Space, shape and measurement: Calculate the surface area and volume of two- and three-dimensional shapes

# Unit 1: Finding the volume of 3-D shapes

Natashia Bearam-Edmunds

### Unit outcomes

By the end of this unit you will be able to:

- Apply the appropriate formulae to find the volume of:
- right pyramids (with square, equilateral triangle or regular hexagonal bases)
- right cones
- spheres.

- Determine and calculate the volume of combinations of these geometric objects.

## What you should know

Before you start this unit, make sure you can:

- calculate perimeter and area of two-dimensional shapes, and surface area and volume of prisms. You can revise this in level 2 subject outcome 3.2.

## Introduction

**Volume** is the amount of three-dimensional (3-D) space taken up by a solid, liquid or gas. Volume is measured in cubes or cubic units. The terms volume and capacity are often used to mean the same thing but there is a slight difference.

**Capacity** is the amount of liquid or other substance a container can hold. Capacity is only used in relation to containers. A can of cooldrink has volume, as it takes up a certain amount of space and has capacity to be filled with liquid. So, when you are filling a container with liquid or other substances you are working with capacity.

We saw in level 2 that we find the volume of right prisms and cylinders by multiplying the area of the base by the height of the 3-D shape. Now, we will find the volume of right pyramids, cones and spheres.

## Pyramids

Pyramids have only one polygon base and the side faces are triangles that meet at a vertex called the apex. There are many types of pyramids, and they are named after the shape of their base. We call a pyramid a “right pyramid” if the line between the apex and the centre of the base is perpendicular to the base.

For example, in the **square based pyramid** the base is a square and the sides are isosceles triangles, which meet at the apex.

A **tetrahedron or triangular pyramid** has a triangular base and sides. It is made up of four equilateral triangles, has six edges and four vertices.

A **hexagonal pyramid** has a hexagonal base and six isosceles triangle faces as its sides.

To find the volume of a pyramid we divide the area of its base by three and multiply by its height.

[latex]\scriptsize \displaystyle \text{Volume of pyramid}=\displaystyle \frac{1}{3}\text{area of base}\times \text{height}[/latex]

### Example 1.1

Find the volume of a regular square pyramid with base sides [latex]\scriptsize \displaystyle \text{5 cm}[/latex] and altitude [latex]\scriptsize \displaystyle 10\text{ cm}[/latex].

*Solution*

First, we find the area of the base, which is a square.

[latex]\scriptsize \displaystyle \begin{align*}\text{Area of square base}&=5\text{ cm}\times 5\text{ cm}\\&=25\text{ c}{{\text{m}}^{2}}\end{align*}[/latex]

Altitude is the same as the height. Now we multiply the area of the base by the height of the pyramid and divide by three to find its volume.

[latex]\scriptsize \displaystyle \begin{align*}\text{Volume }&=\displaystyle \frac{1}{3}(25\text{ c}{{\text{m}}^{2}})\times 10\text{ cm}\\&=83.3\text{ c}{{\text{m}}^{3}}\end{align*}[/latex]

### Example 1.2

A paperweight is in the shape of a hexagonal pyramid with base sides [latex]\scriptsize \displaystyle \text{12 cm}[/latex] and a vertical height of [latex]\scriptsize \displaystyle \text{15 cm}[/latex]. Find the volume of the paperweight.

*Solution*

The area of a hexagon is given by: [latex]\scriptsize \displaystyle \frac{{3\sqrt{3}}}{2}{{\text{s}}^{2}}[/latex].

[latex]\scriptsize \displaystyle \begin{align*}\text{Volume }&=\displaystyle \frac{1}{3}\left[ {\displaystyle \frac{{3\sqrt{3}}}{2}{{{(12\text{ cm})}}^{2}}} \right]\times 15\text{ cm}\\&=1\text{ }870.61\text{ c}{{\text{m}}^{3}}\end{align*}[/latex]

### Exercise 1.1

- The figure below shows a triangular pyramid shaped container with altitude of [latex]\scriptsize 2\text{ m}[/latex]. The side length is [latex]\scriptsize \text{3 m}[/latex] and the height of the triangle is [latex]\scriptsize 2.6\text{ m}[/latex].

- Find the volume of the container.
- The container must be packed into a box. If the box has a volume of [latex]\scriptsize 2.5\text{ }{{\text{m}}^{3}}[/latex] will the container fit into the box?

