SPM 2021 Add Maths Trial Paper (Selangor) – Paper 1 (Question 9)

Question 9:
Diagram 4 shows a circle ADBC, with centre O and diameter 12 cm while sector ADC with the centre A and radius AD = 10 cm.

Diagram 4

By using π = 3.142,
find
(a) ∠BAC in radians, (2 marks)
(b) area, in cm², of the shaded region. (3 marks)


Solution:
(a)

$\begin{array}{l}\text{By using the cosine rule,}\\ 6^2={10}^2+6^2-2\left(10\right)\left(6\right)\cos \angle BAC\\ \cos \angle BAC=\frac{{10}^2+6^2-6^2}{2\left(10\right)\left(6\right)}\\ \cos \angle BAC=0.8333\\ \angle BAC={33.56}^o\times \frac{\pi }{{180}^o}\\ =0.5857\text{ rad}\end{array}$


(b)

$\begin{array}{l}\text{Area of sector }CAD=\frac{1}{2}{\left(10\right)}^2\left(0.5857\times 2\right)\\ =58.57{\text{ cm}}^{\text{2}}\\ \\ \text{Area of yellow region}\\ =58.57-2\left(\frac{1}{2}\left(10\right)\left(6\right)\sin {33.56}^o\right)\leftarrow \boxed{\frac{1}{2}ab\sin C}\\ =58.57-33.17\\ =25.40{\text{ cm}}^{\text{2}}\\ \\ \angle COD=\left(0.5857\times 2\right)\times 2\\ =2.3428\text{ rad}\\ \\ \text{Area of shaded region}\\ =\text{Area of sector }CODB-\text{Area of yellow region}\\ =\frac{1}{2}{\left(6\right)}^2\left(2.3428\right)-25.40\\ =16.77{\text{ cm}}^{\text{2}}\end{array}$