Area, volume, surface area, arcs and sectors – Year 10 GCSE maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. \(9\times6=\) 54 cm² The area of a parallelogram is base times perpendicular height, not the slanted side. The area is 54 cm².
   2. \(\tfrac12\times14\times8=\) 56 cm² A triangle is half of a parallelogram with the same base and height, so we halve \(14\times8=112\). The area is 56 cm².
   3. \(\tfrac12(7+13)\times5=\tfrac12\times20\times5=\) 50 cm² Add the parallel sides, halve the total and multiply by the distance between them. The area of the trapezium is 50 cm².
   4. \(\pi\times6^2=36\pi=\) 113.1 cm² Square the radius first, then multiply by \(\pi\): \(36\pi=113.097\ldots\), which rounds to 113.1 cm² to 1 decimal place.

2. Answer to Question 2

   1. \(8\times5-3\times2=40-6=\) 34 m²
   2. The perimeter of an L-shape cut from a corner equals the perimeter of the full rectangle: \(2(8+5)=\) 26 m.
   3. \(34\times14.50=\) £493
   4. Cutting a corner out moves two sides inwards by the same lengths that the other two sides are shortened, so the total length of the boundary does not change.

3. Answer to Question 3

   1. \(12\times5\times4=\) 240 cm³ Multiply the three dimensions together. The volume of the cuboid is 240 cm³.
   2. volume \(=\) area of cross-section \(\times\) length \(=24\times10=\) 240 cm³ Any prism has the same cross-section all the way along, so the volume is 240 cm³.
   3. \(\pi\times4^2\times10=160\pi=\) 502.7 cm³ A cylinder is a prism with a circular cross-section of area \(\pi r^2=16\pi\). The volume is 502.7 cm³ to 1 decimal place.
   4. radius \(=5\) cm, so \(\pi\times25\times6=150\pi=\) 471.2 cm³ Halve the diameter before using the formula: a common mistake is to use 10 as the radius. The volume is 471.2 cm³.

4. Answer to Question 4

   1. \(2(8\times5+8\times3+5\times3)=2(40+24+15)=\) 158 cm² A cuboid has three pairs of identical faces, so we find one face of each pair and double the total. The surface area is 158 cm².
   2. \(2\pi r^2+2\pi rh=2\pi\times9+2\pi\times3\times8=18\pi+48\pi=66\pi=\) 207.3 cm² The two circular ends give \(2\pi r^2\) and the curved surface opens out into a rectangle of length \(2\pi r\) and height \(h\). The surface area is 207.3 cm².
   3. \(6\times49=\) 294 cm² A cube has 6 identical square faces, each of area \(7^2=49\) cm². The surface area is 294 cm².
   4. Two triangles: \(2\times\tfrac12\times3\times4=12\). Three rectangles: \(10\times(3+4+5)=120\). Total \(=\) 132 cm². The 3-4-5 triangle has a right angle between the 3 cm and 4 cm sides, so its area is \(\tfrac12\times3\times4\). The total surface area of the prism is 132 cm².

5. Answer to Question 5

   1. \(2.5\times10\,000=\) 25 000 cm² One square metre is \(100\text{ cm}\times100\text{ cm}\), so we multiply by 10 000, not 100. So \(2.5\text{ m}^2=25\,000\text{ cm}^2\).
   2. \(3\times100=\) 300 mm² One square centimetre is \(10\text{ mm}\times10\text{ mm}\), so we multiply by 100. So \(3\text{ cm}^2=300\text{ mm}^2\).
   3. \(4500\div1000=\) 4.5 litres There are 1000 cm³ in one litre, so we divide by 1000. So \(4500\text{ cm}^3=4.5\) litres.
   4. \(0.6\text{ m}^3=600\,000\text{ cm}^3=\) 600 litres First multiply by 1 000 000 to change m³ into cm³, then divide by 1000 to change into litres. In one step, \(1\text{ m}^3=1000\) litres, so \(0.6\text{ m}^3=600\) litres.

