Question

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

$\left(A\right)\left(PR\right)\left(QR\right)=R{S}^2$
$\left(B\right)Q{S}^2+R{S}^2=Q{R}^2$
$\left(C\right)P{R}^2+Q{R}^2=P{Q}^2$
$\left(D\right)P{S}^2+R{S}^2=P{R}^2$

Answer 1

$Given:In\Delta PQR$
PS = QS = RS …(i)
$In\Delta PSR$
PS = RS [from Eq.(i)]
$\Rightarrow \angle 1=\angle 2...\left(ii\right)$

$Similarly,in\Delta RSQ$
QS = RS (Given)
$\Rightarrow \angle 3=\angle 4$
In $\Delta PQR$,
$\Rightarrow \angle P+\angle Q+\angle R={180}^{\circ }$ (from angle sum property of triangle)
$\Rightarrow \angle 2+\angle 4+\angle 1+\angle 3={180}^{\circ }$
$\Rightarrow \angle 1+\angle 3+\angle 1+\angle 3={180}^{\circ }$
$\Rightarrow 2(\angle 1+\angle 3)={180}^{\circ }$
$\Rightarrow \angle 1+\angle 3=\frac{{180}^{\circ }}{2}={90}^{\circ }$
$\therefore \angle R={90}^{\circ }$
$\Rightarrow$ $\Delta PQR$ is a right angled triangle.

Therefore, by applying Pythagoras theorem in $\Delta PQR$, we get,
$P{R}^2+Q{R}^2=P{Q}^2$