- We need to fit square pyramid ornaments with base side of [latex]\scriptsize 10\text{ mm}[/latex] and height of [latex]\scriptsize \text{5 mm}[/latex] into a rectangular box to be transported from Cape Town to Johannesburg. If the box has a length of [latex]\scriptsize 25\text{ cm}[/latex], a breadth of [latex]\scriptsize 15\text{ cm}[/latex]and a height of [latex]\scriptsize 5\text{ cm}[/latex], how many ornaments will fit into the box?
- The solid below is made of a cube and a square pyramid. Determine the volume of the shape.

The full solutions are at the end of the unit.

## Cones

Cones are similar to pyramids except that their bases are circles instead of polygons.

We find the volume of a cone using a similar formula to that for right pyramids.

[latex]\scriptsize \displaystyle \begin{align*}\text{Volume of cone}&=\displaystyle \frac{1}{3}\text{area of base}\times \text{height of cone}\\ &= \displaystyle \frac{1}{3}\pi {{r}^{2}}H\end{align*}[/latex]

### Example 1.3

Find the volume of the following cone (correct to 1 decimal place).

*Solution*

First, find the area of the base.

[latex]\scriptsize \displaystyle \begin{align*}A&=\pi {{r}^{2}}\\&=\pi {{(5\text{ cm)}}^{2}}\\&=25\pi \text{ c}{{\text{m}}^{2}}\end{align*}[/latex]

Next, calculate the volume of the cone.

[latex]\scriptsize \displaystyle \begin{align*}V&=\displaystyle \frac{1}{3}\text{area of base}\times \text{height of cone}\\ &= \displaystyle \frac{1}{3}(25\pi )\times 15\text{ }\\& =125 \pi \text{ c}{{\text{m}}^{3}}\\&=392.7\text{ c}{{\text{m}}^{3}}\end{align*}[/latex]

## Spheres

Spheres are solids that are perfectly round and look the same from any direction.

### Note

Watch this video to see an example of calculating the volume of a sphere.

### Example 1.4

A world globe is in the shape of a sphere with radius of [latex]\scriptsize 12\text{ cm}[/latex]. Find its volume correct to one decimal place.

*Solution*

[latex]\scriptsize \begin{align*}{{V}_{{\text{sphere}}}}&=\displaystyle \frac{4}{3}\pi {{r}^{3}}\\&=\displaystyle \frac{4}{3}\pi {{(12)}^{3}}\\&=7238.2\text{ c}{{\text{m}}^{3}}\end{align*}[/latex]

### Exercise 1.2

- Tom has [latex]\scriptsize 20[/latex] new marbles. If the marbles are spherical in shape and each one has a diameter of [latex]\scriptsize 4\text{ mm}[/latex] what is the total volume of Tom’s marbles?
- The vertical height of a cone is [latex]\scriptsize 7[/latex] units and the slant height is [latex]\scriptsize 7.28[/latex] units; the radius of the cone is [latex]\scriptsize 2[/latex] units. Calculate the volume of the cone. Round your answer off to two decimal places.
- An open rectangular fish tank, [latex]\scriptsize 80\text{ cm}[/latex] long, [latex]\scriptsize \text{35 cm}[/latex] wide and [latex]\scriptsize \text{45 cm}[/latex] high, is full to the top with water. John puts [latex]\scriptsize 30[/latex] marbles, each with a diameter of [latex]\scriptsize \text{1 cm}[/latex], into the tank resulting in some of the water spilling out of the fish tank.

- Determine the initial volume of the water in the tank.
- Calculate the volume of one marble.
- Calculate the volume of the water that spilled out of the tank after all the marbles have been added in.

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

- How to find the volume of right pyramids.
- How to find the volume of spheres.
- How to find the volume of cones.

# Unit 1: Assessment

#### Suggested time to complete: 30 minutes

- A solid is made up of a cube and a square pyramid. Find its volume if the height of the solid is [latex]\scriptsize 11\text{ cm}[/latex] and the sides of the base are [latex]\scriptsize \text{5 cm}[/latex].

- An insecticide company has mistakenly manufactured cube-shaped moth balls instead of spherical-shaped moth balls. The cube-shaped moth balls have a side length of [latex]\scriptsize 16\text{ mm}[/latex]. The spherical-shaped moth balls must have a diameter of [latex]\scriptsize \text{20 mm}[/latex].

