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. 

 

Answer 1

 

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)]