If perimeter of an isoceles right angled `Delta` is P . Then find the area of this triangle .

Answer

To find the area of an isosceles right-angled triangle given its perimeter \( P \), we can follow these steps:

### Step 1: Understand the properties of the triangle
An isosceles right-angled triangle has two equal sides and one right angle. Let's denote the equal sides as \( a \). The angles opposite these sides are both \( 45^\circ \).

### Step 2: Express the hypotenuse in terms of \( a \)
Using the Pythagorean theorem, the hypotenuse \( AC \) can be expressed as:
\[ AC = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

### Step 3: Write the perimeter equation
The perimeter \( P \) of the triangle can be expressed as the sum of all its sides:
\[ P = a + a + a\sqrt{2} = 2a + a\sqrt{2} \]

### Step 4: Solve for \( a \)
Rearranging the perimeter equation gives:
\[ P = a(2 + \sqrt{2}) \]
From this, we can isolate \( a \):
\[ a = \frac{P}{2 + \sqrt{2}} \]

### Step 5: Calculate the area of the triangle
The area \( A \) of a triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In our case, both the base and height are equal to \( a \):
\[ A = \frac{1}{2} \times a \times a = \frac{1}{2} a^2 \]

### Step 6: Substitute \( a \) into the area formula
Now substituting \( a = \frac{P}{2 + \sqrt{2}} \) into the area formula:
\[ A = \frac{1}{2} \left(\frac{P}{2 + \sqrt{2}}\right)^2 \]
Calculating \( a^2 \):
\[ A = \frac{1}{2} \times \frac{P^2}{(2 + \sqrt{2})^2} \]

### Step 7: Simplify the expression
Now, we need to simplify \( (2 + \sqrt{2})^2 \):
\[ (2 + \sqrt{2})^2 = 4 + 2 + 4\sqrt{2} = 6 + 4\sqrt{2} \]
Thus, the area becomes:
\[ A = \frac{P^2}{2(6 + 4\sqrt{2})} \]

### Step 8: Final expression for the area
So, the final expression for the area of the isosceles right-angled triangle is:
\[ A = \frac{P^2}{12 + 8\sqrt{2}} \]