Volume and surface area of prisms – Year 8 maths worksheet answers to download as PDF

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1. Answer to Question 1

   Volume of a cuboid = length × width × height. Surface area = 2 × (sum of the areas of the three different faces).

   1. \(V=8\times5\times4=\) 160 cm³. \(SA=2(40+32+20)=\) 184 cm².
   2. \(V=6^3=\) 216 cm³. \(SA=6\times36=\) 216 cm².
   3. \(V=2.5\times1.2\times0.8=\) 2.4 m³. \(SA=2(3+2+0.96)=\) 11.92 m².
   4. Side \(=\sqrt[3]{125}=5\) cm. \(SA=6\times25=\) 150 cm².

2. Answer to Question 2

   First find the area of the cross-section using the right formula for its shape, then multiply by the length of the prism.

   1. Area \(=\frac12\times6\times4=12\) cm². \(V=12\times10=\) 120 cm³.
   2. Area \(=\frac12\times3\times4=6\) cm². \(V=6\times12=\) 72 cm³.
   3. Area \(=\frac12(5+9)\times4=28\) cm². \(V=28\times15=\) 420 cm³.
   4. \(360\div24=\) 15 cm

   In (d) we work backwards: length = volume ÷ cross-section area.

3. Answer to Question 3

   1. \(V=\pi r^{2}h=\pi\times5^{2}\times12=300\pi=\)942.5 cm³.
   2. Curved surface \(=2\pi rh=2\pi\times5\times12=120\pi=\)377.0 cm².
   3. Two circular ends \(2\times\pi\times5^{2}=50\pi\). Total \(50\pi+120\pi=170\pi=\)534.1 cm².
   4. Radius \(8\div2=4\). \(h=\dfrac{603}{\pi\times4^{2}}=\dfrac{603}{50.27}=\)12.0 cm.

   The total surface area of a closed cylinder is always the two circles plus the curved rectangle, which unrolls to a width of \(2\pi r\) and a length of h.

4. Answer to Question 4

   1. \(\frac12\times5\times12=\) 30 cm²
   2. \(5\times20=100\), \(12\times20=240\), \(13\times20=260\) cm².
   3. Two ends \(=60\). Rectangles \(=100+240+260=600\). Total \(=\) 660 cm².
   4. \(30\times20=\) 600 cm³

   The three rectangles together have area \((5+12+13)\times20=600\) cm², the perimeter of the triangle times the length.

5. Answer to Question 5

   1. \(50\times40\times30=60000\) cm³. Divide by 1 000: 60 litres.
   2. 1 m³ holds 1 000 litres: \(120\times1000=\)120 000 litres.
   3. 1 litre = 1 000 cm³: \(1.5\times1000=\)1 500 cm³.
   4. \(20^{3}=8000\) cm³, which is 8 litres.

   Converting the volume to cm³ first and then dividing by 1 000 is the safest route to litres.

6. Answer to Question 6

   1. \(10\times6\times h=360\), so \(h=360\div60=\)6 cm.
   2. Cross-section = volume ÷ length: \(150\div10=\)15 cm².
   3. Triangle area \(=252\div12=21\) cm². Then \(\dfrac{1}{2}\times7\times h=21\), so \(3.5h=21\) and \(h=6\) cm.
   4. \(h=\dfrac{500}{\pi\times4^{2}}=\dfrac{500}{50.27}=9.95\ldots\), so 9.9 cm.

   Each part reverses the volume formula: divide the volume by everything you already know.

7. Answer to Question 7

   1. Depth of water \(=40\) cm. \(V=80\times35\times40=\) 112 000 cm³
   2. 112 litres
   3. 112 kg
   4. Base \(80\times35=2800\). Long sides \(2\times80\times45=7200\). Short sides \(2\times35\times45=3150\). Total \(=\) 13 150 cm².
   5. \(13150\times0.012=\) £157.80

   Only the water depth (40 cm) is used for the water volume, but the full height (45 cm) is used for the glass.

8. Answer to Question 8

   1. \(2(2.4\times2+1.8\times2)=2\times8.4=\) 16.8 m²
   2. \(2.4\times1.8=\) 4.32 m²
   3. Total \(=21.12\). Door \(=1.44\), window \(=0.6\). Paint area \(=\) 19.08 m².
   4. \(19.08\div6=3.18\), so 4 litres.
   5. \(4\times14.50=\) £58

   Paint must be bought in whole litres, so 3.18 litres rounds up to 4 litres.

9. Answer to Question 9

   1. Area of triangle \(=\frac12\times2\times1.5=1.5\) m². \(V=1.5\times3=\) 4.5 m³
   2. \(\sqrt{1^2+1.5^2}=\sqrt{3.25}=\) 1.80 m
   3. \(2\times1.80\times3=\) 10.8 m²
   4. Floor \(=2\times3=6\). Ends \(=2\times1.5=3\). Total \(=10.8+6+3=\) 19.8 m².

   The sloping side uses Pythagoras in the right-angled triangle formed by half the base (1 m) and the height (1.5 m).

10. Answer to Question 10

    1. There are 100 cm in a metre: \(8\div100=\)0.08 m.
    2. All lengths in metres: \(12\times1.2\times0.08=\)1.152 m³.
    3. \(1.152\times95=\)£109.44.
    4. 1.152 rounds up to the next quarter, 1.25 m³, costing \(1.25\times95=\)£118.75.

