Question 14
If S is a point on side PQ of a ∆ PQR such that PS = QS = RS, then
PR . QR = RS2 (B) QS2 + RS2 = QR2
(C) PR2 + QR2 = PQ2 (D) PS2 + RS2 = PR2
Given
PS = QS = RS
Now,
We know that
Angles opposite equal sides are equal
In Δ PSR
∠ 1 = ∠ 2
In Δ QSR
∠ 3 = ∠ 4
In Δ PQR
By Angle sum property
∠ P + ∠ Q + ∠ R = 180°
∠ 2 + ∠ 4 + (∠ 1 + ∠ 3) = 180°
Putting ∠ 1 = ∠ 2 and ∠ 3 = ∠ 4
∠ 1 + ∠ 3 + (∠ 1 + ∠ 3) = 180°
2(∠ 1 + ∠ 3) = 180°
(∠ 1 + ∠ 3) = (180° )/2
(∠ 1 + ∠ 3) = 90°
∠ R = 90°
Thus, Δ PRQ is a right angled triangle
where ∠ R = 90°
By Pythagoras theorem
PQ2 = PR2 + QR2
So, the correct answer is (C)

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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