Page 69 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 1 Circular Measure

16
The diagram below shows the area on the rear-view Solution
mirror with the shape of a sector of a circle with centre (a) s = 12
BD
O that can be cleaned by the windshield wiper. r × x = 12
12
r = cm
x

1 1
Minion Pro 10pt = —
Minion Pro 10pt = — 2 rad 52 cm (b) Since BC and DC are tangents, thus ∠ABC
2 2 O 18 cm and ∠ADC are right-angled.
---------------------------------------- By using π = 3.142, calculate the area, in cm , of the BC = DC
----------------------------------------
2

Arial 8.5pt = 1 1 xReserved.
=
Arial
8.5pt

2 2 rear-view mirror that can be cleaned by the windshield 13 2 12 2
–
x
x
1 1 wiper. = 1 2 1 2
Arial 8pt =
Arial 8pt =
2 2
Penerbitan Pelangi Sdn Bhd. All Rights
Solution 169 144
---------------------------------------- 1 = 2 – x 2
----------------------------------------
A large sector = r q
2
2
80
Cth KB A T_Optima 9pt = 80 1 2 = 25
Cth KBAT_Optima 9pt =
1.25
1.25
----------------------------------------
---------------------------------------- = 2 × (18 + 52) × 2 x 2
= 1 × 70 × 2 = 5 cm
2
----------------------------------------
---------------------------------------- 2 x
= 4 900 cm Area of triangle ABC = Area of triangle ADC
2
1
A = r q 1 12 5
2
small sector 2 = × 1 2 1 2
×
1 1 = 1 × 18 × 2 2 x x
2
2
XX
XX 2 = 30 2 cm
x
= 324 cm
2
Penyelesaian
Penyelesaian Area of quadrilateral ABCD
(a) XX Area of the rear-view mirror that can be cleaned = Area of triangle ABC + Area of triangle ADC
(a)

X
X
= A large sector – A small sector
= 4 900 – 324 = 30 2 + 30 2
x
x
= 4 576 cm
2
= 60 cm
2
Try Question 10 in ‘Try This! 1.3’ x 2
1
Area of sector ABD = r q
2
SPM Highlights 2
2
12
= 1 × 1 2 × x
The diagram below shows a circle with centre A. 2 x
= 72 cm
2
x
Form 5
A
Area of the shaded region
x rad D
= Area of quadrilateral ABCD
B C – Area of sector ABD
60 72
BC and DC are tangents to the circle at points B and = x 2 – x
D respectively. Given that the minor arc length BD is
13 60 – 72x 2
12 cm and AC = cm. Express in terms of x, = x 2 cm
x
(a) the radius, r, of the circle,
(b) the area, A, of the shaded region.
230