Circles: circumference and area – Year 8 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. The diameter is twice the radius: \(d=2\times7=\)14 cm.
   2. \(C=\pi d=\pi\times14=43.98\ldots\), which is 44.0 cm to 1 d.p.
   3. \(C=2\pi r=2\times\pi\times7=44.0\) cm. Both formulae agree because \(2r\) is the diameter.
   4. Leave π as a symbol and write the number in front: \(14\pi\) cm, which is the exact answer.

2. Answer to Question 2

   1. \(C=2\pi r=2\times\pi\times3=18.85\ldots\), so 18.8 cm.
   2. With a diameter use \(C=\pi d=\pi\times20=\)62.8 cm.
   3. \(C=2\times\pi\times4.5=28.27\ldots\), so 28.3 m.
   4. \(C=\pi\times12.6=39.58\ldots\), so 39.6 mm.

   Estimate check: each circumference is just over three times the diameter, for example \(3\times20=60\) for part (b).

3. Answer to Question 3

   The area formula is \(A=\pi r^{2}\): square the radius first, then multiply by π. When a diameter is given, halve it to find the radius.

   1. \(A=\pi\times5^2=\pi\times25=\) 78.5 cm²
   2. \(A=\pi\times2.5^2=\pi\times6.25=\) 19.6 m²
   3. radius \(=7\), \(A=\pi\times49=\) 153.9 cm²
   4. radius \(=4.5\), \(A=\pi\times20.25=\) 63.6 mm²

   The units of area are always squared: cm², m² or mm².

4. Answer to Question 4

   Answers in terms of π are exact: do the arithmetic with the numbers and leave π as a symbol at the end.

   1. \(\pi\times36=\) \(36\pi\) cm²
   2. \(2\pi\times6=\) \(12\pi\) cm
   3. radius 5, \(\pi\times25=\) \(25\pi\) m²
   4. \(\pi\times10=\) \(10\pi\) m

5. Answer to Question 5

   To work backwards, put the known value into the formula and use inverse operations: divide by π, then halve or take the square root as needed.

   1. \(d=C\div\pi=31.4\div3.14=\) 10 cm, so \(r=\) 5 cm
   2. \(d=62.8\div3.14=20\), so \(r=\) 10 m
   3. \(r^2=78.5\div3.14=25\), so \(r=\) 5 cm
   4. \(r^2=49\), \(r=7\), so \(d=\) 14 cm

6. Answer to Question 6

   1. \(\frac12\times\pi\times10^2=50\pi=\) 157.1 cm²
   2. Curved edge \(=\frac12\times2\pi\times10=10\pi=31.4\). Straight edge \(=20\). Perimeter \(=\) 51.4 cm
   3. \(\frac14\times\pi\times100=25\pi=\) 78.5 cm²
   4. Curved edge \(=\frac14\times2\pi\times10=5\pi=15.7\). Two radii \(=20\). Perimeter \(=\) 35.7 cm

   A common mistake is to forget the straight edges: the perimeter of a semicircle or quarter circle includes the diameter or the two radii.

7. Answer to Question 7

   1. \(C=\pi d=\pi\times66=207.34\ldots\), so 207.3 cm.
   2. One turn moves the bike one circumference: 207.3 cm ÷ 100 = 2.07 m.
   3. 1 km = 1 000 m. \(1000\div2.0735=482.3\), so about 482 complete turns.
   4. 3.2 km = 3 200 m. \(3200\div2.0735=1543.3\), which is about 1 500 turns to the nearest 100.

8. Answer to Question 8

   1. Small: \(\pi\times10^2=314\) cm². Medium: \(\pi\times14^2=616\) cm². Large: \(\pi\times18^2=1018\) cm².
   2. Small: \(600\div314.2=1.91\)p. Medium: \(950\div615.8=1.54\)p. Large: \(1400\div1017.9=1.38\)p.
   3. The large pizza has the lowest cost per cm² (about 1.38p), so it is the best value.
   4. No. The area of the large pizza is \(1018\div314=3.24\) times the small one, so it is about 3.2 times as big, not 2 times. Doubling the diameter multiplies the area by 4.

9. Answer to Question 9

   1. The radius is half the diameter: \(3\div2=\)1.5 m.
   2. \(A=\pi\times1.5^{2}=\pi\times2.25=\)7.07 m².
   3. The path adds 1 m all round: \(1.5+1=\)2.5 m.
   4. Path = large circle − pond: \(\pi\times2.5^{2}-\pi\times1.5^{2}=19.63-7.07=\)12.6 m².
   5. \(12.57\times18=226.19\), so the paving costs about £226.

10. Answer to Question 10

    1. The two semicircles join to make a circle whose diameter is the width of the pitch: 90 m.
    2. \(C=\pi d=\pi\times90=\)282.7 m for both curved ends together.
    3. Add the two straights: \(2\times100+282.7=482.7\), so about 483 m per lap.
    4. 5 km = 5 000 m. \(5000\div483=10.4\), so she runs 10 complete laps and needs part of an 11th lap to reach 5 km.

