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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.
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Solution
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 .
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