The perimeter of an isosceles right triangle is 2p, its area is

Options

A. (3 - 2`sqrt2)p^(2)`
B. (1 - 2`sqrt2)p^(2)`
C. (3 + 2`sqrt2)p^(2)`
D. (1 + 2`sqrt2)p^(2)`

Answer

To find the area of an isosceles right triangle given that its perimeter is \(2p\), we will follow these steps:

### Step 1: Understand the properties of an isosceles right triangle
An isosceles right triangle has two equal sides and one right angle. Let the lengths of the two equal sides be \(a\). By the Pythagorean theorem, the length of the hypotenuse will be \(a\sqrt{2}\).

### Step 2: Write the perimeter equation
The perimeter \(P\) of the triangle can be expressed as:
\[ P = a + a + a\sqrt{2} = 2a + a\sqrt{2} \]
According to the problem, the perimeter is given as \(2p\). Therefore, we can set up the equation:
\[ 2a + a\sqrt{2} = 2p \]

### Step 3: Solve for \(a\)
We can factor out \(a\) from the left-hand side:
\[ a(2 + \sqrt{2}) = 2p \]
Now, solving for \(a\):
\[ a = \frac{2p}{2 + \sqrt{2}} \]

### 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 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 5: Substitute the value of \(a\) into the area formula
Substituting \(a = \frac{2p}{2 + \sqrt{2}}\) into the area formula:
\[ A = \frac{1}{2} \left(\frac{2p}{2 + \sqrt{2}}\right)^2 \]
Calculating \(a^2\):
\[ A = \frac{1}{2} \times \frac{4p^2}{(2 + \sqrt{2})^2} \]

### Step 6: Simplify \((2 + \sqrt{2})^2\)
Calculating \((2 + \sqrt{2})^2\):
\[ (2 + \sqrt{2})^2 = 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2} \]
Thus, we have:
\[ A = \frac{1}{2} \times \frac{4p^2}{6 + 4\sqrt{2}} = \frac{2p^2}{6 + 4\sqrt{2}} \]

### Step 7: Rationalize the denominator
To rationalize the denominator:
\[ A = \frac{2p^2}{6 + 4\sqrt{2}} \cdot \frac{6 - 4\sqrt{2}}{6 - 4\sqrt{2}} = \frac{2p^2(6 - 4\sqrt{2})}{(6 + 4\sqrt{2})(6 - 4\sqrt{2})} \]
Calculating the denominator:
\[ (6 + 4\sqrt{2})(6 - 4\sqrt{2}) = 36 - 32 = 4 \]
Thus:
\[ A = \frac{2p^2(6 - 4\sqrt{2})}{4} = \frac{p^2(6 - 4\sqrt{2})}{2} \]

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

### Conclusion
The area of the isosceles right triangle is:
\[ A = 3 - 2\sqrt{2} \cdot p^2 \]