(Long Questions) – Question 8

Question 8:
Diagram below shows a cyclic quadrilateral PQRS.


(a) Calculate
(i) the length, in cm, of PR,
(ii) ∠PRQ.
(b) Find
(i) the area, in cm², of ∆ PRS,
(ii) the short distance, in cm, from point S to PR.

Solution:
(a)(i)
$\begin{array}{l}P{R}^2=7^2+8^2-2\left(7\right)\left(8\right)\cos {80}^o\\ P{R}^2=113-19.4486\\ PR=\sqrt{93.5514}\\ PR=9.6722\text{ cm}\end{array}$


(a)(ii)
$\begin{array}{l}\text{In cyclic quadrilateral}\\ \angle PQR+\angle PSR=180\\ \angle PQR+80=180\\ \angle PQR={100}^o\\ \\ \frac{\sin \angle QPR}{3}=\frac{\sin 100}{9.6722}\\ \sin \angle QPR=0.3055\\ \angle QPR={17}^{o}47\text{'}\\ \\ \angle PRQ={180}^o-{100}^o-{17}^{o}47\text{'}\\ ={62}^{o}13\text{'}\end{array}$


(b)(i)
$\begin{array}{l}\text{Area of }△PRS\\ =\frac{1}{2}\times 7\times 8\sin {80}^o\\ =27.5746{\text{ cm}}^2\end{array}$


(b)(ii)


$\begin{array}{l}\text{Area of }△PRS=27.5746\\ \frac{1}{2}\times 9.6722\times h=27.5746\\ h=\frac{27.5746\times 2}{9.6722}\\ =5.7018\text{ cm}\\ \text{Shortest distance}=5.7018\text{ cm}\end{array}$