Calculate:- The volume of each cube-shaped moth ball.
- The volume of one spherical-shaped moth ball.
- The extra amount of material required if the original order was for [latex]\scriptsize 2[/latex] million [latex]\scriptsize \displaystyle (2\text{ }000\text{ }000)[/latex]spherical moth balls.

- A solid golden sphere with a diameter of [latex]\scriptsize \displaystyle 14\text{ mm}[/latex] is to be melted and cast into a piece of jewellery. The piece of jewellery is in the shape of a right pyramid with a square base. The square pyramid has a vertical height of [latex]\scriptsize \displaystyle 10\text{ mm}[/latex] and a side length of [latex]\scriptsize \displaystyle 20\text{ mm}[/latex]. Calculate the excess gold (gold left over) after the piece of jewellery has been made.

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

- .
- .

[latex]\scriptsize \begin{align*}V&=\displaystyle \frac{1}{3}\text{area of base}\times \text{height}\\ \therefore V&= \displaystyle \frac{1}{3}\left( {\displaystyle \frac{1}{2}(3)\times 2.6} \right)\times 2\\&=2.6\text{ }{{\text{m}}^{3}}\end{align*}[/latex] - The container will not fit into the box since its volume of [latex]\scriptsize 2.7\text{ }{{\text{m}}^{3}}[/latex] is greater than the volume of the box.

- .
- Remember to convert lengths from millimetres to centimetres.

[latex]\scriptsize \displaystyle \begin{align*}{{V}_{{\text{pyramid}}}}&=\displaystyle \frac{1}{3}\text{area of base}\times \text{height}\\&=\displaystyle \frac{1}{3}(1\text{ cm}\times 1\text{ cm})\times 0.5\text{ cm}\\ &= \displaystyle \frac{1}{6}\text{ c}{{\text{m}}^{3}}\end{align*}[/latex]

[latex]\scriptsize \displaystyle \begin{align*}{{V}_{{\text{box}}}}&=25\text{ cm}\times 15\text{ cm}\times 5\text{ cm}\\ &= \text{ 1875 c}{{\text{m}}^{3}}\end{align*}[/latex]

Number of ornaments:

[latex]\scriptsize \begin{align*}\displaystyle \frac{{{{V}_{{\text{box}}}}}}{{{{V}_{{\text{ornaments}}}}}}&=1875\div \displaystyle \frac{1}{6}\\&=11\text{ }250\end{align*}[/latex] - .

[latex]\scriptsize \begin{align*}{{V}_{{\text{cube}}}}&={{s}^{3}}\\ &={{6}^{3}}\\&=216\text{ c}{{\text{m}}^{3}}\\ {{V}_{{\text{pyramid}}}}&=\displaystyle \frac{1}{3}{{\left( 6 \right)}^{2}}\times (15-6)\\ &=108\text{ c}{{\text{m}}^{3}}\\ \text{Total volume:}\text{ }\\ 216\text{ c}{{\text{m}}^{3}}+108\text{ c}{{\text{m}}^{3}}&=324\text{ c}{{\text{m}}^{3}} \end{align*}[/latex]

### Exercise 1.2

- Remember to divide [latex]\scriptsize 4\text{ mm}[/latex] by two to get the length of the radius.

[latex]\scriptsize \displaystyle \begin{align*}V&=\displaystyle \frac{4}{3}\pi {{r}^{3}}\\&=\displaystyle \frac{4}{3}\pi {{(2)}^{3}}\\&=\displaystyle \frac{{32}}{3}\pi \text{ m}{{\text{m}}^{3}}\text{ volume of each marble}\\ \text{Total volume:}\\\displaystyle \frac{{32}}{3}\pi \text{ m}{{\text{m}}^{3}}\times 20&=670.2\text{ m}{{\text{m}}^{3}}\end{align*}[/latex] - .

