Ex 9.3

Ex 9.3, 1
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Ex 9.3, 2 Important Deleted for CBSE Board 2023 Exams

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Ex 9.3, 11 Important Deleted for CBSE Board 2023 Exams You are here

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Chapter 9 Class 9 - Areas of Parallelograms and Triangles [Deleted]

Serial order wise

Last updated at March 2, 2017 by Teachoo

Ex 9.3, 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) Ex 9.3, 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