Ex 9.3, 13 - ABCD is a trapezium with AB || DC. A line - Triangles with same base & same parallel lines

Ex 9.3, 13 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

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


Transcript

Question 13 ABCD is a trapezium with AB ∥ DC . A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ADX) = ar(ACY). Given: A trapezium ABCD where AB ∥ DC & AC ∥ XY To prove: ar (ADX) = ar (ACY) Construction: Joining XC Proof : For ΔACX and ΔACY lie on the same base AC and are between parallel lines AC & XY ∴ ar(ACX) = ar(ACY) Also, for ΔACX and ΔADX lie on the same base AX and are between parallel lines AX & DC ∴ ar(ACX) = ar(ADX) From (1) & (2), ar( ACY ) = ar(ADX) 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.