Last updated at Dec. 16, 2024 by Teachoo
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)
About the Author
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