 10}
Calculate the volume of a sphere with a diameter of \(30 \mathrm{~m}\).

\section*{Solution}
The radius is \(15 \mathrm{~m}\).

\[
\begin{aligned}
V & =\frac{4}{3} \pi r^{3} \\
& =\frac{4}{3} \times \pi \times 15^{3} \\
& =4500 \pi \mathrm{m}^{3}
\end{aligned}
\]

\section*{Example 11}
A sphere has volume \(2800 \mathrm{~cm}^{3}\). Find the radius of the sphere, correct to the nearest millimetre.

\section*{Solution}
\[
\begin{aligned}
V & =\frac{4}{3} \pi r^{3} \\
2800 & =\frac{4}{3} \pi r^{3} \\
\text { so } \quad r^{3} & =\frac{2800 \times 3}{4 \pi} \\
r & =\sqrt[3]{\frac{2800 \times 3}{4 \pi}} \\
& \approx 8.7 \mathrm{~cm} \quad \text { (Correct to one decimal place.) }
\end{aligned}
\]

The radius is approximately \(87 \mathrm{~mm}\).

\section*{Spheres}
\begin{itemize}
  \item The surface area of a sphere of radius \(r\) is given by: \(S=4 \pi r^{2}\), where \(r\) is the radius of the sphere.
  \item The volume of a sphere of radius \(r\) is given by:
\end{itemize}

\[
V=\frac{4}{3} \pi r^{3}, \text { where } r \text { is the radius of the sphere. }
\]

\section*{Exercise 6D}
1 Calculate the surface area, correct to two decimal places, of:

a a sphere of radius \(8 \mathrm{~cm}\)

b a sphere of radius \(15 \mathrm{~cm}\)

c a sphere of diameter \(14 \mathrm{~cm}\)

d a sphere of diameter \(21 \mathrm{~cm}\)

2 For the solid hemisphere shown opposite, find:

a the area of the flat surface

b the surface area of the curved part of the hemisphere

c the total surface area of the solid hemisphere

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24}
\end{center}

3 A sphere has a surface area of \(500 \mathrm{~cm}^{2}\). Find its radius, correct to four decimal places.

4 A hemispherical tent is made using \(28 \mathrm{~m}^{2}\) of material. Find, correct to two decimal places, the radius of the tent if:

a the tent does not have a material floor

b the tent does have a material floor

5 Calculate the surface area of each object.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(1)}
\end{center}

\(10 \mathrm{~cm}\) b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(10)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(5)}
\end{center}

\(10 \mathrm{~cm}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(4)}
\end{center}

6 Calculate the volume of each solid. Give your answers correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(2)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(6)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(7)}
\end{center}

\(\mathbf{e}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(3)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(9)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-24(8)}
\end{center}

7 a Calculate, correct to two decimal places, the radius of a sphere with a volume of \(1000 \mathrm{~cm}^{3}\).

b Calculate, correct to the nearest millimetre, the diameter of a sphere with a volume of \(3000 \mathrm{~cm}^{3}\).

c Calculate, correct to one decimal place, the diameter of a hemispherical bowl with a volume of \(250 \mathrm{~cm}^{3}\).

8 Calculate the volume of each solid. Where appropriate, give your answers correct to two decimal places.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-25}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-25(3)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-25(2)}
\end{center}

9 A spherical soccer ball of diameter \(22 \mathrm{~cm}\) is packaged in a box that is in the shape of a cube with edges of length \(24 \mathrm{~cm}\). Find, correct to the nearest \(\mathrm{cm}^{3}\), the volume of unused space inside the box.

10 Tennis balls are packaged in cylindrical canisters. A tennis ball can be considered to be a sphere of diameter \(70 \mathrm{~mm}\). If the canister has base diameter \(75 \mathrm{~mm}\) and height \(286 \mathrm{~mm}\), and each canister holds four balls, find the volume of unused space inside the canister, correct to two decimal places.

\section*{Enlargement}
In this section, we will investigate what happens to the area and volume of figures under enlargement.

In the diagram below, \(\triangle A^{\prime} B^{\prime} C^{\prime}\) is an enlargement of \(\triangle A B C\). The sides of the larger triangle are three times the lengths of those of the smaller one.

