Ex 9.3, 11 - In figure, ABCDE is a pentagon. A line through - Triangles with same base & same parallel lines

Ex 9.3, 11 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

Go Ad-free

Transcript

Question 11 In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar (ACB) = ar (ACF) Given: A pentagon ABCDE where BF AC To prove: ar (ACB) = ar (ACF) Proof : ACB and ACF lie on the same base AC and are between the same parallels AC and BF. ar( ACB) = ar( ACF) Question 11 In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (ii) ar (AEDF) = ar (ABCDE) In part(i), we proved that ar( ACB) = ar( ACF) Adding ar(AEDC) both sides ar( ACB) + ar(AEDC) = ar( ACF) + ar(AEDC) Area (ABCDE) = Area (AEDF) 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo