1. Teach GCSE Maths
Volumes of Prisms

2. Volumes of Prisms Wednesday, 19 April 2017 © Christine Crisp
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© Christine Crisp

3. triangular prism hexagonal prism triangular prism
These diagrams show prisms.
triangular prism
If we slice them like this . . .
hexagonal prism
triangular prism
the cut surfaces are always the same size and shape as the ends.

4. triangular prism hexagonal prism triangular prism
These diagrams show prisms.
The shape of the cut surface is called a cross-section.
triangular prism
hexagonal prism
triangular prism
The volume of any prism is given by
volume of prism = area of cross-section  length

5. e.g. 1 Find the volume of this hexagonal prism.
200 cm2
20 cm
Solution:
volume of prism = area of cross-section  length
So, volume =
200  20
= 4000 cm3
Reminder: The capacity in litres is found by dividing by 1000.
4000 ÷ 1000 = 4 litres.

6. e.g. 2 Find the volume of this triangular prism.
6 cm
10 cm
4 cm
Solution:
Volume of prism = area of cross-section  length
base  height 2 =
4  6 2
Area of triangle =
= 12 cm2
So, volume =
12  10 = 120 cm3

7. We can use the same method to find the volume of any solid with a cross-section that doesn’t change.
e.g.3 Find
(a) the area of the shaded face of the shape shown and
(b) find the volume
80cm
20cm
40cm
50cm

8. 3 2 1 50  20 = 1000 cm2 80  30 = 2400 cm2 4400 cm2 Solution: 20cm 30
(a) Find the area of the shaded face.
Split the face into 3 rectangles:
Area of 1st rectangle =
50  20 =
1000 cm2
Area of 2nd rectangle =
80 
30 =
2400 cm2
Area of 3rd rectangle = Area of 1st = 1000 cm2
Area of face = =
4400 cm2

9. = 4400 cm2 = 4400  40 = 176 000 cm3 Solution: 20cm 50 50cm 50 80cm
Area of shaded face
= 4400 cm2
(b) Find the volume.
Volume = area of cross-section  length
= 4400  40
= cm3

10. A cylinder has a cross-section that is a circle.
e.g. 4 Find the volume of this cylinder. Give your answer correct to 3 significant figures.
0·5 m
1·5 m
A cylinder has a cross-section that is a circle.
Solution:
Area of circle =
p r2 =
p (1·5)2 =
7·069 m2
So, volume =
7·069  0·5 =
3·53 m3
( 3 s.f. )

11. 1 litre = 1000 millilitre (ml)
When we measure the volume of a liquid we often use the word capacity and measure in litres or millilitres.
( 1 litre is about 1¾ pints. )
1 litre = 1000 millilitre (ml)
A capacity of 1 ml is the same as a volume of 1 cm3.
Tell your partner how you would change from ml to litres.
Ans: Litres are bigger units so we need fewer of them. We divide by 1000.

12. 20 = 40  30  24 000cm3 24 000 ml = 24 000 ÷ 1000 litres = 24 litres
e.g. Find the capacity of the following cuboid giving your answer in litres.
40cm
30cm
20cm
Solution:
Vol =
20 =
40 
30 
24 000cm3
Capacity = ml
= ÷ 1000 litres
= 24 litres

13. Volume = area of cross-section  length
SUMMARY
A prism has the same cross-section throughout its length.
The cross-section of a prism can be any polygon.
The volume of a prism is given by
Volume = area of cross-section  length
We can find the volume of other solids with the same cross-section in the same way.
Reminder:
Capacity is the volume of liquid a vessel can hold. It is measured in litres.
To change from cm3 to litres we divide by 1000.
( Litres are bigger units so we need fewer of them. )

14. EXERCISE
1. Find the volumes of the following solids:
(a)
(b)
15 m
10 m
1 m
2 m
4 cm
8 cm
5 cm
For (b), find also the capacity in litres.

15. 4  5 = = 10 cm2 10  8 = 80 cm3 EXERCISE Solution: (a) 5 cm 8 cm 4 cm
Volume of prism = area of cross-section  length
base  height 2 =
4  5 2
Area of triangle =
= 10 cm2
So, volume =
10  8 = 80 cm3

16. = average of parallel sides  distance apart
EXERCISE
Solution:
(b)
15 m
10 m
1 m
2 m
Cross-section:
Area of trapezium
= average of parallel sides  distance apart
= ½(2 + 1)  10
= 15 m2
So, volume = 15  15 = 225 m3
Changing to cm3, volume = 225  100  100  100 cm3
= cm3
Capacity = ÷ 1000 = litres

17. p r2 = p (0·5)2 = 0·785 m2 0·785  1 = 0·785 m3 ( 3 s.f. ) EXERCISE
2. Find the volume of the cylinder giving the answer correct to 3 significant figures.
Solution:
1 m
50 cm
We can work with metres or centimetres.
Using metres, the radius of the base is 0·5 m.
Area of circle =
p r2 =
p (0·5)2 =
0·785 m2
So, volume =
0·785  1 =
0·785 m3
( 3 s.f. )
In cm3, the answer is cm3

18. EXERCISE
3. Find
(a) the area of the shaded face of the solid giving the answer in m2 and
(b) the volume of the solid, giving the answer in m3.
(b) 2m 1m
50cm

19. 50 cm = ½ m EXERCISE Solution: (b) 2m 50cm 1 3 2 1m 50cm 1m 50cm
(a) The area of the shaded face of the solid giving the answer in m2:
Area 1 = 2  ½ = 1 m2
Area 2 = Area 3 = ½  ½ = ¼ m2
Area of shaded face = 1 + ¼ + ¼ = 1·5 m2

20. 50 cm = ½ m = 1·5  1 = 1·5 m3 EXERCISE Solution: (b) 2m 50cm 1 3 2 1m
(b) Volume of the solid of the solid:
Volume = area of cross-section  height
= 1·5  1
= 1·5 m3