Question : ABC is an isosceles right-angled triangle with $\angle$B = 90°. On the sides AC and AB, two equilateral triangles ACD and ABE have been constructed. The ratio of the area of $\triangle$ABE and $\triangle$ACD is:
Option 1: $1 : 3$
Option 2: $2 : 3$
Option 3: $1 : 2$
Option 4: $1 : \sqrt{2}$
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Correct Answer: $1 : 2$
Solution : Given: $\angle$ABC = 90°, AB = BC In $\triangle$ABC, AC 2 = AB 2 + BC 2 = AB 2 + AB 2 = 2AB 2 Since, $\triangle$ACD $\sim$ $\triangle$ABE, $\frac{\text{area of} \triangle ABE}{\text{area of} \triangle ACD}=\frac{AB^2}{AC^2}$ ⇒ $\frac{\text{area of} \triangle ABE}{\text{area of} \triangle ACD}=\frac{AB^2}{2AB^2}$ $\therefore \frac{\text{area of} \triangle ABE}{\text{area of} \triangle ACD}=\frac{1}{2}$ Hence, the correct answer is $1 : 2$.
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