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.

Explanation

Given, ABCD is a parallelogram and CE = BC i.e., C is the mid-point of BE.
Also, `" " ar (DeltaDFB) = 3 cm^(2)`
Now, `DeltaADF` and `DeltaDFB` are on the same base DF and between parallels CD and AB. Then,
`ar(DeltaADF) = ar (DeltaDFB) = 3 cm^(2)" "` ...(i)
In `DeltaABE`, by the converse of mid-point theorem,
`EF = AF" "` [since, C is mid-point of BE] ...(ii)
In `DeltaADF` and `DeltaECF`,
`angleAFD = angleCFE" "` [Vertically opposite angles]
`AF = EF" "` [from Eq. (ii)]
and `" " angleDAF = angleCEF`
[since, `BE || AD` and AE is transversal, then alternate interior angles are equal]
`therefore" "` `DeltaADF ~=DeltaECF" "` [by ASA congruence rule]
Then, `" " ar (DeltaADF) = ar (DeltaCFE)" "` [since, congruent figures have equal area]
`therefore" "` `ar (DeltaCFE) = ar (DeltaADF) = 3 cm^(2)" "` [from Eq. (i)] ...(iii)
Now, in `DeltaBFE`, C is the mid-point of BE then CF is median of
`DeltaBFE`,
`therefore" "` `ar (DeltaCEF) = ar (DeltaBFC)" "` [since, median of a triangle divides it into two triangles of equal area]
`rArr" "` `ar (DeltaBFC) = 3 cm^(2)" "` ...(iv)
Now, `" " ar (DeltaBDC) = ar (DeltaDFB) + ar (DeltaBFC)" "` [from Eqs. (i) and (iv)]
We know that, diagonal of a parallelogram divides it into two congruent triangles of equal areas.
`therefore" "` Area of parallelogram `ABCD = 2 xx "Area of " DeltaBDC`
`= 2 xx 6 = 12 cm^(2)`
Hence, the area of parallelogram ABCD is `12 cm^(2)` .