Question 11 - Chapter 9 Class 9 Areas of parallelograms and Triangles (Important Question)

Last updated at Dec. 13, 2024 by

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

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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

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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