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Ecolint Campus des NationsMathématiques
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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 8 · 8.13 Circles and Angles

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.31.4 cm
2.50.2 cm²
3.Acute, right, obtuse, straight, reflex
4.10 cm
5.706.5 cm²
6.Convex
7.(a) 14 cm (b) 44 cm (c) 154 cm²
8.3.142
9.(a) 20π20\pi cm (b) 100π100\pi cm²
10.(a) acute (b) obtuse (c) obtuse (d) reflex
Silver
11.(a) 5 cm (b) 31.4 cm (c) 78.5 cm²
12.10 cm
13.25.7 cm
14.50.24 cm²
15.9π9\pi cm² ≈ 28.3 cm²
16.3π3\pi cm ≈ 9.4 cm
17.≈ 1.88 m
18.≈ 18.85 cm
19.(a) 31.4 m (b) 78.5 m²
20.Convex regular hexagon
Gold
21.(a) 12.6 m² (b) 125 600 cm²
22.x=1x = 1
23.(a) 5 m (b) 31.4 m
24.≈ 35.4 cm
25.(a) 50=52\sqrt{50} = 5\sqrt{2} cm (b) 10 cm
26.≈ 21.5 cm²
27.(a) 39.3 cm² (b) 25.7 cm
28.39π≈122.539\pi \approx 122.5 cm²
29.≈ 62.8 cm (each circle's full perimeter = 10π10\pi; touching only at points, so total is 30π30\pi)
30.45°; area ≈88.4\approx 88.4 cm²
Platinum
31.Area ≈64.3\approx 64.3 m²; perimeter ≈24+9.4=33.4\approx 24 + 9.4 = 33.4 m
32.Perimeter ≈ 285.6 m; Area ≈ 4456 m²
33.6π≈10.66\sqrt{\pi} \approx 10.6 cm
34.2π2\pi cm ≈ 6.28 cm — independent of original radius
35.≈ 4.0 cm²
36.≈ 10% (since 1.21=1.121.21 = 1.1^2)
37.≈ 66.0 m² (21π21\pi)
38.(a) 333\sqrt{3} cm (b) Hexagon ≈ 93.5; circle ≈ 84.8; difference ≈ 8.7
39.≈ 455 revolutions
40.(a) 16π≈50.2716\pi \approx 50.27 cm (b) 16π≈50.2716\pi \approx 50.27 cm²

Pack B — Answers

Bronze
1.50.2 cm
2.113.0 cm²
3.Same.
4.20 cm
5.1256.0 cm²
6.Concave
7.(a) 28 cm (b) 88 cm (c) 616 cm²
8.Same.
9.(a) 12π12\pi cm (b) 36π36\pi cm²
10.(a) acute (b) acute (c) obtuse (d) reflex
Silver
11.(a) 7 cm (b) 44.0 cm (c) 153.9 cm²
12.≈ 14.14 cm
13.36.0 cm
14.78.54 cm²
15.13.5π13.5\pi cm² ≈ 42.4 cm²
16.6π6\pi cm ≈ 18.8 cm
17.≈ 2.2 m
18.≈ 47.12 cm
19.(a) 25.12 m (b) 50.24 m²
20.Convex regular pentagon
Gold
21.(a) 28.3 m² (b) 282 600 cm²
22.x=2x = 2
23.(a) 7 m (b) 44.0 m
24.≈ 44.6 cm
25.(a) 32=42\sqrt{32} = 4\sqrt{2} cm (b) 8 cm
26.≈ 30.9 cm²
27.(a) 77.0 cm² (b) 36.0 cm
28.64π≈201.164\pi \approx 201.1 cm²
29.≈ 75.4 cm
30.60°; area ≈209.4\approx 209.4 cm²
Platinum
31.Area ≈114.3\approx 114.3 m²; perimeter ≈44.6\approx 44.6 m
32.Perimeter ≈ 357.1 m; Area ≈ 6963.5 m²
33.10π≈17.710\sqrt{\pi} \approx 17.7 cm
34.Same.
35.≈ 5.8 cm²
36.≈ 20% (since 1.44=1.221.44 = 1.2^2)
37.≈ 113.1 m² (36π36\pi)
38.Same.
39.≈ 398 revolutions
40.Same.

