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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