11. Answer to Question 11

    1. The cloth hangs 15 cm over on each side: \(80+2\times15=\)110 cm.
    2. Radius 55 cm: \(\pi\times55^{2}=\pi\times3025=\)9 503 cm².
    3. Divide by 10 000 to change to m²: 0.9503 m². Cost \(0.9503\times12=\)£11.40.
    4. The lace goes round the circumference: \(\pi\times110=345.6\), so 346 cm of lace.

12. Answer to Question 12

    1. The semicircle sits on the 120 cm width, so its diameter is 120 cm and its radius is 60 cm.
    2. \(120\times70=\)8 400 cm².
    3. Half of a circle of radius 60: \(\dfrac{1}{2}\times\pi\times60^{2}=\)5 654.9 cm².
    4. \(8400+5654.9=14054.9\) cm². Divide by 10 000: 1.41 m² of glass.
    5. The frame goes up both sides (\(2\times70=140\) cm) and round the curved top (\(\pi\times60=188.5\) cm): 328.5 cm = 3.285 m. Cost \(3.285\times35=\)£114.97.

13. Answer to Question 13

    1. \(C=\pi\times7.4=23.25\ldots\), so 23.2 cm.
    2. Unrolled, the label is a rectangle 23.25 cm by 10.2 cm: \(23.25\times10.2=237.1\), so 237 cm².
    3. Radius \(7.4\div2=3.7\) cm: \(\pi\times3.7^{2}=\)43.0 cm².
    4. 237 cm² is more than 200 cm², so the company’s claim is true.

14. Answer to Question 14

    1. In one hour the tip goes once round a circle of radius 1.2 m: \(2\pi\times1.2=7.54\), so 7.5 m.
    2. 15 minutes is a quarter of an hour: \(7.54\div4=\)1.88 m.
    3. \(\pi\times1.5^{2}=7.07\), so 7.1 m².
    4. 20 minutes is \(\dfrac{20}{60}=\)\(\dfrac{1}{3}\) of a turn. Sector area \(\dfrac{1}{3}\times\pi\times1.2^{2}=\)1.51 m².

15. Answer to Question 15

    1. \(\pi\times8^2=\) 201.1 m²
    2. \(8+2=10\) m. Area \(=\pi\times100=\) 314.2 m²
    3. \(314.2-201.1=\) 113.1 m²
    4. Seed: \(201.1\times1.40=281.49\). Flowers: \(113.1\times6.50=735.13\). Total \(=1016.62\), so about £1017
    5. Half the lawn is \(100.5\) m². The border is \(113.1\) m², which is bigger, so he is not right.

    The border is a ring, so its area is the difference between the two circles, not the area of a circle of radius 2 m.

16. Answer to Question 16

    1. \(\pi\times12^2=\) 452 cm²
    2. \(21^2=\) 441 cm²
    3. Tin A: 452 cm², which is at least 450, so it is big enough. Tin B: 441 cm², which is less than 450, so it is not big enough.
    4. Tin A: \(\pi\times24=75.4\) cm. Tin B: \(4\times21=84\) cm. Tin B needs the longer ribbon (84 cm against 75.4 cm).

17. Answer to Question 17

    1. The sector angle out of the full turn: \(\dfrac{60}{360}=\)\(\dfrac{1}{6}\).
    2. \(\pi\times12^{2}=\pi\times144=\)452.4 cm².
    3. One sixth of the circle: \(452.4\div6=\)75.4 cm².
    4. Full circumference \(2\pi\times12=75.4\) cm; one sixth is 12.6 cm.
    5. The perimeter is the arc plus the two radii: \(12.6+12+12=\)36.6 cm.

18. Answer to Question 18

    1. Back: \(\pi\times1.8=\) 5.65 m. Front: \(\pi\times0.9=\) 2.83 m.
    2. \(100\div5.6549=\) 17.7 turns
    3. \(100\div2.8274=\) 35.4 turns
    4. The front wheel makes twice as many turns. Its diameter is half as big, so its circumference is half as big, so it must turn twice as often to cover the same distance.

19. Answer to Question 19

    1. \(30\times20=\)600 m².
    2. Diameter 20 m, so radius 10 m: \(\dfrac{1}{2}\times\pi\times10^{2}=\)157.1 m².
    3. \(600+157.1=\)757.1 m².
    4. \(757.1\div8=94.6\), and we must have enough, so round up to 95 litres.
    5. \(95\div5=19\) tins exactly: \(19\times22=\)£418.

20. Answer to Question 20

    1. \((2r)^2=4r^2\)
    2. \(\pi r^2\)
    3. \(\dfrac{\pi r^2}{4r^2}=\dfrac{\pi}{4}=\) 0.785, so about 78.5%.
    4. \(4\times36-\pi\times36=144-113.1=\) 30.9 cm²
    5. Four circles: \(4\pi\times36=452.4\) cm². Fraction \(=\dfrac{452.4}{576}=0.785\), which is the same \(\dfrac{\pi}{4}\). Cutting more small circles does not waste less.

    The fraction \(\dfrac{\pi}{4}\) does not depend on r, so every circle cut from its smallest surrounding square wastes about 21.5% of the metal.