ABC is an isosceles right angled triangle with `angleB = 90^(@)`. On the sides AC and AB, two equilateral triangles ACD and ABE have been constructed. The ratio of area of `DeltaABE` and `Delta ACD` is

Options

A. `1:3`
B. `2:3`
C. `1:2`
D. `1:sqrt(2)`

Answer

To solve the problem, we need to find the ratio of the areas of the two equilateral triangles ABE and ACD constructed on the sides AB and AC of the isosceles right triangle ABC, where angle B is 90 degrees.

### Step-by-Step Solution:

1. **Identify the sides of triangle ABC**:
Since triangle ABC is an isosceles right triangle with angle B = 90 degrees, let the lengths of sides AB and AC be equal to \( a \). Therefore, we have:
\[ AB = AC = a \]
The length of the hypotenuse BC can be calculated using the Pythagorean theorem:
\[ BC = \sqrt{AB^2 + AC^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

2. **Determine the sides of the equilateral triangles**:
The equilateral triangle ABE is constructed on side AB, and the equilateral triangle ACD is constructed on side AC. Therefore, the sides of these equilateral triangles are:
- For triangle ABE: side = \( a \)
- For triangle ACD: side = \( a \)

3. **Calculate the area of triangle ABE**:
The formula for the area \( A \) of an equilateral triangle with side length \( s \) is given by:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
For triangle ABE, substituting \( s = a \):
\[ \text{Area of } \triangle ABE = \frac{\sqrt{3}}{4} a^2 \]

4. **Calculate the area of triangle ACD**:
Similarly, for triangle ACD, since it also has side length \( a \):
\[ \text{Area of } \triangle ACD = \frac{\sqrt{3}}{4} a^2 \]

5. **Find the ratio of the areas**:
Now, we can find the ratio of the areas of triangles ABE and ACD:
\[ \text{Ratio of areas} = \frac{\text{Area of } \triangle ABE}{\text{Area of } \triangle ACD} = \frac{\frac{\sqrt{3}}{4} a^2}{\frac{\sqrt{3}}{4} a^2} = 1 \]

### Final Answer:
The ratio of the area of triangle ABE to the area of triangle ACD is:
\[ \text{Ratio} = 1:1 \]