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Question
ABC is a triangle in which ∠A =90°, AN⊥ BC, BC = 12 cm and AC = 5cm. Find the ratio of the areas of ΔANC and ΔABC.
Solution
In ΔANC and ΔABC
∠C = ∠C [Common]
∠ANC = ∠BAC [Each 90°]
Then, ΔANC ~ ΔBAC [By AA similarity]
By area of similarity triangle theorem
`("Area"(triangleANC))/("Area"(triangleBAC))="AC"^2/"BC"^2`
`=5^2/12^2`
`=25/144`
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