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If S is a point on side PQ of a triangle PQR such that PS = QS = RS, then

$(A)$ $PR .QR = RS^{2}$ $(B)$ $QS^{2} + RS^{2} = QR^{2}$

$(C)$ $PR^{2} + QR^{2} = PQ^{2}$ $(D)$ $PS^{2} + RS^{2} = PR^{2}$

Answers (1)

Answer : [C]

Given: In triangle PQR

$PS = QS = RS$

$\Rightarrow PS = RS$

$\therefore \angle 1=\angle 2 ...(1)$

(Q If a right angle triangle have equal length of base and height then their acute angles are also equal)

$\text{Similarly}$ $\angle 3=\angle 4 ...(2)$

( Q corresponding angles of equal sides are equal)

$\text{In }\Delta PQR$

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

$\because$ sum of angles of a triangle is 180°

$\angle 2+\angle 4+\angle 1+\angle 3=180^{o} ....(3)$

Using equation (1) and (2) in (3)

$\angle 1+\angle 3+\angle 1+\angle 3=180$

$2\left ( \angle 1+\angle 3 \right )=180^{o}$

$\Rightarrow \angle R=90^{o}$

$\text{Here }\angle R=90^{o}\therefore \Delta PQR$ $\text{is right angle triangle }$

$PR^{2} + QR^{2} = PQ^{2}$

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