**Volume of 3D Shapes Maths ** *Revision and Worksheets*

## What you need to know

The **volume** of a 3D shape is how much space it takes up. You are expected to know how to work out the volumes of a bunch of different shapes, and some of these will come with a formula attached to the question, and some won’t:

- You will have to remember how to find the volumes of:
**prisms**and**pyramids**. - You will be given the formulae for finding the volumes of:
**cones**and**spheres**.

Here, we’ll go through a series of examples of how to work out (and work with) different volumes, clearly outlining the formulae that you don’t have to remember, and the ones you don’t.

**Example: **Below is a triangular prism. The triangular face has base 6cm and perpendicular height 5cm. The prism has length 3.5cm. Work out the volume of the shape.

The formula for the **volume of a prism** is

\text{volume of prism }=\text{ area of cross section }\times\text{ length}

This is for **any prism**, including **cuboids **&** cylinders**, and you do **have to remember this formula**. In this case, the cross section is a triangle, so we need to multiply the area of the triangle by the length. We get:

\text{area of cross section }=\dfrac{1}{2}\times5\times6=15\text{cm}^2

Therefore,

\text{volume of the prism }=15\times3.5=52.5\text{cm}^3.

**Example: **Below is a square-based pyramid. The base has side-length 14mm and the pyramid has perpendicular height 25mm. Work out the volume of the pyramid.

The formula for the **volume of a pyramid **is

\text{volume of pyramid }=\dfrac{1}{3}\text{ area of base }\times\text{ perpendicular height}

This is for **any pyramid** with **any shape of base**, and you do **have to remember this formula**. In this case, the base is a square, so we have to find one third of the area of the square times by the height. We get:

\text{area of base }= 14^2=196\text{mm}^2

Therefore,

\text{volume of the pyramid }=\dfrac{1}{3} \times 196\times25=1633.3 \text{mm}^3

Technically, a cone is a type of pyramid, and can use this method to work out its volume, but you *are* given the formula for the volume of a cone (if you need it in the question).

In this next example, we’ll use a volume formula to find a missing dimension – a common question.

**Example: **Below is a sphere with volume 2,100\text{cm}^3. Work out the radius to 2sf. The formula for the volume of a sphere is \dfrac{4}{3}\pi r^3.

So, we are given the formula, as you would be in an exam question. Now, we know the volume is 2,100, so to find r (the **radius**) we must set the formula we’ve been given equal to 2,100 and rearrange for r. So, we get

\dfrac{4}{3}\pi r^3=2,100

Firstly, multiply both sides by 3 to get

4\pi r^3=6,300

Then, we will divide both sides by 4\pi to get

r^3=\dfrac{6,300}{4\pi}

Finally, to find r, we will get rid of the power of 3 by cube rooting both sides. Putting this into a calculator, we get

r=\sqrt[3]{\dfrac{6,300}{4\pi}}=7.9440...=7.9\text{cm (2sf)}

The final example will be a **composite shape**, which is a shape that is a combination of multiple different shapes. In reality, if you look around you’ll see that most objects aren’t just one shape, they’re a combination. As a result, this is a popular style of question.

**Example: **The shape below was made by attaching a cone to the top of a cylinder. The base of the cylinder has radius 4mm, the height of the cylinder portion is 3mm, and the height of the cone portion is 5.5mm. The formula for the volume of a cone is \dfrac{1}{3}\pi r^2 h. Work out the volume of the whole shape to 1dp.

So, to work out the volume of the shape, we need to work out the two volume separately.

Firstly, the cylinder is a type of prism, so we know we need to multiply the area of the cross section by its length. Here, the cross section is a circle with radius 4mm, and the length of the cylinder is 3mm, so we get

\text{volume of cylinder}=\pi\times4^2\times3=48\pi

We’ll worry about the rounding at the end. Next, we have to work out the volume of the cone part, and fortunately the person who wrote the question (me) has been kind enough to give us the formula for a cone/it’s there because the exam boards decided it would be.

So, h is the height of the cone, which we know to be 5.5, and r is the radius of the cone, which is the same as that of the base: 4mm. Therefore, we get

\text{volume of cone }=\dfrac{1}{3}\pi\times4^2\times5.5=\dfrac{88}{3}\pi

Then, the volume of the shape is the sum of these two answers:

\text{volume of whole shape }=48\pi+\dfrac{88}{3}\pi=242.9498...=242.9\text{mm}^3\text{ (1dp)}

## Example Questions

1) Below is a prism whose cross section is a trapezium. The trapezium’s parallel sides have lengths 45cm and 60cm and its perpendicular height is 20cm. The length of the prism is 80cm. Work out the volume of the prism.

So, to work out the volume of a prism we must multiply the **area of the cross section** by the **length**. In this case, the cross section is a trapezium, and the area of the trapezium is

\text{area of cross section }=\dfrac{1}{2}\times(45+60)\times20=1,050\text{ cm}^2

The length of the prism is 80cm, so we get

\text{volume of prism }=1,050\times80=84,000\text{ cm}^3

2) Below is a triangle-based pyramid. The base of the pyramid has area 18\text{cm}^2 and the height of the pyramid is x+5\text{ cm}. The volume of the pyramid is 54\text{cm}^2. Work out the value of x. (HINT: start by writing the volume of the shape in terms of x)

This may seem a little different, but it actually comes up a lot. As the hint says, we must write the volume in terms of x. The volume of a pyramid is **one third** of the **area of the base** times by the **perpendicular height**. We know the area of the base is 18\text{cm}^2, and the expression we’re given for the height is x+5, so the volume is

\dfrac{1}{3}\times18\times(x+5)=6(x+5)

Now, the question also gave us the volume: 54\text{cm}^3, so we can equate this value to the expression we got above, and voila, we have an equation:

6(x+5)=54

Now, we solve this equation to find x. First, divide both sides by 6 to get

x+5=54\div6=9

Then, subtracting 5 from both sides we get the answer to be

x=9-5=4\text{cm}.

3) The shape below was made by attaching a hemisphere* to the top of a cylinder. The base of the cylinder is 2.3m and the height of the cylinder portion is 5.6m. The formula for the volume of a sphere is \dfrac{4}{3}\pi r^3 . Work out the volume of the whole shape to 3sf.

*A hemisphere is half a sphere.

We will work out the volume of the cylinder first, then the hemisphere, and add together the values.

So, a cylinder is a prism which means that to find the volume, we must multiply the area of the circular cross section by its length. The radius of the circle is 2.3m and the length is 5.6m, so we get

\text{volume of cylinder }=\pi \times (2.3)^2\times5.6=93.0665...

Keep the full answer stored in the calculator for adding it to the other value at the end.

Next, we are given the formula for the volume of a sphere, so to find the volume of the hemisphere we will use this formula and then half the result. The radius of the hemisphere is the same as that of the cylinder, 2.3, so we get

\text{volume of hemisphere }=\dfrac{1}{2}\times\left(\dfrac{4}{3}\pi\times(2.3)^3\right)=25.4825...

Therefore, the total volume of the shape is

93.0665...+25.4825...=118.549...=119\text{ m}^3\text{ (3sf)}

## Volumes of 3D shapes Revision and Worksheets

## Volumes of 3D shapes Teaching Resources

Looking for volumes of shapes questions? Are you searching for revision materials for volumes of shapes? Well you are on the right site. Maths Made Easy has pulled together many expert resources to help students, teachers and tutors all access exceptional volume of shapes revision materials. So whether you are looking for the formula for the volume of a sphere or you want revision questions for a student, you will be able to find it here. If you want a full GCSE Maths formula page, you can view these on a Maths Made Easy formula PDF.