Mathematics
In the adjoining figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of ∆DFB = 3 cm2, find the area of parallelogram ABCD.
Theorems on Area
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Answer
From figure,
BE is a straight line.
Since, BC || AD,
∴ CE || AD.
Hence, ACED is a parallelogram.
Diagonals AE and DC of || ACED bisect each other, so F is mid-point of DC.
So, BF is median of △BDC.
Since, median divides triangle into two triangles with equal areas.
∴ area of △BFC = area of △DFB = 3 cm2.
area of △BDC = area of △BFC + area of △DFB = 6 cm2.
From figure,
⇒ area of △BDC = area of || gm ABCD (As || gm ABCD and △BDC lie on same base CD and between same parallel lines AB and CD.)
⇒ 6 = area of || gm ABCD
⇒ area of || gm ABCD = 12 cm2.
Hence, area of || gm ABCD = 12 cm2.
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