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

Last updated at March 2, 2017 by

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

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

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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

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