ABC is an isosceles right angle `Delta ` . If Perimeter is P find Area

Answer

To find the area of an isosceles right triangle ABC with a given perimeter P, we can follow these steps:

### Step 1: Understand the triangle properties
An isosceles right triangle has two equal sides and a right angle (90 degrees) between them. Let's denote the lengths of the equal sides as \( a \) and the hypotenuse as \( AC \).

### Step 2: Express the perimeter
The perimeter \( P \) of triangle ABC can be expressed as:
\[ P = AB + BC + AC = a + a + AC = 2a + AC \]

### Step 3: Use the Pythagorean theorem
According to the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Since \( AB = a \) and \( BC = a \):
\[ AC^2 = a^2 + a^2 = 2a^2 \]
Thus, we can find \( AC \):
\[ AC = \sqrt{2a^2} = a\sqrt{2} \]

### Step 4: Substitute \( AC \) into the perimeter equation
Now substitute \( AC \) back into the perimeter equation:
\[ P = 2a + a\sqrt{2} \]
This can be rearranged to find \( a \):
\[ P = a(2 + \sqrt{2}) \]
So,
\[ a = \frac{P}{2 + \sqrt{2}} \]

### Step 5: Rationalize the expression for \( a \)
To simplify \( a \), we can rationalize the denominator:
\[ a = \frac{P(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{P(2 - \sqrt{2})}{2^2 - (\sqrt{2})^2} = \frac{P(2 - \sqrt{2})}{4 - 2} = \frac{P(2 - \sqrt{2})}{2} \]

### Step 6: Calculate the area of the triangle
The area \( A \) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case, both the base and height are equal to \( a \):
\[ A = \frac{1}{2} \times a \times a = \frac{1}{2} a^2 \]

### Step 7: Substitute \( a \) into the area formula
Now substitute the expression for \( a \):
\[ A = \frac{1}{2} \left(\frac{P(2 - \sqrt{2})}{2}\right)^2 \]
Calculating \( a^2 \):
\[ a^2 = \left(\frac{P(2 - \sqrt{2})}{2}\right)^2 = \frac{P^2(2 - \sqrt{2})^2}{4} \]
Thus,
\[ A = \frac{1}{2} \cdot \frac{P^2(2 - \sqrt{2})^2}{4} = \frac{P^2(2 - \sqrt{2})^2}{8} \]

### Step 8: Expand \( (2 - \sqrt{2})^2 \)
Calculating \( (2 - \sqrt{2})^2 \):
\[ (2 - \sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2} \]
Thus, the area becomes:
\[ A = \frac{P^2(6 - 4\sqrt{2})}{8} \]

### Final Result
The area of the triangle ABC is:
\[ A = \frac{P^2(6 - 4\sqrt{2})}{8} \]