Ex 9.4, 4 (Optional) - In Fig, ABCD is a parallelogram and BC is

Ex 9.4, 4 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2
Ex 9.4, 4 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 4 In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.] Given: ABCD is a parallelogram, AD = CQ To prove: ar (BPC) = ar (DPQ) Construction: Join AC. Proof: Here, ∆ ACP and ∆BPC are on the same base PC and between same parallels AB and CP. ∴ ar (ACP) = ar (BPC) (AB ∥ CD as opposite sides of parallelogram are parallel) Now, AD = CQ and AD ∥ QC Since one pair of opposite side is equal and parallel, ∴ ACQD is a Parallelogram Now, Diagonals of parallelogram ACQD bisect each other ∴ CP = DP and, AP = QP (Given) (AD ∥ BC as opposite sides of parallelogram are parallel) In ∆CPA and ∆DPQ CP = DP ∠CPA = ∠DPQ AP = QP ∴ ∠CPA = ∠DPQ ∴ ar (CPA) = ar (DPQ) From (1) & (2) ar (BPC) = ar (DPQ) Hence Proved (Proved earlier) (Vertically opposite Angles) (Proved earlier) (SAS Congruency) (Area of Congruent triangles is equal) …(2) From (1) & (2) ar (BPC) = ar (DPQ) Hence Proved

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.