Q12 of 42 Page 220

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.

Given, ABC is an isosceles triangle in which ∠B = 90°


Need to find the ratio between the areas of Δ ABE and Δ ACD


⇒ AB = BC


⇒ By Pythagoras theorem, we have AC2 = AB2 + BC2


⇒ since AB = BC


⇒ AC2 = AB2 + AB2


⇒ AC2 = 2 AB2 …..eq(1)


⇒ it is also given that Δ ABE ∼ Δ ACD


(ratio of areas of similar triangles is equal to ratio of squares of their corresponding sides)


⇒ =


⇒ = from 1


⇒ =


∴ ar(Δ ABC):ar(Δ ACD) = 1:2


Hence the ratio is 1:2


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