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(b) Bearing of C from B = 190° (b) Total surface area
BC = 12.2 cm [2] = 2π(12) + 2π(12)(27)
2
2 Accurate plan with 3 arcs seen. = 2940.91 cm 2 [2]
∠PTS = 120° (c) Volume of cylinder A + Volume of cylinder of
∠RST = 130° B
TS = 5 cm [7] = π(8) (18) + π(12) (27)
2
2
(6 – 2) × 180° = 15835.68 cm 3
3 (a) Interior angle = = 120° (shown)
6 [2] Total mass =15835.68 × 11.3 g
(b) Accurate plan with 4 arcs seen with each = 178943.18 g [3]
= 178.94 kg
interior angle of 120°. [3] 2 2 2
4 (a) Correct triangle shape ABC with 2 arcs seen 5 (a) OU = UV – OV
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= 13 – 12
2
2
∠CAB = 64° OU = √25
∠CBA = 33° [3] = 5 cm [2]
(b) Correct triangle shape ABC with 2 arcs seen (b) (i) RU = OR – OU 2
2
2
∠CAB = 133° = 7 – 5 2
2
∠ACB = 17° [3]
RU = 4.899 cm
SR = 9.798 cm [2]
5 5 Mensuration (ii) Area = [ RQ × OU ] × 5
2
[ 9.798 × 5 ]
Mensuration = 2 × 5
1 (a) 82 [1] = 122.475 cm 2 [2]
(b) 164.9 [1] (c) Volume = 1 × Base area × Height
(c) 30300 [1] 3
(d) 8.76 [1] = 1 × 122.475 × 12
(e) 5290 [1] 3
(f) 6100 [1] = 489.9 cm 3
(g) 450 [1] Price = 489.9 × 19.28 × $55
(h) 38800 [1] = $519470 [3]
(i) 4010 [1] 6 (a) Perimeter of the butterfly
(j) 22.7; 22700 [1] = (AB + BXC + CD + DE + EF + AF) × 2
)
[
5
65
2 (a) ∠POR = 2 × tan –1 ( ) = 3 + 2π × 1.5 + 3 + ( 360 × 2π × 3 + 3 +
2
4.8
= 92.38° ( 1 )]
Reflex angle of ∠POR = 267.661 [2] 4 × 2π × 6 × 2
(b) (i) OP = 4.8 +5 2 13
2
2
OP = 6.931 = 6 + 3π + 6 + 6 π + 6 + 6π
Perimeter = 267.661 × 2π × 6.931 + 10 = 53.09 cm [4]
360 (b) Area of the butterfly
= 42.38 cm [3] = (Area of ACF − Area of BXC +
(ii) Area = 267.661 × π × 6.931 + 10 × 4.8 Area of FED) × 2
2
360 2 ( 1 2 1 2 65 2 )
= 136.21 cm 2 [2] = 4 × π × 6 – 2 π × 1.5 + 360 × π × 3 × 2
3 PQ = x – 8 2 = 59.69 cm 2 [4]
2
2
PQ = √x – 64 7 (a) Curved surface area
2
Area of trapezium = 25 × Area of ΔPQS = π × 5 × 13
(PQ + SR)PS = 25 × PQ × PS = 204.2 cm 2 [2]
2 2 (b) PP' = 2 × π × 5
2
2
(√x – 64 + SR)8 = 25 × √x – 64 × 8 = 31.416 cm
2 2 ∠POP' × 2 × π × 13 = 31.416
SR + √x – 64 = 25√x – 64 360
2
2
∠POP'= 138.462°
2
SR = 24 √x – 64 [5] 138.462 2
Height of cylinder B 12 Area = 360 × π × 13
4 (a) =
18 8 = 204.2 cm 2 [4]
Height of cylinder B = 27 cm [2]
Cambridge IGCSE
TM
170 Ace Your Mathematics

Answers.indd 170 15/03/2022 11:08 AM