(Long Questions) – Question 10

Question 11 (10 marks):
Solution by scale drawing is not accepted.
Diagram shows a quadrilateral PQRS on a horizontal plane.
VQSP is a pyramid such that PQ = 12 m and V is 5 m vertically above P.
Find
(a) ∠QSR,
(b) the length, in m, of QS,
(c) the area, in m², of inclined plane QVS.


Solution: 
(a)
$\begin{array}{l}\frac{\sin \angle QSR}{20.5}=\frac{\sin {64}^o}{22}\\ \sin \angle QSR=\frac{\sin {64}^o}{22}\times 20.5\\ \sin \angle QSR=0.8375\\ \angle QSR={56}^{o}52\text{'}\end{array}$


(b)
$\begin{array}{l}\angle QRS={180}^o-{64}^o-{56}^{o}52\text{'}\\ ={59}^{o}8\text{'}\\ \frac{QS}{\sin {59}^{o}8\text{'}}=\frac{22}{\sin {64}^o}\\ QS=\frac{22}{\sin {64}^o}\times \sin {59}^{o}8\text{'}\\ QS=21.01\text{ m}\end{array}$


(c)


$\begin{array}{l}Q{V}^2=P{Q}^2+V{P}^2\\ QV=\sqrt{{12}^2+5^2}\\ QV=13\text{ m}\\ \\ S{V}^2=P{S}^2+V{P}^2\\ SV=\sqrt{{10}^2+5^2}\\ SV=\sqrt{125}\text{ m}\\ \\ QS=21.01\text{ m}\\ \\ {21.01}^2={13}^2+{\left(\sqrt{125}\right)}^2-2\left(13\right)\left(\sqrt{125}\right)\cos \theta \\ 26\left(\sqrt{125}\right)\cos \theta =169+125-441.42\\ \cos \theta =\frac{169+125-441.42}{26\left(\sqrt{125}\right)}\\ \cos \theta =-0.5071\\ \theta ={120}^{o}28\text{'}\\ \\ \text{Area of }QVS\\ =\frac{1}{2}\times 13\times \sqrt{125}\times \sin {120}^{o}28\text{'}\\ =62.64{\text{ m}}^2\end{array}$