Page 70 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 1 Circular Measure
SPM Highlights
Solution
The diagram below shows a rhombus KLMN inscribed s
in sector KLM with centre L and radius h cm. (a) ∠KOL = r
L = 10
K λ rad M 15
N 2
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= rad
3
Given the area of sector KLM is 12.5 cm , express in 1
2
2 2
1 2
terms of h (b) Area of sector KOL = 2 (15 ) 3
(a) l,
2
(b) the perimeter, in cm, of the shaded region. = 75 cm
Solution 1 2 3.142
(a) A = 12.5 Area of sector LOM = (15 ) 1 2
KLM 2 4
1 2
2
h l = 12.5 = 88.37 cm
2
l = 25 rad tan ∠NOM = MN
h 2 OM
(b) s KLM = rq tan 45° = MN
= hl 15
= h × 25 MN = 15 tan 45°
h 2 = 15 cm
= 25 cm
h 1
Area of triangle NOM = (15)(15)
Perimeter of the shaded region 2
25 = 112.5 cm 2
= h + h +
h Area of the shaded region
2
= 2h + 25 cm
h = Area of sector KOL + (Area of triangle NOM –
Area of sector LOM)
= 75 + (112.5 – 88.37)
= 99.13 cm
2
SPM Highlights

The diagram below shows sector KOM with Try This! 1.3
centre O.
2
O 1. Calculate the area, in cm , for each of the following
15 cm
K sectors.
45 Form 5
(a) (b)
1.9 cm
L O 3.76 rad
2.44 rad O
N M 1.7 cm
Given that the arc length of KL is 10 cm.
Calculate
(a) ∠KOL, in radian, (c)
(b) the area, in cm , of the shaded region.
2
[Use π = 3.142] O 2 rad
2.22 cm

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