6. Answer to Question 6

   1. \(\dfrac{75}{360}\times2\pi\times12=5\pi=\) 15.7 cm The fraction \(\tfrac{75}{360}\) tells us what part of the full circle the sector is. The arc length is 15.7 cm.
   2. \(\dfrac{75}{360}\times\pi\times144=30\pi=\) 94.2 cm² Use the same fraction of the full circle's area \(\pi\times12^2=144\pi\). The sector area is 94.2 cm².
   3. \(\dfrac{144}{360}\times2\pi\times10=8\pi=\) 25.1 cm Here \(\tfrac{144}{360}=0.4\), so the arc is 40% of the circumference \(20\pi\). The arc length is 25.1 cm.
   4. \(\dfrac{144}{360}\times\pi\times100=40\pi=\) 125.7 cm² The sector is 40% of the circle's area \(100\pi\), so the area is 125.7 cm².

7. Answer to Question 7

   1. \(25\times10\times1.5=\) 375 m³, and \(375\times1000=\) 375 000 litres. There are 1000 litres in 1 m³. The pool holds 375 m³, which is 375 000 litres.
   2. \(375\,000\div500=750\) minutes \(=\) 12.5 hours. Divide the minutes by 60 to change to hours: \(750\div60=12.5\). The pool takes 12.5 hours to fill.
   3. Floor: \(25\times10=250\). Long walls: \(2\times25\times1.5=75\). Short walls: \(2\times10\times1.5=30\). Total \(=250+75+30=\) 355 m². The top is open water, so only five faces are tiled. The area to be tiled is 355 m².
   4. \(355\times18=\) £6390. Multiply the area by the price per square metre. The tiles cost £6390.

8. Answer to Question 8

   1. \(\pi\times3.5^2\times10=122.5\pi=384.8\), so 385 cm³.
   2. \(2\pi\times3.5\times10=70\pi=\) 219.9 cm².
   3. Each can is 7 cm across. Length \(=4\times7=28\) cm, width \(=3\times7=21\) cm, height \(=10\) cm. Volume \(=28\times21\times10=\) 5880 cm³.
   4. Cans: \(12\times122.5\pi=1470\pi=4618.1\) cm³. Empty space \(=5880-4618.1=\) 1261.9 cm³ (about 1262 cm³).

9. Answer to Question 9

   1. \(360\div8=\) 45° The 8 slices are equal, so the full turn of \(360^\circ\) is shared equally between them. Each slice has an angle of \(45^\circ\).
   2. radius 15 cm: \(\dfrac{45}{360}\times\pi\times225=28.125\pi=\) 88.4 cm² The radius is half of 30 cm, and one slice is \(\tfrac18\) of the whole pizza. The area of one slice is 88.4 cm².
   3. \(\dfrac{45}{360}\times2\pi\times15=3.75\pi=\) 11.8 cm The crust is the arc of the sector, which is \(\tfrac18\) of the circumference \(30\pi\). The crust of one slice is 11.8 cm long.
   4. Box: \(32^2=1024\). Pizza: \(225\pi=706.9\). Uncovered: \(1024-706.9=\) 317.1 cm². Subtract the area of the circle from the area of the square base. The uncovered area is 317.1 cm².

10. Answer to Question 10

    1. Rectangle \(3\times2.2=6.6\). Triangle \(\tfrac12\times3\times0.8=1.2\). Total \(=\) 7.8 m².
    2. \(7.8\times2.5=\) 19.5 m³. The volume of a prism is the area of the cross-section times its length, so the volume of the shed is 19.5 m³.
    3. Half the width is 1.5 m, so slope \(=\sqrt{1.5^2+0.8^2}=\sqrt{2.89}=\) 1.7 m. The slope is the hypotenuse of a right-angled triangle with sides 1.5 m and 0.8 m. The slope length is 1.7 m.
    4. Roof area \(=2\times1.7\times2.5=8.5\) m². Cost \(=8.5\times4.20=\) £35.70. There are two panels, each 1.7 m by 2.5 m. The felt costs £35.70.

