Question

Plzzzz help me in this question????

Answer 1

ATQ,
AC is parallel to BF

1)
ar (ACF) = ar (ABC)
THEY LIE ON THE SAME BASE AC AND BETWEEN SAME PARALLELS.

2)
ADDING COMMONS ar(AEDC) TO BOTH SIDES WE GET,
ar(ACF) + ar ( AEDC) = ar ( ABC) = ar(AEDC)
ar (AEDF) = ar (ABCDE)

Answer 2

Given:

ABCDE is a Pentagon & BF||AC.

To prove :

(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
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