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) C ≈ 25.1 m, A ≈ 50.2 m² (b) Path area ≈ 28.3 m² (c) ≈ 565 chf
Full working
(a) C = 2π×4=25.122\pi \times 4 = 25.12 m. A = π×16=50.24\pi \times 16 = 50.24 m².

(b) Outer radius 5. Outer area π×25=78.5\pi \times 25 = 78.5. Path = 78.5 - 50.24 = 28.27 m² ≈ 9π9\pi.

(c) 28.27×20=565.428.27 \times 20 = 565.4 chf.
2Problem 2
Answer
(a) ≈ 314, 707, 1256 cm² (b) 3.18, 2.55, 2.23 c/cm² (c) 16" pizza
Full working
(a) 8": π×100=314\pi \times 100 = 314 cm². 12": π×225=706.5≈707\pi \times 225 = 706.5 \approx 707. 16": π×400=1256\pi \times 400 = 1256.

(b) 8": 10/314≈3.1810/314 \approx 3.18 c/cm². 12": 2.552.55. 16": 2.232.23.

(c) 16" pizza is best value per cm² — doubling diameter quadruples area, but price doesn't quadruple.
3Problem 3
Answer
(a) 10 cm (b) 78.5 cm² (c) 21.5 cm² (d) ≈ 78.5%
Full working
(a) Diameter = side = 10 cm.

(b) Radius 5. Area = π×25=78.5\pi \times 25 = 78.5 cm².

(c) Square area 100. Not covered = 100 - 78.5 = 21.5 cm².

(d) 78.5/100=78.5%78.5/100 = 78.5\%. Also = π/4≈0.7854\pi/4 \approx 0.7854.
4Problem 4
Answer
(a) ≈ 36.6 cm (b) (64+8π)(64 + 8\pi) cm²
Full working
(a) Three square sides (24 cm) + arc of semicircle (π×4≈12.57\pi \times 4 \approx 12.57). Total ≈ 36.57.

(b) Square 64 + semicircle (1/2)π×16=8π(1/2) \pi \times 16 = 8\pi. Total (64+8π)(64 + 8\pi) cm² ≈ 89.1 cm².
5Problem 5
Answer
(a) 120° (b) ≈ 41.6 m² (c) ≈ 1456 chf
Full working
(a) Sum of interior angles =(6−2)×180°=720°= (6-2) \times 180° = 720°. Each angle 720°/6=120°720°/6 = 120°.

(b) Hexagon = 6 equilateral triangles of side 4. Equilateral area =(3/4)×16=43≈6.93= (\sqrt{3}/4) \times 16 = 4\sqrt{3} \approx 6.93 m². Total =41.57= 41.57 m².

(c) 41.57×35≈145541.57 \times 35 \approx 1455 chf.
6Problem 6
Answer
(a) 9π9\pi ≈ 28.3 cm² (b) 3π+123\pi + 12 ≈ 21.4 cm (c) 36−9π36 - 9\pi ≈ 7.7 cm²
Full working
(a) Quarter of πr2=9π\pi r^2 = 9\pi.

(b) Quarter of circumference = (1/4)(2πr)=3π(1/4)(2\pi r) = 3\pi; plus 2 radii of 6 cm each = 12 cm. Total 3π+12≈21.423\pi + 12 \approx 21.42 cm.

