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Question
ABC is an isosceles triangle, right-angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ΔABE and ΔACD.
Solution
We have, ABC as an isosceles triangle, right angled at B.
Now, AB = BC
Applying Pythagoras theorem in right-angled triangle ABC, we get:
`AC^2=AB^2+BC^2=2AB^2 (∵ AB=AC)` .............(1)
∵ Δ ACD ∼ Δ ABE
We know that ratio of areas of 2 similar triangles is equal to squares of the ratio of their corresponding sides.
`ar(Δ ABE)/ar(ΔACD)=(AB^2)/(AC^2)=(AB^2)/(2AB^2)` [𝑓𝑟𝑜𝑚 (𝑖)]
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