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Question
Prove that the area of Δ BCE described on one side BC of a square ABCD is one half the area of the similar Δ ACF described on the diagonal AC.
Solution
In right angled triangle ABC ,
By Pythagoras Theorem , AB2 + BC2 = AC2
Given , Δ BCE ∼ Δ ACF
`("Ar" triangle "BCE")/("Ar" triangle "ACF") = "BC"^2/"AC"^2`
[The ration of areas of two similar triangle is equal to the ratio of square of their corresponding sides.]
`= "BC"^2/"AC"^2`
`= 1/2`
Required ratio is 1 : 2.
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