(c) Square area 36 - quarter-circle area 9π≈28.279\pi \approx 28.27. Outside = 36−28.27≈7.7336 - 28.27 \approx 7.73 cm².
7Problem 7
Answer
(a) π>3\pi > 3 (since perim < circumference 2π2\pi, but here perim 6 < 2π2\pi, so π>3\pi > 3) (b) π>3.105\pi > 3.105 (c) π≈3.14159\pi \approx 3.14159
Full working
(a) Inscribed polygon has perimeter < circumference of circle. For a regular hexagon: 6 < 2π2\pi → π>3\pi > 3.

(b) 12-gon: 6.21 < 2π2\pi → π>3.105\pi > 3.105.

(c) Modern: π≈3.14159\pi \approx 3.14159. Archimedes' bounds tightened with more sides — the 96-gon gave π\pi within ±0.001.
8Problem 8
Answer
(a) ≈ 219.9 cm (b) ≈ 219.9 m (c) ≈ 100
Full working
(a) C=2π×35=70π≈219.9C = 2\pi \times 35 = 70\pi \approx 219.9 cm.

(b) 100×219.9=21990100 \times 219.9 = 21990 cm = 219.9 m.

(c) 22000/219.9≈10022000/219.9 \approx 100.
9Problem 9
Answer
(a) Inner 25π25\pi ≈ 78.5; outer 64π64\pi ≈ 201.1 (b) 39π39\pi ≈ 122.5 (c) 25 : 64
Full working
(a) Inner = 25π25\pi; outer = 64π64\pi.

(b) Annulus = 64π−25π=39π≈122.564\pi - 25\pi = 39\pi \approx 122.5 cm².

(c) Ratio 25π:64π=25:6425\pi : 64\pi = 25 : 64.
10Problem 10
Answer
(a) ≈ 25.1 cm² (b) ≈ 6.3 cm (c) ≈ 22.3 cm
Full working
(a) Area =(45/360)×π×64=8π≈25.13= (45/360) \times \pi \times 64 = 8\pi \approx 25.13 cm².

(b) Arc length =(45/360)×2π×8=2π≈6.28= (45/360) \times 2\pi \times 8 = 2\pi \approx 6.28 cm.

(c) Perimeter = arc + 2 radii = 6.28+16=22.286.28 + 16 = 22.28 cm.
11Problem 11
Answer
(a) Straight: ≈ 86 m (b) Radius 30 m (c) 2×86+2×π×30≈172+188=3602 \times 86 + 2 \times \pi \times 30 \approx 172 + 188 = 360 — doesn't hit 400. Adjust radius.
Full working
Total perimeter = 2 × straight + 2 × semicircle = 2L + 2π r = 400. The track must fit in 100 m × 60 m. End radius is at most 30 m (so circle width is at most 60 m). Total: 2L+60π=400⇒2L=400−60π≈211.52L + 60\pi = 400 \Rightarrow 2L = 400 - 60\pi \approx 211.5. So L≈105.7L \approx 105.7. But L can't exceed 100. So with radius 30 m, fit constraint violated — must reduce r. Compromise: try r=25r = 25: 2L+50π=400⇒2L≈242.9⇒L≈121.52L + 50\pi = 400 \Rightarrow 2L \approx 242.9 \Rightarrow L \approx 121.5 — also too long. The straight track length needs to be > 100 m → the design constraint conflicts. **Conclusion**: a strict 400 m oval track can't fit inside 100 m × 60 m; need larger field.
12Problem 12
Answer
(a) 3.2, 3.125, 3.1, 3.17, 3.13 (b) ≈ 3.145 (c) An estimate of π ≈ 3.142
Full working
(a) C/d: 16/5=3.216/5 = 3.2; 25/8=3.12525/8 = 3.125; 31/10=3.131/10 = 3.1; 38/12≈3.1738/12 \approx 3.17; 47/15≈3.1347/15 \approx 3.13.

(b) Mean ≈ 3.145.

(c) The experiment estimates π≈3.142\pi \approx 3.142, close to the true value 3.14159. The experimental estimate has small errors from measurement, but works for any circular object — a beautiful demonstration of the constant ratio C/d = π.