The perimeter of an isosceles right angled triangle is 2p cm. Its area is -

Options

A. `(2+sqrt2)p^2 sq cm`
B. `(3+2sqrt2) p sq cm`
C. `(3-2sqrt2)p^2 sq cm`
D. `(2-sqrt2)p sq cm`

Answer

To find the area of an isosceles right-angled triangle with a given perimeter of \(2p\) cm, we can follow these steps:

### Step 1: Understand the triangle's dimensions
In an isosceles right-angled triangle, let the lengths of the two equal sides be \(x\). The length of the hypotenuse can be calculated using the Pythagorean theorem, which gives us:
\[ \text{Hypotenuse} = \sqrt{x^2 + x^2} = \sqrt{2x^2} = x\sqrt{2} \]

### Step 2: Write the perimeter equation
The perimeter \(P\) of the triangle is the sum of all its sides:
\[ P = x + x + x\sqrt{2} = 2x + x\sqrt{2} \]
Given that the perimeter is \(2p\), we can set up the equation:
\[ 2p = 2x + x\sqrt{2} \]

### Step 3: Rearrange the equation to solve for \(x\)
We can factor out \(x\) from the right side:
\[ 2p = x(2 + \sqrt{2}) \]
Now, solve for \(x\):
\[ x = \frac{2p}{2 + \sqrt{2}} \]

### Step 4: Calculate the area of the triangle
The area \(A\) of a triangle is given by the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In our case, both the base and height are equal to \(x\):
\[ A = \frac{1}{2} \times x \times x = \frac{x^2}{2} \]

### Step 5: Substitute \(x\) into the area formula
Now we substitute the value of \(x\) we found earlier:
\[ A = \frac{1}{2} \left(\frac{2p}{2 + \sqrt{2}}\right)^2 \]
Calculating \(x^2\):
\[ x^2 = \left(\frac{2p}{2 + \sqrt{2}}\right)^2 = \frac{4p^2}{(2 + \sqrt{2})^2} \]
Now we simplify \((2 + \sqrt{2})^2\):
\[ (2 + \sqrt{2})^2 = 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2} \]
Thus, we have:
\[ A = \frac{1}{2} \cdot \frac{4p^2}{6 + 4\sqrt{2}} = \frac{2p^2}{6 + 4\sqrt{2}} \]

### Step 6: Rationalize the denominator
To simplify further, we can multiply the numerator and denominator by the conjugate of the denominator:
\[ A = \frac{2p^2(6 - 4\sqrt{2})}{(6 + 4\sqrt{2})(6 - 4\sqrt{2})} \]
Calculating the denominator:
\[ (6 + 4\sqrt{2})(6 - 4\sqrt{2}) = 36 - 32 = 4 \]
Thus, we have:
\[ A = \frac{2p^2(6 - 4\sqrt{2})}{4} = \frac{p^2(6 - 4\sqrt{2})}{2} \]

### Final Result
The area of the isosceles right-angled triangle is:
\[ A = \frac{p^2(3 - 2\sqrt{2})}{1} \text{ cm}^2 \]