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To prove ar(ACB) = ar(ACF) Since, the triangles between the same parallels and on the same base are equal in area. i.e., ΔACB and ΔACF are on the same base AC and between same parallels AC and BF ∴ ar(ACB) = ar(ACF)
ii)To prove ar(AEDF) = ar(ABCDE) since, ar(ACB) = ar(ACF) [ Proved] Adding ar(AEDC) to both sides we get ⇒[ar(ACB) + ar(AEDC)] = [ar(ACF) + ar(AEDC)] ⇒ ar(AEDF) = ar(ABCDE)
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∆ABC and ∆ACF being on the same base AC and between the same parallel AC and BF
therefore,ar(∆ABC)=ar(∆ACF)
NOW, ar(AEDF)=ar(∆ACF)
=ar (AEDC)+ar(∆ABC)
=ar(ABCDE)
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