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

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1. Answer to Question 1 (4 marks)

   1. Circumference \(C=2\pi r\) (1 mark for the formula). With \(r=9\): \(C=2\times\pi\times9=18\pi=56.548\ldots\), which is 56.5 cm to 1 decimal place (1 mark).
   2. The area formula uses the radius, so halve the diameter first: \(r=14\div2=7\) cm (1 mark). Then \(A=\pi r^{2}=\pi\times7^{2}=49\pi=153.938\ldots\), which is 153.9 cm² (1 mark).

   A common mistake is using the diameter 14 in the area formula, which gives \(196\pi\approx615.8\) cm², four times too big. Another is working out \((\pi\times7)^{2}\) instead of \(\pi\times7^{2}\). Estimate: \(3\times49\approx150\), close to 153.9.

2. Answer to Question 2 (4 marks)

   1. The circumference is \(C=\pi d\), so the diameter is \(d=C\div\pi=37.68\div3.14=\)12 cm (1 mark). The radius is half of the diameter: \(r=\)6 cm (1 mark).
   2. The area is \(A=\pi r^{2}\), so divide by \(\pi\) first: \(r^{2}=200.96\div3.14=64\) (1 mark). Then take the square root: \(r=\sqrt{64}=\)8 cm (1 mark).

   Working backwards means undoing each operation in reverse order. Check (a): \(3.14\times12=37.68\). Check (b): \(3.14\times8^{2}=3.14\times64=200.96\). Remember to take the square root at the end; stopping at 64 gives the square of the radius, not the radius.

3. Answer to Question 3 (4 marks)

   1. The diameter is 10 m, so the radius is 5 m (1 mark). A semicircle is half a circle: \(A=\frac12\times\pi\times5^{2}=12.5\pi=39.27\ldots\), so the area is 39.3 m² (1 mark).
   2. The curved edge is half of a full circumference: \(\frac12\times\pi\times10=5\pi=15.71\ldots\) m (1 mark). The perimeter also includes the straight edge, which is the diameter of 10 m, so the perimeter is \(15.71+10=\)25.7 m (1 mark).

   Forgetting the straight edge is the most common error and gives only 15.7 m. A gardener edging the flower bed needs both parts.

4. Answer to Question 4 (4 marks)

   1. The tarmac is a ring: the large circle minus the grass island. Large circle \(\pi\times5^{2}=25\pi\approx78.5\) m²; island \(\pi\times3^{2}=9\pi\approx28.3\) m² (1 mark for both areas). Tarmac area \(=16\pi=50.265\ldots\), which is 50.3 m² (1 mark).
   2. Multiply by the price per square metre, using the unrounded area: \(16\pi\times28=1407.43\ldots\) (1 mark). To the nearest pound the cost is £1 407 (1 mark).

   Using the rounded area gives \(50.3\times28=£1\,408.40\), which is also accepted with full working. Subtracting the radii first, \(\pi\times2^{2}\), is wrong because a ring is not a circle of radius 2 m.

5. Answer to Question 5 (4 marks)

   1. Change the diameter to metres: 70 cm = 0.7 m. The circumference is \(\pi d=\pi\times0.7=2.199\ldots\) m (1 mark for the formula), which is 2.20 m to 2 decimal places (1 mark).
   2. The journey is 2.5 km = 2 500 m. Each turn of the wheel moves the bicycle one circumference forward, so the number of turns is \(2500\div2.1991\ldots=1136.8\ldots\) (1 mark). The question asks for complete turns, so round down: 1 136 turns (1 mark).

   The 1 137th turn is not finished when Zane arrives, so it does not count as complete. Use the unrounded circumference: dividing by 2.20 gives 1 136.4, which leads to the same number of complete turns.

Total : 20 marks