Page 74 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 1 Circular Measure

Alternative Method Solution
Obtuse angle POR = 360° – 100° (a)
= 260° A
1
= 260° × 3.142 2 rad
180°
= 4.538 rad O
C B
(b) Arc length PQR = rq
= 12.5 × 4.538 Arc length of 3 semicircles = 15π cm
= 56.725 cm Arc length of 1 semicircle = 5π cm
Perimeter = 56.725 + 12.5 + 12.5 + 38.2 r × π = 5π
= 119.925 cm semicircle
r semicircle = 5 cm O
(c) For ∆POR, d semicircle = 5 × 2 120 r
PR = OP + OR – 2(OP)(OR) cos ∠POR = 10 cm C B
2
2
2
= (12.5) + (12.5) – 2(12.5)(12.5)(cos 100°) 10 cm
2
2
PR = 19.15 cm In ∆OCB, CB = 10 cm
1 r logo 10
Area of trapezium APRB = (19.15 + 38.2)(8) =
1
2 sin 180° – 120° 2 sin 120°
2
= 229.4 cm 2
1
Area of segment = r (q – sin q) r logo = 10 × sin 30°
sin 120°
2
2 = 5.774 cm
= 1 (12.5) (1.746 – sin 100°)
2
2 (b)
= 59.47 cm 2
Area of picture holder
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
X
= Area of trapezium APRB – Area of segment
= 229.4 – 59.47 θ
= 169.93 cm Y Z
2

Try Question 4 in ‘Try This! 1.4’
Area of equilateral triangle XYZ,
1
SPM Highlights A = 2 (5)(5) sin 60°
1
= 10.83 cm
2
STEM Club of SMK Ceria organised a competition to 3.142
1
design a logo for the club. The diagram below shows q = 60° × 180° 2 rad
the circular logo designed by Fauziah.
= 1.047 rad Form 5
Area of segment XY,
A = 1 (5 )(1.047 – sin 60°)
2
2 2
= 2.262 cm
2
Total area of the shaded region
= 3A + (6 – 3)A
2
1
The three shaded regions are congruent. Given that = 3(10.83) + 3(2.262)
the perimeter of the shaded region is 15π cm. Calculate
2
(a) the radius, in cm, of the logo, = 39.276 cm
(b) the total area, in cm , of the shaded region.
2
[Use π = 3.142]
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