SSC CGL Model Paper 16

∠QPT = 30°
∠QPS = 2 × 30° = 60°
∠SPR = ∠QPS= 60°
⇒ ∠TPR = ∠TPS + ∠SPR
∠TPR = 30°+ 60° = 90° In ΔPTR,
PT = 9 cm
TR = 15 cm
In right angle ΔPTR, Pythagoras theorem,
$\mathit{T}{\mathit{R}}^2=\mathit{P}{\mathit{T}}^2+\mathit{P}{\mathit{R}}^2$
⇒ ${\left(15\right)}^2={\left(9\right)}^2=\mathit{P}{\mathit{R}}^2$
⇒ $225=81+\mathit{P}{\mathit{R}}^2$
⇒ $\mathit{P}{\mathit{R}}^2=144$
∴ PR = 12 cm
Area of ΔPTR $=\frac{1}{2}\times 12\times 9$
$=54{\mathrm{cm}}^2$