Long Questions (Question 5)

Question 5:
Diagram below shows a semicircle PTS, centre O and radius 8 cm. PTR is a sector of a circle with centre P and Q is the midpoint of OS.

[Use π = 3.142]
Calculate
(a) ∠TOQ, in radians,
(b) the length , in cm , of the arc TR ,
(c) the area, in cm² ,of the shaded region.

Solution:
(a)
$\begin{array}{l}\cos \angle TOQ=\frac{4}{8}=\frac{1}{2}\\ \angle TOQ={60}^{\text{o}}\\ =60\times \frac{\pi }{180}\\ =1.047\text{ radians}\end{array}$

(b)


$\begin{array}{l}\angle TPO={30}^{\text{o}}\\ =30\times \frac{\pi }{180}\\ =0.5237\\ P{T}^2=8^2+8^2-2\left(8\right)\left(8\right)\cos 120\\ P{T}^2=192\\ PT=\sqrt{192}\\ PT=13.86\text{ cm}\\ \\ \text{Length of arc }TR=13.86\times 0.5237\\ =7.258\text{ cm}\end{array}$

(c)
$\begin{array}{l}\text{Area of sector }PTR\\ =\frac{1}{2}\times {13.86}^2\times 0.5237\\ =50.30{\text{ cm}}^2\\ \\ \text{Length }TQ\\ =\sqrt{P{T}^2-P{Q}^2}\\ =\sqrt{{13.86}^2-{12}^2}\\ =6.935\text{ cm}\\ \\ \text{Area of }△PTQ\\ =\frac{1}{2}\times 12\times 6.935\\ =41.61{\text{ cm}}^2\\ \\ \text{Area of shaded region}\\ =50.30-41.61\\ =8.69{\text{ cm}}^2\end{array}$