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Question
In the given figure ABC is a triangle with ∠EDB = ∠ACB. Prove that Δ ABC ~ Δ EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm. And area of Δ BED = 9 cm2. Calculate the
(1) length of AB
(2) area of Δ ABC
Solution
In Δ ABC and Δ EBD,
∠ACB = ∠EDB (given)
∠ABC = ∠EBD (common)
∠ABC ~ ∠EBD (by AA- similarity).
1) We have `(AB)/(BE) = (BC)/(BD) => AB = (6 xx 10)/5 = 12 cm`
2)`"Area of Δ ABC"/"Area of Δ BED" = ((AB)/(BE))^2`
=> Area of Δ ABC =`(12/6)^2 xx 9 cm = 4 xx 9 cm^2 = 36 cm^2`
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