We say that the enlargement factor is 3 .
\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-25(1)}

\section*{Plane figures and enlargements}
Notice what happens when we compare the areas of the two triangles. The area of \(\triangle A B C\) is \(6 \mathrm{~cm}^{2}\), and the area of \(\triangle A^{\prime} B^{\prime} C^{\prime}\) is \(54 \mathrm{~cm}^{2}\). In this case, the area of the larger triangle is 9 times the area of the smaller one. Since \(9=3^{2}\), we see that the area of the smaller triangle is multiplied by the square of the enlargement factor.

\section*{Example 12}
By what factor does the area of a circle with radius \(2 \mathrm{~cm}\) change when we enlarge the radius by a factor of 5 ?

\section*{Solution}
A circle with radius \(2 \mathrm{~cm}\) has area \(\pi \times 2^{2}=4 \pi \mathrm{cm}^{2}\).

Enlarging by a factor of 5 gives a radius of \(10 \mathrm{~cm}\).

A circle with radius \(10 \mathrm{~cm}\) has area \(\pi \times 10^{2}=100 \pi \mathrm{cm}^{2}\).

The area has been multiplied by a factor of \(\frac{100 \pi}{4 \pi}=25\).

(Notice that this is the square of the enlargement factor.)

In general, if each of the dimensions of a plane figure is enlarged by a factor of \(k\), then the area of the figure is multiplied by a factor of \(k^{2}\).
\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-26}

\section*{Example 13}
A regular pentagon has area \(45 \mathrm{~cm}^{2}\). If the sides of the pentagon are enlarged by a factor of 8 , what is the area of the resulting pentagon?

\section*{Solution}
The enlargement factor is 8 , so the area is multiplied by a factor of \(8^{2}=64\).

Hence, the area of the resulting pentagon is \(64 \times 45=2880 \mathrm{~cm}^{2}\).

\section*{Solids and enlargements}
The diagram to the right enables us to draw the following general conclusions. If the dimensions of a solid are enlarged by a factor of \(k\), then the surface area of the figure is multiplied by a factor of \(k^{2}\).

If the dimensions of a solid are enlarged by a factor of \(k\), then the volume of the figure is multiplied by a factor of \(k^{3}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-27}
\end{center}

surface area \(=6\) volume \(=1\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-27(1)}
\end{center}

surface area \(=6 k^{2}\) volume \(=k^{3}\)

\section*{Example 14}
A spherical balloon has radius \(10 \mathrm{~cm}\). It is inflated so that the radius becomes \(15 \mathrm{~cm}\). By what factor has the following changed?
a surface area
b volume

\section*{Solution}
\section*{a Method 1 - Find the two areas}
For a sphere, \(A=4 \pi r^{2}\)

Hence, surface area of the balloon \(=4 \times \pi \times 10^{2}=400 \pi \mathrm{cm}^{2}\)

If the radius becomes \(15 \mathrm{~cm}\), then the surface area becomes \(4 \times \pi \times 15^{2}=900 \pi \mathrm{cm}^{2}\),

so the surface area has been multiplied by a factor of \(\frac{900 \pi}{400 \pi}=\frac{9}{4}\)

Method 2 - Enlargement factor method

Enlargement factor for the radius is \(\frac{15}{10}=\frac{3}{2}\).

Hence, enlargement for the surface area is \(\left(\frac{3}{2}\right)^{2}=\frac{9}{4}\).

\section*{b Method 1 - Find the two volumes}
For a sphere, \(V=\frac{4}{3} \pi r^{3}\).

Hence, volume of the balloon \(=\frac{4}{3} \times \pi \times 10^{3}=\frac{4000}{3} \pi \mathrm{cm}^{3}\).

If the radius becomes \(15 \mathrm{~cm}\), then the volume becomes \(\frac{4}{3} \times \pi \times 15^{3}=4500 \pi \mathrm{cm}^{3}\), so the volume has been multiplied by a factor of \(4500 \pi \div \frac{4000 \pi}{3}=\frac{27}{8}\).

\section*{Method 2 - Enlargement factor method}
The enlargement factor for the radius is \(\frac{3}{2}\).

Hence, the enlargement factor for the volume is \(\left(\frac{3}{2}\right)^{3}=\frac{27}{8}\).

\section*{Scale drawings and scale models}
A designer, architect or engineer will often build a scale model of the object being designed. The real object is an enlargement of the model. A scale of, for example, \(1: 40\) means that the dimensions of the real object are 40 times the dimensions of the model.

