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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.
Solution
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 = 120cm2.
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