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# in fig. DEparallelCA and D is a point on BC such that BD/DC=2/3.the ratio of area of triangle ABC to area of triangle BDE is

Asked by deepika babbar 23rd September 2013, 6:30 PM
DE is parallel to CA
So, angle BDE = angle BCA
angle BED = angle BAC
Hence, triangles BDE and BCA are similar (by AA similarity)
We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
So,
ar (ABC) / ar(EBD) = BC2 / BD2         ... (1)

Given, BD / DC = 2/3
So, BD/BD+DC = 2/2+3
This gives BD/BC = 2/5

From (1),
ar (ABC) / ar(EBD) = 52 / 22 = 25/4
Answered by Expert 23rd September 2013, 10:08 PM
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