[latex]\scriptsize \displaystyle \begin{align*}\text{Volume of cone}&=\displaystyle \frac{1}{3}\text{area of base}\times \text{height of cone}\\ &= \displaystyle \frac{1}{3}\pi {{(2)}^{2}}\times 7\\&=29.32\text{ cubic units}\end{align*}[/latex] - .
- Volume of the water in the tank:

[latex]\scriptsize \begin{align*}V&=l\times b\times h\\&=80\text{ cm}\times 35\text{ cm}\times 45\text{ cm}\\&=126\text{ }000\text{ c}{{\text{m}}^{3}}\end{align*}[/latex] - Volume of one marble:

Radius is [latex]\scriptsize 0.5\text{ cm}[/latex]

[latex]\scriptsize \displaystyle \begin{align*}V&=\displaystyle \frac{4}{3}\pi {{r}^{3}}\\&=\displaystyle \frac{4}{3}\pi {{(0.5\text{ cm})}^{3}}\\&=\displaystyle \frac{1}{6}\pi \text{ c}{{\text{m}}^{3}}\\&=0.52\text{ c}{{\text{m}}^{3}}\end{align*}[/latex] - Total volume of marbles will give the amount of water that spilled over:

[latex]\scriptsize \displaystyle \displaystyle \frac{1}{6}\pi \text{ c}{{\text{m}}^{3}}\times 30=15.71\text{ c}{{\text{m}}^{3}}[/latex]

- Volume of the water in the tank:

### Unit 1: Assessment

- .

[latex]\scriptsize \begin{align*}{{V}_{{\text{cube}}}}&={{s}^{3}}\\&={{5}^{3}}\\&=125\text{ c}{{\text{m}}^{3}}\\{{V}_{{\text{pyramid}}}}&=\displaystyle \frac{1}{3}{{\left( 5 \right)}^{2}}\times (11-5)\\&=50\text{ c}{{\text{m}}^{3}}\\\text{Total volume:}\text{ }\\125\text{ c}{{\text{m}}^{3}}+50\text{ c}{{\text{m}}^{3}}&=175\text{ c}{{\text{m}}^{3}}\end{align*}[/latex] - .
- .

[latex]\scriptsize \begin{align*}{{V}_{{\text{cube}}}}&={{s}^{3}}\\&={{16}^{3}}\\&=4\text{ }096\text{ m}{{\text{m}}^{3}}\end{align*}[/latex] - .

Diameter of [latex]\scriptsize \text{20 mm}[/latex][latex]\scriptsize \therefore[/latex]radius of [latex]\scriptsize 10\text{ mm}[/latex].

[latex]\scriptsize \displaystyle \begin{align*}V&=\displaystyle \frac{4}{3}\pi {{r}^{3}}\\&=\displaystyle \frac{4}{3}\pi {{(10)}^{3}}\\&=\displaystyle \frac{{4\text{ }000}}{3}\pi \text{ m}{{\text{m}}^{3}}\\&=4188.79\text{ m}{{\text{m}}^{3}}\text{ correct to 2 decimal places}\end{align*}[/latex] - .

[latex]\scriptsize \displaystyle 2\text{ }000\text{ }000\times (\displaystyle \frac{{4000}}{3}\pi \text{ m}{{\text{m}}^{3}}-4\text{ }096\text{ m}{{\text{m}}^{3}})=185\text{ }580\text{ }409\text{ m}{{\text{m}}^{3}}[/latex]

- .
- .

[latex]\scriptsize \displaystyle \begin{align*}{{V}_{{\text{sphere}}}}&=\displaystyle \frac{4}{3}\pi {{r}^{3}}\\&=\displaystyle \frac{4}{3}\pi {{(7)}^{3}}\\&=\displaystyle \frac{{1\text{ 372}}}{3}\pi \text{ m}{{\text{m}}^{3}}\\\\{{V}_{{\text{pyramid}}}}&=\displaystyle \frac{1}{3}{{(20)}^{2}}\times 10\\&=\displaystyle \frac{{4\text{ }000}}{3}\text{ m}{{\text{m}}^{3}}\\\text{Excess gold:}\\\displaystyle \frac{{1\text{ 372}}}{3}\pi \text{ m}{{\text{m}}^{3}}-\displaystyle \frac{{4\text{ }000}}{3}\text{ m}{{\text{m}}^{3}}&=103.42\text{ m}{{\text{m}}^{3}}\end{align*}[/latex]

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- Square pyramid © kids-flashcards.com is licensed under a All Rights Reserved license
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- Ex 1.1 Q1 © kids-flashcards.com is licensed under a All Rights Reserved license
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- Ex 1.2 Q3 © DHET is licensed under a CC BY (Attribution) license
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- Assess Q2 © DHET is licensed under a CC BY (Attribution) license