11. Answer to Question 11

    1. Radius \(7.4\div2=3.7\) cm. \(V=\pi\times3.7^{2}\times10.8=464.5\), about 464 cm³.
    2. 1 cm³ = 1 ml, so 464 ml.
    3. The label is the curved surface: \(\pi\times7.4\times10.8=251.1\), about 251 cm².
    4. \(25\times12=300\) tins. \(300\times464=139200\) ml, about 139 litres.

12. Answer to Question 12

    1. \(\frac12\times3.5\times3=5.25\) cm². \(V=5.25\times20=\) 105 cm³
    2. \(3\times3\times20=\) 180 cm³
    3. The cuboid holds more. Difference \(=75\) cm³ and \(\dfrac{75}{180}=\) 41.7% of the cuboid.
    4. \(105\times0.9=94.5\)p, so about 95p.

    Both bars have the same length, so comparing the cross-section areas (5.25 cm² and 9 cm²) gives the same conclusion.

13. Answer to Question 13

    1. \(\frac12(40+30)\times25=\) 875 cm²
    2. \(875\times200=175000\) cm³ \(=\) 175 litres
    3. \(175\div10=17.5\) min \(=\) 17 min 30 s
    4. \(175\div60=2.9\), so 2 cows for a full day.

    Remember to change 2 m into 200 cm before multiplying, so that all lengths are in centimetres.

14. Answer to Question 14

    1. Radius = half the diameter: \(2.4\div2=\)1.2 m.
    2. \(V=\pi\times1.2^{2}\times0.25=1.131\), so 1.13 m³.
    3. \(1.131\div0.04=28.3\). 28 bags are not quite enough, so 29 bags.
    4. \(29\times3.20=\)£92.80.

    Working in metres throughout keeps the volume in m³, the same unit as the bags.

15. Answer to Question 15

    1. Whole \(12\times10=120\) minus the missing corner \(6\times5=30\): 90 cm².
    2. \(90\times20=\) 1800 cm³
    3. Perimeter \(=12+10+6+5+6+5=\) 44 cm.
    4. Two ends \(=180\). Sides \(=44\times20=880\). Total \(=\) 1060 cm².

    The perimeter of the L-shape is the same as that of the full 12 by 10 rectangle, because the cut-out corner replaces two edges with two equal edges.

16. Answer to Question 16

    1. A: \(10\times7\times20=\)1 400 cm³. B: radius 4 cm, \(\pi\times4^{2}\times20=320\pi=\)1 005 cm³.
    2. Container A holds more, by \(1400-1005=\)395 ml.
    3. A: \(2(10\times7+10\times20+7\times20)=2(70+200+140)=\)820 cm². B: ends \(2\times16\pi=32\pi\), curved \(2\pi\times4\times20=160\pi\); total \(192\pi=\)603 cm².
    4. A: \(820\times0.4=328\)p. B: \(603\times0.4=241\)p. Container B is cheaper to make, by 87p.
    5. Cardboard cost per ml: A \(328\div1400\approx0.234\)p, B \(241\div1005\approx0.240\)p. Container A is slightly better value, because it holds much more juice for its extra cardboard.

17. Answer to Question 17

    1. \(\frac12(1+2.5)\times20=\) 35 m²
    2. \(35\times10=\) 350 m³
    3. \(350\times1000=\) 350 000 litres
    4. \(350000\div6000=58.3\) min, about 58 min 20 s, so just under 1 hour.
    5. Floor length \(=\sqrt{20^2+1.5^2}=20.06\) m, floor area \(=200.6\) m². Two side walls \(=2\times35=70\). End walls \(=10\times1+10\times2.5=35\). Total \(\approx200.6+70+35=305.6\) m², cost \(\approx\) £6 723.

    The trapezium cross-section is the side view of the pool, so the volume is that area times the 10 m width.

18. Answer to Question 18

    1. \(12\times9\times5=\)540 cm³; melting does not change the volume.
    2. \(\pi\times3^{2}\times h=540\), so \(h=\dfrac{540}{9\pi}=19.09\ldots\), 19.1 cm.
    3. \(2\pi\times3\times19.1=360.0\), about 360 cm².
    4. Mass = density × volume: \(540\times7.8=4212\) g, which is 4.212 kg.

19. Answer to Question 19

    1. Area \(=\frac12\times8\times5=20\) cm². \(V=20\times12=\) 240 cm³.
    2. Ali
    3. Bea forgot the \(\frac12\). Cal added instead of multiplying (\(8+5+12\)) or used the wrong numbers. Dev found only the triangle area times 3 (60), not the length.

    The volume of a triangular prism must include the half in the triangle’s area; checking that a prism is half of a 8 by 5 by 12 cuboid (480 cm³) confirms 240 cm³.

20. Answer to Question 20

    1. \(r=3\): \(h=17.7\), SA \(=2\pi\times9+2\pi\times3\times17.7=\)389.9. \(r=4\): \(h=9.9\), SA 350.5. \(r=5\): \(h=6.4\), SA 357.1. \(r=6\): \(h=4.4\), SA 392.9 (cm and cm²).
    2. \(r=4\) gives the smallest surface area in the table; trying more radii shows the true minimum is at about 4.3 cm.
    3. At \(r\approx4.3\) the height is about 8.6 cm, equal to the diameter: the best tin is as tall as it is wide (\(h\approx2r\)).
    4. Tins are also chosen for stacking, gripping, shelf space and label size, so the shape using the least metal is not always the most practical.