ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.

If `ar (DeltaDFB) = 3 cm^(2)`, then find the area of the parallelogram ABCD.

Answer

`28 cm^(2)`

Explanation

`triangleMDA congMCP" " [therefore angleDMA = angleCMP, angleMDA=angleMCP, AD = CP " since " AD = BC and CP = BC]`
`therefore DM = MC` (c.p.c.t) and so BM is a median of `triangleBDC`.
Thus, `ar(DMB)=(1)/(2)ar(BDC)`.
But, `ar(BDC)=(1)/(2)ar(ABCD)` [`therefore` BD is a diagonal of ||gm ABCD].
`therefore ar(DMB)=(1)/(4)ar(ABCD)` and so `ar(ABCD)= 28 cm^(2).