Choose the correct answer from the given four options:
If $ S $ is a point on side $ P Q $ of a $ \triangle P Q R $ such that $ P S=Q S=R S $, then
(A) $ \mathrm{PR}, \mathrm{QR}=\mathrm{RS}^{2} $
(B) $ \mathrm{QS}^{2}+\mathrm{RS}^{2}=\mathrm{QR}^{2} $
(C) $ \mathrm{PR}^{2}+\mathrm{QR}^{2}=\mathrm{PQ}^{2} $
(D) $ \mathrm{PS}^{2}+\mathrm{RS}^{2}=\mathrm{PR}^{2} $

Given:

\( S \) is a point on side \( P Q \) of a \( \triangle P Q R \) such that \( P S=Q S=R S \).

To do:

We have to choose the correct answer.

Solution:

Let $\angle PRS=\angle 1, \angle RPS=\angle 2, \angle SRQ=\angle 3$ and $\angle RQS =\angle 4$

$P S=Q S=R S$

In $\triangle P S R$,

$P S=R S$

This implies,

$\angle 1=\angle 2$
Similarly,

In $\triangle R S Q$,

$\angle 3=\angle 4$     (Corresponding angles)

In $\triangle P Q R$,

$\angle P+\angle Q+\angle R=180^{\circ}$

$\angle 2+\angle 4+\angle 1+\angle 3=180^{\circ}$

$\angle 1+\angle 3+\angle 1+\angle 3 =180^{\circ}$

$2(\angle 1+\angle 3) =180^{\circ}$

$\angle 1+\angle 3=\frac{180^{\circ}}{2}$

$=90^{\circ}$

$\angle R =90^{\circ}$

Therefore,

In $\triangle P Q R$, by Pythagoras theorem,

$P R^{2}+Q R^{2}=P Q^{2}$.

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Updated on: 10-Oct-2022

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