Question

In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF: FE = 5:3 and area of  ΔADF is 60 cm²; find
(i) area of ΔADE.
(ii) if AE: EB = 4:5, find the area of  ΔADB.
(iii) also, find the area of parallelogram ABCD.

Answer 1

ΔADF and ΔAFE have the same vertex A and their bases are on the same straight line DE.
∴ `"A(ΔADF)"/"A(ΔAFE)" = "DF"/"FE"`

⇒ `60/"A(ΔAFE)" = 5/3 `

⇒ A(ΔAFE) = `( 60 xx 3 )/5` = 36cm².

Now, A(ΔADE) = A(ΔADF) + A(ΔAFE) = 60 + 36 = 96 cm².

ΔADE and ΔEDB have the same vertex D and their bases are on the same straight line AB.\

∴ `"A( ΔADE )"/"A( ΔEDB )" = "AE"/"EB"`

⇒ `96/"A( ΔEDB )" = 4/5`

⇒ A( ΔEDB ) = `( 96 xx 5 )/4` = 120 cm² .

Now, A( ΔADB ) and ||^m ABCD are on the same base AB and between the same parallels AB and DC.

∴ A( ΔADB ) = `1/2` A( ||^m ABCD ) 

⇒ 216 = `1/2` A( ||^m ABCD ) 

⇒ A( ||^m ABCD ) = 2 x 216 = 432 cm² .