\section*{Example 15}
A farmer builds a scale model of a silo, as shown opposite. The scale of the model is \(1: 100\).

a Find the volume of the model.

b What is the volume of the silo, in cubic metres?

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-28}
\end{center}

\section*{Solution}
a Volume of cone \(=\frac{1}{3} \pi r^{2} h\)

\[
\begin{aligned}
& =\frac{1}{3} \times \pi \times 5^{2} \times 3 \\
& =25 \pi \mathrm{cm}^{3}
\end{aligned}
\]

Volume of cylinder \(=\pi r^{2} h\)

\[
\begin{aligned}
& =\pi \times 5^{2} \times 3 \\
& =75 \pi \mathrm{cm}^{3}
\end{aligned}
\]

\[
\text { Volume of model }=25 \pi+75 \pi
\]

\[
=100 \pi \mathrm{cm}^{3}
\]

b The enlargement factor is 100 , so the volume of the silo is \(100^{3} \times\) volume of the model Hence, the volume of the silo is:

\[
\begin{aligned}
V & =\left(100 \pi \times 100^{3}\right) \mathrm{cm}^{3} \\
& =100 \pi \mathrm{m}^{3} \quad\left(100^{3} \mathrm{~cm}^{3}=1 \mathrm{~m}^{3}\right)
\end{aligned}
\]

\section*{Enlargement}
Let \(k>0\).

\begin{itemize}
  \item If a plane figure is enlarged by a factor of \(k\), then the area of the figure is multiplied by a factor of \(k^{2}\).
  \item If a solid is enlarged by a factor of \(k\), then the surface area of the solid is multiplied by a factor of \(k^{2}\).
  \item If a solid is enlarged by a factor of \(k\), then the volume of the solid is multiplied by a factor of \(k^{3}\).
\end{itemize}

\section*{Exercise 6E}
1 Find what factor the area of the smaller figure must be multiplied by to find the area of the larger figure. Work this out by:

i calculating the areas

ii using the enlargement factor method

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-29(8)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-29(2)}
\end{center}

\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-29(5)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-29(1)}
\end{center}

\(12 \mathrm{~cm}\)
\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-29(4)}

2 i Find what factor the total surface area of the smaller solid must be multiplied by in order to obtain the surface area of the larger solid. In parts \(\mathbf{a}, \mathbf{c}\) and \(\mathbf{e}\), do this by finding the surface areas; in parts \(\mathbf{b}\) and \(\mathbf{d}\), use the enlargement factor method.

ii Find what factor the volume of the smaller solid must be multiplied by in order to obtain the volume of the larger solid. In parts \(\mathbf{b}\) and \(\mathbf{d}\), do this by finding the volumes; in parts \(\mathbf{a}, \mathbf{c}\) and \(\mathbf{e}\), use the enlargement factor method.

\(\mathbf{a}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-29}
\end{center}

\(4 \mathrm{~cm}\)

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-29(6)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-29(3)}
\end{center}

\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-29(7)}
\(\mathbf{c}\)

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-30(4)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-30(2)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-30(3)}
\end{center}

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-30}
\end{center}

3 A cube has a volume of \(343 \mathrm{~cm}^{3}\). The cube is enlarged by a factor of 5 .

a What is the volume of the resulting cube?

b Find the surface area of the resulting cube.

4 By what factor must the radius of a spherical balloon be multiplied if the volume is to be increased from \(760 \mathrm{~cm}^{3}\) to \(389120 \mathrm{~cm}^{3}\) ?

5 A cylindrical container holds \(125 \mathrm{~cm}^{3}\) of liquid when full, and requires \(240 \mathrm{~cm}^{2}\) of material to manufacture. The company wants to increase the height and radius by the same enlargement factor to produce a cylindrical container that can hold \(1000 \mathrm{~cm}^{3}\) of liquid. How much material will be required to produce the new container?

6 A solid, \(A\), is enlarged to form a new solid, \(B\). If the surface area of \(B\) is twice the surface area of \(A\), by what factor is the volume of \(A\) multiplied to give the volume of \(B\) ?

7 A model car has scale \(1: 24\); that is, \(1 \mathrm{~cm}\) on the model represents \(24 \mathrm{~cm}\) on the actual car.

a If the model car requires \(300 \mathrm{~cm}^{2}\) of material to be made, how much material is required to make the actual car?

b If the volume of the actual car's interior is \(4.5 \mathrm{~m}^{3}\), find the volume of the model's interior, in \(\mathrm{cm}^{3}\) and correct to one decimal place.

