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


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


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

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 3:
Solution:

Question 4:
Solution:
(a) s = r θ
$=\frac{1}{2}{\left(8\right)}^2\left(0.75\right)-\frac{1}{2}{\left(3\right)}^2\left(1.2\right)$

Question 1:


Solution:
(a)
$\begin{array}{l}s=r\theta \\ 5.2=13\left(\angle COB\right)\\ \angle COB=0.4\text{ radian}\end{array}$

(b)
$\begin{array}{l}\cos \angle COB=\frac{OA}{OC}\\ \cos 0.4=\frac{OA}{13}\text{ (change calculator to Rad mode)}\\ OA=11.97\text{ cm}\\ \therefore AB=13-11.97=1.03\text{ cm}\\ \\ CA=\sqrt{{13}^2-{11.97}^2}\\ CA=5.07\text{ cm}\end{array}$

Question 2:

Solution:
(a) Arc AB = 6cm
$\begin{array}{l}\text{Area of sector }AOB\\ =\frac{1}{2}{r}^2\theta =\frac{1}{2}{\left(5\right)}^2\left(\frac{6}{5}\right)=15c{m}^2\end{array}$
$\begin{array}{l}\text{Area of shaded region}\\ \text{=}\frac{1}{2}{r}^2\left(\theta -\sin \theta \right)\left(\text{change calculator to Rad mode}\right)\\ \text{=}\frac{1}{2}{\left(5\right)}^2\left(\frac{6}{5}-\sin \frac{6}{5}\right)\\ =3.35{\text{ cm}}^2\end{array}$

(B) Length of an Arc of a Circle




Example 1:
An arc, AB, of a circle of radius 5 cm subtends an angle of 1.5 radians at the centre.  Find the length of the arc AB.
$\begin{array}{l}=12\times {30}^{\circ }\times \frac{\pi }{{180}^{\circ }}\\ =6.283\text{ cm}\end{array}$



Example 3:

8.1 Radians
(A) Terminology: