Long Questions (Question 3)

Question 3:
Diagram below shows a sector QPR with centre P and sector POQ, with centre O.

It is given that OP = 17 cm and PQ = 8.8 cm.
[Use π = 3.142]
Calculate
(a) angle OPQ, in radians,
(b) the perimeter, in cm, of sector QPR,
(c) the area, in cm², of the shaded region.

Solution:
$\begin{array}{l}\left(a\right)\\ \angle OPQ=\angle OQP\\ x+x+30=180\\ 2x=150\\ x=75\\ \angle OPQ=\frac{75\times 3.142}{180}\\ =1.3092\text{ radians}\end{array}$


$\begin{array}{l}\left(b\right)\\ \text{Length of arc }QR=r\theta \\ =8.8\times 1.3092\\ =11.52\text{ cm}\\ \\ \text{Perimeter of sector }QPR\\ =11.52+8.8+8.8\\ =29.12\text{ cm}\end{array}$


$\begin{array}{l}\left(c\right)\\ {30}^o=\frac{30\times 3.142}{180}=0.5237\text{ rad}\\ \text{Area of segment }PQ\\ =\frac{1}{2}{r}^2\left(\theta -\sin \theta \right)\\ =\frac{1}{2}\times {17}^2\times \left(0.5237-\sin 30\right)\\ =\frac{1}{2}\times 289\times \left(0.5237-0.5\right)\\ =3.4247{\text{ cm}}^2\\ \\ \text{Area of sector }QPR\\ =\frac{1}{2}{r}^2\theta \\ =\frac{1}{2}\times {8.8}^2\times 1.3092\\ =50.692{\text{ cm}}^2\\ \\ \text{Area of shaded region}\\ =3.4247+50.692\\ =54.1167{\text{ cm}}^2\end{array}$