8 A barn is in the shape of a triangular prism on top of a rectangular prism, as shown below.

a Find the volume of the barn.

b If a model of scale \(1: 50\) is made of the barn, find the volume of the model, in \(\mathrm{cm}^{3}\).

c Find the outside surface area of the barn (not including the floor).

d How much material, in \(\mathrm{cm}^{2}\) and correct to one decimal place, is needed to make the model?

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-30(1)}
\end{center}

9 a The base of a triangle is increased by a factor of 7, while the height is kept the same. What happens to the area of the triangle?

b The sides of the square base of a cube are each increased in length by a factor of 6 , but the height is kept the same. This produces a square prism. By what factor has the volume changed?

c The radius of a cylinder is trebled, but the height is kept the same. By what factor has the volume changed?

d The radius of a cone is multiplied by 9 , but the height is kept the same. By what factor has the volume changed?

\section*{Review exercise}
1 For each solid, calculate:

i the volume

ii the surface area

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-31(2)}
\end{center}

c
\includegraphics[max width=\textwidth, center]{2024_02_22_71b86f97ab17bc51fbe1g-31(6)}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-31(5)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-31(4)}
\end{center}

f

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-31(1)}
\end{center}

2 For each solid, calculate:

i the volume

ii the surface area

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-31(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-31}
\end{center}

3 A solid cylinder has a shaft with a square cross-section through it, as shown in the diagram below. The volume of the solid is \((500 \pi-80) \mathrm{cm}^{3}\). Calculate the length of a side of the square cross-section.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-32(1)}
\end{center}

4 Calculate the volume of each solid.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-32(6)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-32(3)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-32(5)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-32}
\end{center}

5 A triangular pyramid has all its edges equal in length. This is called a regular tetrahedron. Calculate the length of each edge, given that the surface area is \(64 \sqrt{3} \mathrm{~cm}^{2}\).

6 For each solid shown, calculate:

i the volume

ii the surface area

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-32(4)}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-32(2)}
\end{center}

7 A hemispherical bowl is carved out of a solid block of marble, as shown.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-33(4)}
\end{center}

After the bowl is carved, the volume of marble remaining is \(\left(6-\frac{144}{125} \pi\right) \mathrm{m}^{3}\). Calculate the radius of the hemisphere.

8 As part of a major development, an architect designed a building and had a model made to a scale of \(1: 400\); that is, \(1 \mathrm{~cm}\) on the model is \(4 \mathrm{~m}\) on the building.

a The external surface area of the building is \(10600 \mathrm{~m}^{2}\). Calculate the external surface area of the model, in \(\mathrm{cm}^{2}\).

b The volume of the building is \(60000 \mathrm{~m}^{3}\). Calculate the volume of the model, in \(\mathrm{cm}^{3}\).

9 In the pyramid below, \(F\) is the midpoint of \(A B\). \(V F\) is vertical and is \(5 \mathrm{~m}\) in length.

\(A B C D\) is horizontal and rectangular, with \(A B=6 \mathrm{~m}\) and \(B C=8 \mathrm{~m}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-33(3)}
\end{center}

Find:

a the volume of this solid

b the surface area of this solid

10 Find the volume and surface area of the solids.

a

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-33(2)}
\end{center}

c

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-33}
\end{center}

b

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-33(5)}
\end{center}

d

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-33(1)}
\end{center}

\section*{Challenge exercise}
1 A parallelepiped is a six-faced polyhedron, each face of which is a parallelogram. A certain parallelepiped with a square base has volume \(192 \mathrm{~cm}^{3}\). Each side of its base is one-third of its height. Find the length of each side of the base.

2 The figure at the right shows a sphere of radius \(r\) fitting exactly into a cylinder. The sphere touches the cylinder at the top, bottom and curved surface. Show that the surface area of the sphere is equal to the area of the curved surface of the cylinder.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-34(2)}
\end{center}

3 The surface area of a cube is \(x \mathrm{~cm}^{2}\) and its volume is \(y \mathrm{~cm}^{3}\). If \(x=y\), find the length of the side of the cube.

4 A cone has radius \(R\), base area \(A\) and height \(h\). A horizontal slice is taken at a distance \(x\) units from the vertex, as shown.

Let \(A_{x}\) be the area of the circular slice and \(r\) be the radius.

a Use similar triangles to show that \(\frac{r}{R}=\frac{x}{h}\).

b Show that the ratio \(A_{x}: A=x^{2}: h^{2}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-34}
\end{center}

5 A square pyramid has base length \(2 a\), base area \(A\) and height \(h\). A horizontal slice is taken at a distance \(x\) units from the vertex, as shown. Let \(A_{x}\) be the area of the square slice and \(2 b\) be the side length of the square slice. Use the method of question 4 to show that \(A_{x}: A=x^{2}: h^{2}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-34(1)}
\end{center}

6 The portion of a right cone remaining after a smaller cone is cut off it is called a frustum. Suppose that the top and bottom are circles of radius \(r\) and \(R\), respectively. Also, suppose that the height of the frustum is \(h\) and the height of the original cone is \(H\).

a Show that the volume \(V\) of the

frustum is \(\frac{1}{3} \pi\left[H\left(R^{2}-r^{2}\right)+r^{2} h\right]\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-34(3)}
\end{center}

b Use similar triangles to show that \(H=\frac{h R}{R-r}\).

c Deduce that \(V=\frac{1}{3} \pi h\left(R^{2}+r^{2}+r R\right)\).

7 a Use the method of question 6 to show that the volume of a truncated square pyramid with height \(h\) and square base and square top of side lengths \(x\) and \(y\), respectively, is given by \(\frac{1}{3} h\left(x^{2}+y^{2}+x y\right.\) ). (A truncated pyramid is formed in a similar way to a frustum.)
b If the top square has a side length that is half that of the bottom square, what is the ratio of the volume of the truncated pyramid to that of the whole pyramid?

8 A cone has height \(h\) and base radius \(r\).

a Show that the surface area of the cone is \(\pi r\left(r+\sqrt{r^{2}+h^{2}}\right)\).

b Suppose that the height and radius are equal. Show that the surface area is \(\pi r^{2}(1+\sqrt{2})\).

9 The frustum of a cone has base radii \(r\) and \(R\), and the slant height is \(s\).

a Let the slant height of the full cone be \(\ell\).

Show that \(\ell=\frac{s R}{R-r}\).

b Hence, show that the surface area of the curved section is \(\pi(r+R) s\) and that the total surface area is \(\pi\left(r^{2}+R^{2}\right)+\pi(r+R) s\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-35}
\end{center}

c Show that the latter can be written as \(\pi(r+R) \sqrt{(R-r)^{2}+h^{2}}+\pi\left(r^{2}+R^{2}\right)\), where \(h\) is the height of the frustum.

10 Cavalieri's principle states that if we have two solids of the same height and the cross-sections of each solid taken at the same distance from the base have the same area, then the solids have the same volume. Take a hemisphere of radius \(r\) and look at the area of a typical cross-section at height \(h\) above the base.

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-35(3)}
\end{center}

Also, consider a cylinder of height \(r\) and radius \(r\), with a cone cut out of it, also of height \(r\) and base radius \(r\). We also take a cross-section at height \(h\).

a Show that the radius of the circular cross-section of the sphere at height \(h\) is \(\sqrt{r^{2}-h^{2}}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-35(2)}
\end{center}

b Deduce that the area of the cross-section is \(\pi\left(r^{2}-h^{2}\right)\).

c Draw a diagram of the cross-section of the cylinder with the cone removed, and show that the area of the cross-section at height \(h\) is also \(\pi\left(r^{2}-h^{2}\right)\).

d We now conclude that the two solids have the same volume, by Cavalieri's principle. Find the volume of the cylinder minus the cone.

e Deduce the formula for the volume of the sphere.

11 Consider a sphere of radius \(r\) split up into very small pyramids, as shown. The volume of each pyramid \(=\frac{1}{3} A r\), where \(A\) is the area of the base of the pyramid on the surface of the sphere. The base of each pyramid is considered to be a plane surface.

Consider the sum of the volumes of these small pyramids to show that the surface area of the sphere is \(4 \pi r^{2}\).

\begin{center}
\includegraphics[max width=\textwidth]{2024_02_22_71b86f97ab17bc51fbe1g-35(1)}
\end{center}


\end{document}