Check sibling questions
Chapter 8 Class 9 Quadrilaterals
Serial order wise

Ex 8.2, 2 - ABCD is a rhombus P, Q, R and S are mid-points - Ex 8.2

Ex 8.2, 2 - Chapter 8 Class 9 Quadrilaterals - Part 2
Ex 8.2, 2 - Chapter 8 Class 9 Quadrilaterals - Part 3
Ex 8.2, 2 - Chapter 8 Class 9 Quadrilaterals - Part 4
Ex 8.2, 2 - Chapter 8 Class 9 Quadrilaterals - Part 5

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Ex 8.2, 2 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. Given: ABCD is rhombus where P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively To prove: PQRS is a rectangle Construction: Join A & C Proof: A rectangle is a parallelogram with one angle 90° First we will prove PQRS is a parallelogram, and prove one angle 90 ° From (1) & (2) PQ ∥ RS and PQ = RS In PQRS, one pair of opposite side is parallel and equal. Hence, PQRS is a parallelogram. Now we prove have to prove PQRS is a rectangle Since AB = BC 1/2AB = 1/2BC So, PB = BQ Now, in Δ BPQ PB = BQ ∴ ∠ 2 = ∠ 1 In ∆ APS & ∆ CQR AP = CQ AS = CR PS = QR ∴ ∆ APS ≅ ∆ CQR ∠ 3 = ∠ 4 Now, AB is a line So, ∠ 3 + ∠ SPQ + ∠ 1 = 180° Similarly, for line BC ∠ 2 + ∠ PQR + ∠ 4 = 180° ∠ 1 + ∠ PQR + ∠ 3 = 180° From (5) & (6) ∠ 1 + ∠ SPQ + ∠ 3 = ∠ 1 + ∠ PQR + ∠ 3 ∴ ∠ SPQ = ∠ PQR Now, PS ∥ QR & PQ is a transversal So, ∠ SPQ + ∠ PQR = 180° ∠ SPQ + ∠ SPQ = 180° 2∠ SPQ = 180° ∠ SPQ = 180"°" /2 = 90° So, PQRS is a parallelogram with one angle 90° ∴ PQRS is a rectangle Hence proved

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.