If S is a point on side PQ of $Δ P Q R$ , such that $P S = Q S = R S,$ then:

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

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

P R 2 + Q R 2 = P Q 2

{PQ}^{2} - {PR}^{2} = {QR}^{2}

answer is C.

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Detailed Solution

It is given that, S is a point on side PQ of

Δ P Q R

, in such a way that

P S = Q S = R S .

Let us draw the figure according to the given data.
Question Image

We shall use the properties of right-angled triangles for solving this.
Now in ΔPSR,
PS=SR (given)
∠1=∠P [angles opposite to equal sides in a triangle are equal]
Similarly, in ΔSRQ,
RS=SQ(given)
∠Q=∠2
Now, if we consider, ΔPQR,
As we know that the sum of the angles of a triangle is 180°.
So, ∠P+∠Q+∠R=

180^{°}

According to the AAA criterion if two angles are equal of two angles of another triangle, the third angle is also same.
Therefore, ∠1+∠2=∠PRQ
∠1+∠2+∠1+∠2=