Page 46 - ACE YR IGCSE A TOP APPR' TO ADD MATH

Perimeter 5. Q
= SP + PR + RS
1
π + (8 – 2√13)
= (8)(0.805) + 6 1 2 T
3
= 13.51 cm
[5]
(c) Area of sector TSP O P
1
2
= (PT) (q)
2
= 1 (8) (0.805) S
2
2
= 25.76 cm 2 R
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Area of sector ORP Let radius of circle = r
= 1 (OP) (∠ROQ) sin ∠TOP = TP
2
2 OP
1 2 1 π r
= (6) 1 2 sin =
π
2 3 6 30 – r
= 18.85 cm 30 – r = 2r
2
Area of triangle 3r = 30
r = 10 cm
= 1 (RT)(OT) sin q
2π
2 Arc TS = 10 1 2
= 1 (2√13)(2) sin 0.805 20π 3
2 = cm
= 5.198 cm 3
2
OT = OP – PT
2
2
2
Area of shaded region = 20 – 10
2
2
= (25.76 − 18.85 − 5.198) cm OT = 10√3 cm
2
= 1.712 cm 2
[4] Perimeter of shaded region
1 = 10√3 + 10√3 + 20π
2
3. (a) 60 = (8) (q) 3
2 = 55.58 cm
q = 1.875 rad Area of sector TPS
[1]
1
1 2
(b) Perimeter = AO + OB + BA = (10) 2 2π
3
2
1
= 8 + 8 + (8) (π − 1.875) 100
2
2 = π cm
2
= 56.531 cm 3
[3] Area of OTPS
4. ∠QOR = 2 × 30° = Area of triangle OTP × 2
1
= 60° = × 10√3 × 10 × 2
2
(Angle at the centre is twice the angle at = 100√3 cm
2
circumference)
Area of shaded region
Area of sector QOR 100
1
1 2
= (5) 2 1 π = 100√3 – 3 π
2 3 = 68.49 cm
2
= 25 π cm [7]
2
6
Area of triangle QOR 6. (a) Major arc PQ
3
2
1
1 1 = 8 2π – π
3 1 24
= (5)(5) sin π 7
2 3 = 39.49 cm
= 25√3 cm [1]
2
4
Area of shaded region (b) Area of sector POQ
2 3
1
π
= 25 π – 25√3 = 2 (8) 1 2
7
6 4 2
= 2.26 cm 2 = 43.08 cm
[4]
Cambridge IGCSE
TM
186 Ace Your Additional Mathematics
Answers Add Math.indd 186 14/03/2022 12:29 PM