11. Answer to Question 11

    1. \(\pi\times0.36\times1.4=0.504\pi=1.583\) m³ \(=\) 1583 litres. The volume of a cylinder is \(\pi r^2h\) and 1 m³ is 1000 litres. The water butt holds 1.583 m³, which is 1583 litres.
    2. \(1583\div20=79.2\), so 79 minutes. Divide the volume in litres by the rate. It takes about 79 minutes to fill.
    3. \(0.9\text{ m}^3\div(\pi\times0.36)=0.9\div1.131=\) 0.80 m (0.7958). Change 900 litres into 0.9 m³, then divide by the area of the base, \(\pi\times0.6^2\). The water is 0.80 m deep.
    4. \(\pi\times0.36\times0.01=0.01131\) m³ \(=\) 11 litres. Change 1 cm into 0.01 m, then multiply by the base area. One centimetre of rain adds about 11 litres.

12. Answer to Question 12

    1. The two semicircles make one full circle: \(2\pi\times36.5=73\pi=229.3\). Lap \(=200+229.3=\) 429.3 m.
    2. \(5000\div429.3=11.65\), so she needs 12 laps to run at least 5000 m, which is \(12\times429.3=5151.6\) m.
    3. Rectangle \(100\times73=7300\). Circle \(\pi\times36.5^2=4185.4\). Total \(=11485.4\), so 11 485 m². The inside of the track is a rectangle 100 m by 73 m plus two semicircles that make one circle.
    4. \(11\,485\times3.20=\) £36 752. So it costs about £36 752 to turf the inside of the track.

13. Answer to Question 13

    1. \(2\times(5+4)\times2.5=\) 45 m². Two walls are 5 m by 2.5 m and two are 4 m by 2.5 m, so the total is the perimeter of the floor times the height.
    2. Window \(=1.5\times1.2=1.8\). Painted area \(=45-2-1.8=\) 41.2 m². The door and the window are not painted, so their areas are subtracted.
    3. Two coats: \(82.4\) m². Litres \(=82.4\div12=\) 6.9 litres (6.87). Two coats means the area is doubled: \(2\times41.2=82.4\) m². About 6.9 litres of paint are needed.
    4. \(6.87\div2.5=2.7\), so 3 tins. Cost \(=3\times22=\) £66. We must round up because 2 tins would not be enough. So 3 tins are needed and the cost is £66.

14. Answer to Question 14

    1. A: \(\pi\times16\times10=160\pi=503\) cm³. B: \(\pi\times25\times7=175\pi=550\) cm³.
    2. A: \(89\div5.03=17.7\)p per 100 cm³ (18p). B: \(95\div5.50=17.3\)p per 100 cm³ (17p).
    3. Tin B is slightly better value (about 17.3p against 17.7p per 100 cm³). The smaller cost per 100 cm³ is the better buy, so tin B is better value.
    4. \(\pi\times25\times h=160\pi\), so \(h=160\div25=\) 6.4 cm.

15. Answer to Question 15

    1. \(\dfrac{140}{360}\times\pi\times64=24.89\pi=\) 78.2 m². The sector is \(\tfrac{140}{360}\) of a circle of radius 8 m. The watered area is 78.2 m².
    2. \(\dfrac{140}{360}\times2\pi\times8=\dfrac{140}{360}\times16\pi=19.5\), so 19.5 m. The fence follows the arc, which is \(\tfrac{140}{360}\) of the circumference \(16\pi\). The fence is 19.5 m long.
    3. \(19.5+8+8=\) 35.5 m. The two straight edges are radii of 8 m each. The perimeter of the watered region is 35.5 m.
    4. \(\dfrac{\theta}{360}\times64\pi=100\), so \(\theta=\dfrac{100\times360}{64\pi}=179.0\), which is 179°. Rearrange the sector area formula to make \(\theta\) the subject. The sprinkler must turn through about \(179^\circ\).

