Question : In the given figure, in $\triangle \operatorname{STU}$, $\mathrm{ST }= 8$ cm, $\mathrm{TU} = 9$ cm and $\mathrm{SU} = 12$ cm. $\mathrm{QU }= 24$ cm, $\mathrm{SR} = 32$ cm and $\mathrm{PT }= 27$ cm. What is the ratio of the area of $\triangle \operatorname{PUQ}$ and the area of $\triangle\operatorname{PTR}$?

Option 1: 1 : 1

Option 2: 4 : 9

Option 3: 2 : 3

Option 4: 5 : 2

Correct Answer: 4 : 9

Solution : Using the Sine rule,
In $\triangle \operatorname{PUQ}$,
Area of $\triangle \operatorname{PUQ}=\frac{1}{2}× \sin (\operatorname{\angle PUQ})×\operatorname{PU}×\operatorname{QU}$
⇒ Area of $\triangle \operatorname{PUQ}=\frac{1}{2}× \sin (\operatorname{\angle PUQ})×18×24$
⇒ Area of $\triangle \operatorname{PUQ}=18× \sin (\operatorname{\angle PUQ}) ×12$
In $\triangle \operatorname{PTR}$,
Area of $\triangle \operatorname{PTR}=\frac{1}{2}× \sin (\operatorname{\angle PTR})×\operatorname{PT}×\operatorname{TR}$
⇒ Area of $\triangle \operatorname{PTR}=\frac{1}{2}× \sin (\operatorname{\angle PTR})×27×24$
⇒ Area of $\triangle \operatorname{PTR}=27× \sin (\operatorname{\angle PTR})×12$
In $\triangle \operatorname{SUT}$,
$\frac{1}{2}× \sin \operatorname{(180^{\circ}–\angle PTR)}×8×9=\frac{1}{2}× \sin \operatorname{(180^{\circ}–\angle PUQ)}×12×9$
⇒ $4× \sin (\operatorname{\angle PTR})=6× \sin (\operatorname{\angle PUQ})$
⇒ $\frac{\sin (\operatorname{\angle PTR})}{\sin (\operatorname{\angle PUQ})}=\frac{3}{2}$
$\therefore\frac{\operatorname{Area of \triangle PUQ}}{  \operatorname{Area of \triangle PTR}}=\frac{18× \sin (\operatorname{\angle PUQ})×12}{27× \sin (\operatorname{\angle PTR})×12}=\frac{2×2}{3×3}=\frac{4}{9}$
Hence, the correct answer is 4 : 9.