(Long Questions) – Question 4

Question 4:
Diagram below shows a quadrilateral PQRS.


(a) Find
(i) the length, in cm, of QS.
(ii) ∠QRS.
(iii) the area, in cm², of the quadrilateral PQRS.
(b)(i) Sketch a triangle S’Q’R’ which has a different shape from triangle SQR such that S’R’ = SR, S’Q’ = SQ and ∠S’Q’R’ = ∠SQR.
(ii) Hence, state ∠S’R’Q’.

Solution:
(a)(i)
$\begin{array}{l}\angle P=180-76-34=70\\ \frac{QS}{\sin 70}=\frac{8}{\sin 34}\\ QS=\frac{8\times \sin 70}{\sin 34}\\ =13.44\text{ cm}\end{array}$

(a)(ii)
$\begin{array}{l}{13.44}^2=6^2+9^2-2\left(6\right)\left(9\right)\cos \angle QRS\\ 108\cos \angle QRS=6^2+9^2-{13.44}^2\\ \cos \angle QRS=\frac{6^2+9^2-{13.44}^2}{108}\\ \angle QRS={\cos }^{-1}\left(-0.5892\right)\\ ={126}^{o}6\text{'}\end{array}$

(a)(iii)
$\begin{array}{l}\text{Area of }PQRS\\ =\text{Area of }PQS+\text{Area of }QRS\\ =\left(\frac{1}{2}\times 8\times 13.44\times \sin 76\right)+\left(\frac{1}{2}\times 6\times 9\times \sin {126}^{o}6\text{'}\right)\\ =52.16+21.82\\ =73.98{\text{ cm}}^2\end{array}$


(b)(i)



(b)(ii)
$\begin{array}{l}\angle S\text{'}R\text{'}Q\text{'}=\angle S\text{'}RR\text{'}\\ =180-{126}^{o}6\text{'}\\ ={53}^{o}54\text{'}\end{array}$