1.5.1 Circular Measure Long Questions (Question 1 & 2)

Question 1:

Solution:
(a)
$\begin{array}{l}\text{For triangle }OPQ\\ \sin {30}^{\circ }=\frac{8}{OP}\\ OP=\frac{8}{\sin {30}^{\circ }}=16\text{ cm}\\ \\ \text{Radius of sector }SPT=16+8=24\text{ cm}\\ \\ \angle SPT=60\times \frac{3.142}{180}=1.047\text{ radian}\\ \\ \text{Length of arc }ST=24\times 1.047=25.14\text{ cm}\end{array}$


(b)
$\begin{array}{l}\text{For triangle }OPQ:\\ \tan {30}^{\circ }=\frac{8}{QP}\\ PQ=\frac{8}{\tan {30}^{\circ }}=13.86\text{ cm}\\ \\ \angle QOR=2\left({60}^{\circ }\right)={120}^{\circ }\\ \text{Reflex angle }\angle QOR={360}^{\circ }-{120}^{\circ }={240}^{\circ }\\ {240}^{\circ }=\frac{3.142}{{180}^{\circ }}\times {240}^{\circ }=4.189\text{ radian}\\ \\ \text{Area of shaded region}\\ \text{=}\left(\begin{array}{l}\text{ Area of }\\ \text{sector }SPT\end{array}\right)-\left(\begin{array}{l}\text{Area of major }\\ \text{ sector }OQR\end{array}\right)-\left(\begin{array}{l}\text{Area of triangle }\\ OPQ\text{ and }OPR\end{array}\right)\\ =\frac{1}{2}{\left(24\right)}^2\left(1.047\right)-\frac{1}{2}{\left(8\right)}^2\left(4.189\right)-2\left(\frac{1}{2}\times 8\times 13.86\right)\\ =301.54-134.05-110.88\\ =56.61{\text{ cm}}^2\end{array}$

Question 2:
Diagram below shows a semicircle PTQ, with centre O and quadrant of a circle RST, with centre R.

[Use π = 3.142]
Calculate
(a) the value of θ, in radians,
(b) the perimeter, in cm, of the whole diagram,
(c) the area, in cm², of the shaded region.


Solution:
$\begin{array}{l}\left(a\right)\\ \sin \angle ROT=\frac{2.5}{5}\\ \angle ROT={30}^o\\ \theta ={180}^o-{30}^o={150}^o\\ =150\times \frac{\pi }{180}\\ =2.618\text{ rad}\end{array}$


$\begin{array}{l}\left(b\right)\\ \text{Length of arc }PT=r\theta \\ =5\times 2.618\\ =13.09\text{ cm}\\ \text{Length of arc }ST=\frac{\pi }{2}\times 2.5\\ =3.9275\text{ cm}\\ \\ O{R}^2+{2.5}^2=5^2\\ O{R}^2=5^2-{2.5}^2\\ OR=4.330\\ \\ \text{Perimeter}=13.09+3.9275+2.5+4.330+5\\ =28.8475\text{ cm}\end{array}$


$\begin{array}{l}\left(c\right)\\ \text{Area of shaded region}\\ =\text{Area of quadrant }RST-\text{Area of quadrant }RQT\\ \\ \text{Area of quadrant }RQT\\ =\text{Area of }OQT-\text{Area of }OTR\\ =\frac{1}{2}{\left(5\right)}^2\times \left(30\times \frac{\pi }{180}\right)-\frac{1}{2}\left(4.33\right)\left(2.5\right)\\ =1.1333{\text{ cm}}^2\\ \\ \text{Area of shaded region}\\ =\text{Area of quadrant }RST-\text{Area of quadrant }RQT\\ =\frac{1}{2}{\left(2.5\right)}^2\times \left(90\times \frac{\pi }{180}\right)-1.1333\\ =3.7661{\text{ cm}}^2\end{array}$