(Q21) ABC is an isosceles triangle right angled at B. Equilateral triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ∆ABE and ∆ACD.

Solution :

Given : ∆ABC, AB = BC and ∠B = 90°

∆ABE on AB; ∆ACD on AC are equiangular triangles.

Let equal sides of the isosceles right triangle, AB = BC = a (say)

Then, in ∆ABC, ∠B = 90°

AC² - AB² + BC²

[hypotenuse² = side² + side² -- Pythagoras theorem]

  = a² + ² = 2²

Since, ∆ABE ∼ ∆ACD

[∵ Ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides]

∆ABE : ∆ACD = : .