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Question
In the given figure ABC is a triangle with ∠EDB = ∠ACB.
(i) Prove that ΔABC ∼ ΔEBD.
(ii) If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of ΔBED = 9 cm2. Calculate the length of AB and area of ΔABC.
Solution
(i) In ΔABC and ΔEBD,
∠ACB = ∠EDB ...(given)
∠ABC = ∠EBD ...(common)
∴ ΔABC ∼ ΔEBD
Hence Proved.
(ii) We have, `"AB"/"BE"= "BC"/"BD"`
⇒ AB = `(6 xx 10)/(5) = 12 "cm"`.
(iii) `"Area of ΔABC"/"Area of ΔBED" = ("AB"/"BE")^2`
Area of ΔABC = `(12/16)^2 xx 9 "cm"^2`
= 4 x 9 cm2
= 36 cm2.
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