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Question
ABCD is a parallelogram, prove that Area (ΔPAB) = Area (quadrilateral ACPD)

Sum
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Solution
We have,

We know that,
If a triangle and parallelogram are drawn on the same base and between the same parallel lines, then the area of the triangle is half of the area of the parallelogram.
Since, ΔABC and || gm ABCD are on same base BC and between same parallels AD and BC.
∴ ar(ΔABC) = `1/2` ar(|| gm ABCD) ...(i)
Also, ΔPAD and || gm ABCD are on same base AD and between same parallels AD and BC.
∴ ar(ΔPAD) = `1/2` ar(|| gm ABCD) ...(ii)
From (i) and (ii),
Area (ΔABC) = Area (ΔPAD)
Adding Area (ΔACP) to both sides, we get
Area (∆ABC) + Area (∆ACP) = Area (∆PAD) + Area (∆ACP)
Area (∆PAB) = Area (quadrilateral ACPD)
Hence proved.
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