Question

"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) (ii) ar (AEDF) = ar (ABCDE)

Class 9 - Math - Areas of Parallelograms and Triangles Page 163"

Answer 1

Two Triangles on the same base and between the same parallels are equal in area.

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Given:

ABCDE is a Pentagon & BF||AC.

To show:

(i) ar (ACB) = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)

Proof:

(i) △ACB and △ACF lie on the same base AC and between the same parallels AC and BF.

∴ ar(△ACB) = ar(△ ACF)

(ii)ar(△ACB) = ar(△ACF)

ar(△ACB)+ar(△ACDE) =ar(△ACF) + ar(△ACDE)

[ On adding ar(△ACDE) on both sides]

 ar(ABCDE) = ar(△AEDF)

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Hope this will help you...