Q28 of 95 Page 5

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 is right angled i.e. 90°

⇒ AB = BC


In right angled ∆ABC, by Pythagoras theorem, we have


(Perpendicular)2 + (Base)2 = (Hypotenuse)2


⇒ (AB)2 + (BC)2 = (AC)2


⇒ (AB)2 + (AB)2 = (AC)2


[∵ABC is an isosceles triangle, AB =BC]


⇒ 2(AB)2 = (AC)2


⇒ (AC)2 = 2(AB)2 …(i)


It is also given that ∆ABE ~ ∆ADC


And we also know that, the ratio of similar triangles is equal to the ratio of their corresponding sides.



[from (i)]



∴ ar(∆ABE) : ar(∆ADC) = 1 : 2


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