Question

In the figure4, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD).

Answer 1

FIGURE IS IN THE ATTACHMENT

SOLUTION:
Given :
ABCD and AEFD are two parallelograms. Join PD.
Now ∆PAD & ∆FAD both lie on the same base AD and between the same parallels AD & EQ.
ar (∆PAD) = ar(∆FAD).............(1)
Now, ∆PAD & ||gm EPDA both lie on the same base AD and between the same parallels AD & EQ.
ar (∆PAD) = ar(EPDA).............(2)
Similarly , ar (∆FAD) = ar(∆FADQ).............(1)
ar (∆PAD) = ar(∆FADQ). (From eq 1).............(3)
From eq 2 & 3
ar (EPDA) = ar(∆FADQ).............(1)
ar (∆PEA) + ar(∆PAD) =ar (∆QFD) + ar(∆AFD)
ar (∆PEA) + ar(∆PAD) =ar (∆QFD) + ar(∆PAD) [From eq 1]
ar(∆PEA) = ar(∆QFD)

HOPE THIS WILL HELP YOU....

Answer 2

ABCD and AEFD are parallelogram as shown in figure .

when ABCD is parallelogram,
BC || AD
AB || CD so, PA || QD { because PA is a part of BA and QD is a part of CD that's why PA || QD if AB || CD }
=> PA || QD -----------(1)

when AEFD is a parallelogram,
EA || FD
EF || AD so, PQ || AD { because AE and PQ lies on same line as shown in figure. }
=> PQ || AD ------------(2)

from equations (1) and (2) it is clear that ,
APQD is also a parallelogram because both the pair of opposite sides are parallel .

now, parallelograms AEFD and APQD are on the same base e.g., AD and they lies on the parallels AD and EQ .
so, ar ( AEFD ) = ar (AQPD)
$=> ar(AEP) + \cancel{ar( APFD)} =\cancel{ar(APFD)} + ar(QFD)$
so, ar(PEA) = ar(QFD)