In an isosceles triangle, the lengh of each equal side is 1.5 times the length of the third side. The ratio of areas of isoscees triangle and an equilateral triangle with same perimeter is

Options

A. `3sqrt(6) :8`
B. `9:8sqrt(6)`
C. `9sqrt(6) : 8`
D. `9sqrt(2) : 8`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the sides of the isosceles triangle
Let the length of the third side of the isosceles triangle be \( x \). Since each of the equal sides is 1.5 times the length of the third side, the lengths of the equal sides will be:
\[ \text{Length of equal sides} = 1.5x \]

### Step 2: Calculate the perimeter of the isosceles triangle
The perimeter \( P \) of the isosceles triangle is the sum of all its sides:
\[ P = x + 1.5x + 1.5x = 4x \]

### Step 3: Define the side of the equilateral triangle
Let the side length of the equilateral triangle be \( y \). The perimeter of the equilateral triangle is:
\[ P = 3y \]
Setting the perimeters equal gives:
\[ 3y = 4x \implies y = \frac{4x}{3} \]

### Step 4: Calculate the area of the isosceles triangle
The area \( A \) of the isosceles triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \( x \) and we need to find the height \( h \).

Using the Pythagorean theorem, we can find the height. The height divides the base into two equal parts of \( \frac{x}{2} \). The length of each equal side is \( 1.5x \). Therefore, we can set up the equation:
\[ h^2 + \left(\frac{x}{2}\right)^2 = (1.5x)^2 \]
This simplifies to:
\[ h^2 + \frac{x^2}{4} = 2.25x^2 \]
\[ h^2 = 2.25x^2 - \frac{x^2}{4} = \frac{9x^2}{4} - \frac{x^2}{4} = \frac{8x^2}{4} = 2x^2 \]
Thus, the height \( h \) is:
\[ h = \sqrt{2}x \]
Now substituting back into the area formula:
\[ A = \frac{1}{2} \times x \times \sqrt{2}x = \frac{\sqrt{2}}{2} x^2 \]

### Step 5: Calculate the area of the equilateral triangle
The area \( A_e \) of the equilateral triangle is given by:
\[ A_e = \frac{\sqrt{3}}{4} y^2 \]
Substituting \( y = \frac{4x}{3} \):
\[ A_e = \frac{\sqrt{3}}{4} \left(\frac{4x}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{16x^2}{9} = \frac{4\sqrt{3}}{9} x^2 \]

### Step 6: Find the ratio of the areas
Now we need to find the ratio of the area of the isosceles triangle to the area of the equilateral triangle:
\[ \text{Ratio} = \frac{A}{A_e} = \frac{\frac{\sqrt{2}}{2} x^2}{\frac{4\sqrt{3}}{9} x^2} \]
Cancelling \( x^2 \) from numerator and denominator:
\[ \text{Ratio} = \frac{\frac{\sqrt{2}}{2}}{\frac{4\sqrt{3}}{9}} = \frac{\sqrt{2} \cdot 9}{2 \cdot 4\sqrt{3}} = \frac{9\sqrt{2}}{8\sqrt{3}} \]

### Step 7: Simplify the ratio
To express this ratio in a more standard form, we can multiply the numerator and denominator by \( \sqrt{3} \):
\[ \text{Ratio} = \frac{9\sqrt{6}}{24} = \frac{3\sqrt{6}}{8} \]

### Final Answer
The ratio of the areas of the isosceles triangle to the equilateral triangle is:
\[ \frac{3\sqrt{6}}{8} \]