Question

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

If ar.(∆DFB) = 30 cm²; find the area of parallelogram.

Answer 1

BC = CE                     .....( given )
Also, in parallelogram ABCD, BC = AD
⇒ AD = CE
Now, in ΔADF and ΔECF, We have
AD = CE
∠ADF = ∠ECF           .....( Alternate angles )
∠DAF = ∠CEF           ......( Alternate angles )
∴ ΔADF ≅ ΔECF       ......( ASA Criterion )
⇒ Area( ΔADF ) = Area( ΔECF )     ....(1)

Also, in ΔFBE, FC is the median     ....( Since BC = CE )
⇒ Area( ΔBCF ) = Area( ΔECF )      .....(2)

From (1) and (2)
Area( ΔADF ) = Area( ΔBCF )         ......(3)
Again, ΔADF and ΔBDF are on the base DF and between parallels DF and AB.
⇒ Area( ΔBDF ) = Area( ΔADF )    ........(4)

From (3) and (4),
Area( ΔBDF ) = Area( ΔBCF ) = 30 cm²
Area( ΔBCD ) = Area( ΔBDF ) + Area( ΔBCF ) = 30 + 30 = 60 cm²
Hence, Area of parallelogram ABCD = 2 x Area( ΔBCD ) = 2 x 60 = 120cm^2.