Volume of 3D Shapes Worksheets, Questions and Revision (2024)

Question 1:The diagram shows the outline of a cuboid, ABCDEFGH. Given that: AB=3 cm, BC=12 cm, and CF=16 cm. Find the volume of the cuboid.

[3 marks]

Level 4-5GCSEKS3AQAEdexcelOCRCambridge iGCSE

To find the volume we simply have to multiply all three lengths together:

\text{volume}= 3\times12 \times16=576\text{cm}^3

Question 2: The diagram shows the outline of a square based pyramid, ABCDE. The vertical height of the pyramid is 12 m and the base length, CD, is 5 m. Find the volume of the square based pyramid.

[3 marks]

Level 4-5GCSEAQAEdexcelOCRCambridge iGCSE

The volume of a square base pyramid is given by the following formula:

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

Substituting the values given in the question into the equation above, we find that,

\text{volume}= \dfrac{1}{3} \times 5^{2} \times 12 =100 \text{m}^3

Question 3: Below is a trapezoidal prism. The trapezium’s parallel sides have lengths 45 cm and 60 cm and its perpendicular height is 20 cm. The length of the prism is 80 cm. Work out the volume of the prism.

[3 marks]

Level 4-5GCSEKS3AQAEdexcelOCRCambridge iGCSE

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 80 cm, so we get:

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

Question 4: Below is a triangle-based pyramid. The base of the pyramid has area 18 cm^2 and the height of the pyramid is x+5 cm^2. The volume of the pyramid is 54 cm^2. Work out the value of x.

[4 marks]

Level 4-5GCSEAQAEdexcelOCRCambridge iGCSE

This may seem a little different, but it actually comes up a lot.

First, 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 found 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}

Question 5: The composite shape below was made by attaching a hemisphere (half of a sphere) to the top of a cylinder. The base of the cylinder is 2.3 m and the height of the cylinder is 5.6 m. The formula for the volume of a sphere is \dfrac{4}{3}\pi r^3 . Work out the volume of the whole shape to 3 significant figures.

[4 marks]

Level 4-5GCSEAQAEdexcelOCRCambridge iGCSE

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

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 \approx 93.07 \text{m}^3

Note: 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)\approx 25.48\text{m}^3

Therefore, the total volume of the shape is:

93.0665...+25.4825...=119\text{ m}^3 \text{ (3s.f.)}