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Question
In figure , DEF is a right -angled triangle with ∠ E = 90 °.FE is produced to G and GH is drawn perpendicular to DE = 8 cm , DH = 8 cm ,DH = 6 cm and HF = 4 cm , find `("Ar" triangle "DEF")/("Ar" triangle "GHF")`
Solution
ln Δ DEF and Δ GHF,
∠ DEF = ∠ GHF (90 ° each)
∠ DEF = ∠ GHF ...(common)
Δ DEF = Δ GHF ....(AA corollary)
`therefore ("Ar" triangle "DEF")/("Ar" triangle "GHF") = "EF"^2/"HF"^2` ......(1)
[The ration of areas of two similar triangle is equal to the ratio of square of their corresponding sides.]
In right Δ DEF, (By Pythagoras theorem)
DE2 + EF2 = DF2
EF2 = 102 - 82
EF2 = 36
EF = 6
From ( 1),
`("Ar" triangle "DEF")/("Ar" triangle "GHF") = (6/4)^2 = 9/4`
i.e. 9 : 4
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