(Long Questions) – Question 9

Question 9 (10 marks):
Solution by scale drawing is not accepted.
Diagram shows a transparent prism with a rectangular base PQRS. The inclined surface PQUT is a square with sides 12 cm and the inclined surface RSTU is a rectangle. PTS is a uniform cross section of the prism. QST is a green coloured plane in the prism.


It is given that ∠PST = 37^o and ∠TPS = 45^o.
Find
(a) the length, in cm, of ST,
(b) the area, in cm², of the green coloured plane.
(c) the shortest length, in cm, from point T to the straight line QS.


Solution:
(a)



$\begin{array}{l}\frac{ST}{\sin {45}^o}=\frac{12}{\sin {37}^o}\\ ST=\frac{12}{\sin {37}^o}\times \sin {45}^o\\ ST=14.1\text{ cm}\end{array}$


(b)
$\begin{array}{l}Q{T}^2=Q{P}^2+P{T}^2\\ Q{T}^2={12}^2+{12}^2\\ QT=\sqrt{{12}^2+{12}^2}=16.97\text{ cm}\\ \\ \angle PTS={180}^o-{45}^o-{37}^o={98}^o\\ \\ \frac{PS}{\sin {98}^o}=\frac{12}{\sin {37}^o}\\ PS=\frac{12}{\sin {37}^o}\times \sin {98}^o\\ PS=19.75\text{ cm}\\ \\ Q{S}^2=Q{P}^2+P{S}^2\\ QS=\sqrt{Q{P}^2+P{S}^2}\\ QS=\sqrt{{12}^2+{19.75}^2}\\ QS=23.11\text{ cm}\end{array}$




$\begin{array}{l}Q{S}^2=Q{T}^2+S{T}^2-2\left(QT\right)\left(ST\right)\cos \angle QTS\\ {23.11}^2={16.97}^2+{14.1}^2-2\left(16.97\right)\left(14.1\right)\cos \angle QTS\\ \cos \angle QTS=\frac{{16.97}^2+{14.1}^2-{23.11}^2}{2\left(16.97\right)\left(14.1\right)}\\ \cos \angle QTS=-\frac{47.28}{478.55}\\ \angle QTS={95.67}^o\\ \\ \text{Thus, area of green coloured plane }QTS\\ =\frac{1}{2}\left(16.97\right)\left(14.1\right)\sin {95.67}^o\\ =119.05{\text{ cm}}^2\end{array}$


(c)



$\begin{array}{l}\text{Let the shortest length from point }T\\ \text{to the straight line }QS\text{ is }h.\\ \\ \text{Area of }\Delta QTS=119.05\\ \frac{1}{2}\left(h\right)\left(23.11\right)=119.05\\ h=10.3\text{ cm}\end{array}$