Question

Question 12
If S is a point on side PQ of a $\Delta$ PQR such that PS = QS = RS, then

(A) $PR.QR=R{S}^2$
(B) $Q{S}^2+R{S}^2=Q{R}^2$
(C) $P{R}^2+Q{R}^2=P{Q}^2$
(D)​​​​​​​ $P{S}^2+R{S}^2=P{R}^2$​​​​​​​

Answer 1

(C) $P{R}^2+Q{R}^2=P{Q}^2$

Given, in $\Delta PQR$

PS = QS = RS ………(i)
In $\Delta PSR$ PS = RS [from Eq.(i)]
$\Rightarrow \angle 1=\angle 2$ ……..(ii)
Similarly, in $\Delta RSQ$
QS = RS
$\Rightarrow \angle 3=\angle 4$
In $\Delta PQR$,
$\Rightarrow \angle P+\angle Q+\angle R={180}^{\circ }$
$\Rightarrow \angle 2+\angle 4+\angle 1+\angle 3={180}^{\circ }$
$\Rightarrow \angle 1+\angle 3+\angle 1+\angle 3={180}^{\circ }$
$\Rightarrow 2(\angle 1+\angle 3)={180}^{\circ }$
$\Rightarrow \angle 1+\angle 3=\frac{{180}^{\circ }}{2}={90}^{\circ }$
$\therefore \angle R={90}^{\circ }$
i.e; $\Delta PQR$ is a right angled triangle.
In $\Delta PQR$, by Pythagoras theorem,
$P{R}^2+Q{R}^2=P{Q}^2$