16. Answer to Question 16

    1. \(l=\sqrt{5^2+12^2}=\sqrt{169}=13\) cm. \(V=\tfrac13\pi\times25\times12=100\pi=\) 314.2 cm³. The radius, the height and the slant height form a right-angled triangle, so we use Pythagoras. The slant height is 13 cm and the volume is 314.2 cm³.
    2. \(\pi\times5\times13=\) \(65\pi\) cm² \(=\) 204.2 cm². Use the slant height \(l=13\), not the vertical height. The curved surface area is \(65\pi\) cm², about 204.2 cm².
    3. \(V=\tfrac43\pi\times216=288\pi=\) 904.8 cm³. \(A=4\pi\times36=144\pi=\) 452.4 cm². Cube the radius for the volume and square it for the area. The sphere has volume 904.8 cm³ and surface area 452.4 cm².
    4. \(\tfrac13\times36\times9=\) 108 cm³. The base area is \(6\times6=36\) cm². The volume of the pyramid is 108 cm³.

17. Answer to Question 17

    1. \(\pi\times2^2\times5=20\pi\) m³. Use \(V=\pi r^2h\) with \(r=2\) and \(h=5\). The volume of the cylinder is \(20\pi\) m³.
    2. \(\tfrac12\times\tfrac43\pi\times8=\tfrac{16}{3}\pi\) m³. A sphere of radius 2 m has volume \(\tfrac43\pi\times8=\tfrac{32}{3}\pi\), and the hemisphere is half of this. Its volume is \(\tfrac{16}{3}\pi\) m³.
    3. \(20\pi+\tfrac{16}{3}\pi=\tfrac{76}{3}\pi=79.6\) m³ \(=\) 79 587 litres (about 79 600). Add the two exact volumes, then multiply by 1000 to change m³ into litres. The total volume is 79.6 m³.
    4. Cylinder: \(2\pi\times2\times5=20\pi\). Hemisphere: \(\tfrac12\times4\pi\times4=8\pi\). Total \(=28\pi=87.96\), so 88 m². The flat base and the circle where the two parts join are not painted. The area painted is about 88 m².

18. Answer to Question 18

    1. \(6^3=\) 216 cm³.
    2. \(\pi\times4\times h=216\), so \(h=216\div(4\pi)=\) 17.2 cm. The volume does not change when the metal is melted, so the cylinder also has volume 216 cm³.
    3. One sphere: \(\tfrac43\pi=4.189\). \(216\div4.189=51.57\), so 51 spheres. We round down because only complete spheres count. So 51 spheres can be made.
    4. Some metal is always wasted or lost when it is melted, and the spheres would never use up the volume perfectly, so fewer would be made.

19. Answer to Question 19

    1. \(\pi\times10^2=\) \(100\pi\) cm².
    2. A square with diagonal \(d\) has area \(\tfrac12d^2\). With \(d=20\): \(\tfrac12\times400=\) 200 cm².
    3. \(100\pi-200=114.2\), so \(100\pi-200\) cm² \(\approx\) 114.2 cm². The shaded area is the area of the circle minus the area of the inner square.
    4. The outer square has side 20 cm and area 400 cm². The circle covers \(\dfrac{100\pi}{400}=\dfrac\pi4=78.5\%\) (1 d.p.).

20. Answer to Question 20

    1. 30 cm: \(\pi\times15^2=225\pi=706.9\) cm². 20 cm: \(\pi\times10^2=100\pi=314.2\) cm².
    2. 30 cm: \(12\div706.9=\textbf{0.0170}\) £ per cm². 20 cm: \(7\div314.2=\textbf{0.0223}\) £ per cm².
    3. The 30 cm pizza is better value. Two 20 cm pizzas give \(628.3\) cm² for £14, which is less pizza for more money than the 30 cm pizza (\(706.9\) cm² for £12).
    4. Area needed \(=2\times100\pi=200\pi\). Then \(\pi r^2=200\pi\), so \(r=\sqrt{200}=14.14\) and the diameter is 28.3 cm (so a 28.3 cm pizza equals two 20 cm pizzas).