Question : $PQR$ is a triangle, whose area is 180 cm2. $S$ is a point on side $QR$ such that $PS$ is the angle bisector of $\angle QPR$. If $PQ: PR = 2:3$, then what is the area (in cm2) of triangle $PSR$?
Option 1: 90
Option 2: 108
Option 3: 144
Option 4: 72
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Correct Answer: 108
Solution :
Given: $ar(\triangle PQR)=180 \, \operatorname{ cm^2}$. S is a point on side $QR$ such that $PS$ is the angle bisector of $\angle QPR$ and $PQ: PR = 2:3$.
$\angle QPS=\angle RPS$ (because $PS$ is the angle bisector of $\angle QPR$)
By the Angle bisector theorem,
⇒ $\frac{PQ}{PR}=\frac{QS}{SR}$
⇒ $\frac{2}{3}=\frac{QS}{SR}$
⇒ $\frac{QS}{SR}=\frac{ar(\triangle PQS)}{ar(\triangle PSR)}$
⇒ $\frac{ar(\triangle PQS)}{ar(\triangle PSR)}=\frac{2}{3}$
⇒ $\frac{ar(\triangle PSR)}{ar(\triangle PQR)}=\frac{3}{5}$
⇒ $ar(\triangle PSR)=\frac{3}{5}×ar(\triangle PQR)$
⇒ $ar(\triangle PSR)=\frac{3}{5}×180=108 \, \operatorname{ cm^2}$
Hence, the correct answer is 108.
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