1.

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that(i) ar(ΔACB)=ar(ΔACF)(ii) ar(AEDF)=ar(ABCDE)

Answer»

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

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)


Hope this will help you...


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