Volume of a Prism

Worked example

1. 1

   $A=8\times 5$ — The end face is a rectangle 8 cm by 5 cm. A rectangle's area is length times width.

2. 2

   $A=40{\text{ cm}}^2$ — 8 × 5 = 40. Area is in square units, cm².

3. 3

   $V=40\times 3$ — The prism is 3 cm long. Each 1 cm slice holds 40 cm³, and there are 3 slices, so multiply.

4. 4

   $V=120{\text{ cm}}^3$ — 40 × 3 = 120. cm² × cm gives cm³, so volume is in cubic units.

Worked example

1. 1

   $50÷8=6.25\to 6$ — Count along the length. Only whole boxes fit, so round down. A 7th box would stick out.

2. 2

   $40÷8=5$ — Count across the width. 5 boxes fit exactly.

3. 3

   $30÷8=3.75\to 3$ — Count up the height. Round down again: a 4th layer would not fit under the lid.

4. 4

   $6\times 5\times 3=90\text{ boxes}$ — Boxes along × across × up gives the total, just like length × width × height.

5. 5

   $50\times 40\times 30=60000{\text{ cm}}^3$ — Some questions also ask for the empty space. First find the crate's volume.

6. 6

   $8^3=512{\text{ cm}}^3$ — One box is a cube, so its volume is 8 × 8 × 8 = 512.

7. 7

   $90\times 512=46080{\text{ cm}}^3$ — All 90 boxes together take up 46 080 cm³.

8. 8

   $60000-46080=13920{\text{ cm}}^3$ — Crate minus boxes leaves the empty space.

Worked example

1. 1

   $r=4.5,h=12$ — Read off the radius and the height. If you are given the diameter, halve it first.

2. 2

   $A=\pi \times {4.5}^2=20.25\pi$ — The end is a circle, area πr². 4.5 × 4.5 = 20.25. Keep the π for now so nothing is rounded too early.

3. 3

   $V=20.25\pi \times 12=243\pi$ — Multiply the end area by the height. 20.25 × 12 = 243. If the question says “in terms of π”, stop here.

4. 4

   $V=763{\text{ cm}}^3$ — 243π = 763.40…, which is 763 to 3 significant figures.

Worked example

1. 1

   $\pi \times 7^2\times h=1000$ — Put r = 7 and V = 1000 into $V=\pi r^{2}h$.

2. 2

   $49\pi \times h=1000$ — 7² = 49, so the end area is 49π, about 153.9.

3. 3

   $h=1000÷49\pi$ — h is multiplied by 49π, so divide both sides by 49π to undo it.

4. 4

   $h=6.50\text{ cm}$ — 1000 ÷ 153.93… = 6.496…, which is 6.50 to 3 s.f.

Worked example

1. 1

   $x\times x\times 12=432$ — The base is a square, so each side is x. Volume = length × width × height.

2. 2

   $x^2\times 12=432$ — x × x is written x².

3. 3

   $x^2=432÷12=36$ — x² is multiplied by 12, so divide both sides by 12 to undo it.

4. 4

   $x=\sqrt{36}=6\text{ cm}$ — x is squared, so square root to undo it. A length cannot be negative, so ignore −6.

Worked example

1. 1

   $\pi \times r^2\times 15=850$ — Put h = 15 and V = 850 into $V=\pi r^{2}h$.

2. 2

   $15\pi \times r^2=850$ — Join the numbers: π × 15 is 15π, about 47.12.

3. 3

   $r^2=850÷15\pi =18.04\dots$ — Divide both sides by 15π to leave r² alone. Keep every digit on your calculator.

4. 4

   $r=\sqrt{18.04\dots }=4.25\text{ cm}$ — Square root undoes the square. 4.247… is 4.25 to 3 s.f.

Worked example

1. 1

   $A=\frac{1}{2}\times 6\times 5$ — The end is a triangle with base 6 cm and height 5 cm. Triangle area is half of base × height.

2. 2

   $A=15{\text{ cm}}^2$ — 6 × 5 = 30, and half of 30 is 15.

3. 3

   $V=15\times 14=210{\text{ cm}}^3$ — Multiply the end area by the length of the prism, 14 cm.

Worked example

1. 1

   $A=\frac{1}{2}(5+9)\times 4$ — The end is a trapezium. Add the parallel sides, halve, then times by the gap between them.

2. 2

   $A=\frac{1}{2}\times 14\times 4=28$ — 5 + 9 = 14. Half of 14 is 7, and 7 × 4 = 28 cm².

3. 3

   $28\times L=630$ — Volume = end area × length. The volume is 630, so call the length L.

4. 4

   $L=630÷28=22.5\text{ cm}$ — L is multiplied by 28, so divide by 28 to undo it.

Worked example

1. 1

   $r=10÷2=5$ — The straight edge is the diameter. Halve it to get the radius.

2. 2

   $A=\frac{1}{2}\times \pi \times 5^2$ — The end is half a circle, so its area is half of πr².

3. 3

   $A=12.5\pi {\text{ cm}}^2$ — 5² = 25, and half of 25 is 12.5.

4. 4

   $V=12.5\pi \times 30=375\pi$ — Multiply the end area by the length, 30 cm. 12.5 × 30 = 375.

5. 5

   $V=1180{\text{ cm}}^3$ — 375π = 1178.09…, which is 1180 to 3 s.f.

Worked example

1. 1

   $V=60\times 30\times 40$ — The tank is a cuboid, so volume = length × width × height, all in cm.

2. 2

   $V=72000{\text{ cm}}^3$ — 60 × 30 = 1800, and 1800 × 40 = 72 000.

3. 3

   $72000÷1000=72\text{ litres}$ — 1 litre is 1000 cm³, so divide by 1000 to change cm³ into litres.

4. 4

   $72÷4=18\text{ minutes}$ — The tap gives 4 litres every minute. Dividing by 4 counts how many minutes it needs.

Worked example

1. 1

   $20\times 15\times 10=3000{\text{ cm}}^3$ — Volume of the water in the box: length × width × height.

2. 2

   $\pi \times 8^2\times h=3000$ — In the cylinder the same 3000 cm³ fills a circle of radius 8 up to a depth h.

3. 3

   $64\pi \times h=3000$ — 8² = 64, so the end area is 64π, about 201.1.

4. 4

   $h=3000÷64\pi =14.9\text{ cm}$ — Divide by 64π to undo it. 14.92… is 14.9 to 3 s.f.

Worked example

1. 1

   $5^3=125{\text{ cm}}^3$ — The cube pushes up its own volume of water: 5 × 5 × 5 = 125.

2. 2

   $\pi \times 6^2\times d=125$ — That extra water is a thin cylinder of radius 6 on top. Call its depth d.

3. 3

   $36\pi \times d=125$ — 6² = 36, so the end area is 36π, about 113.1.

4. 4

   $d=125÷36\pi =1.11\text{ cm}$ — Divide by 36π. 1.105… is 1.11 to 3 s.f., so the water rises 1.11 cm.