If the perimeter of an isosceles right triangle is `(16sqrt2+16)` cm, then the area of the triangle is:

Options

A. 76 sq.cm
B. 64 sq.cm
C. 58 sq.cm
D. 66 sq.cm

Answer

To solve the problem, we need to find the area of an isosceles right triangle given its perimeter. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Triangle**:
- We have an isosceles right triangle, which means two sides are equal, and the angles opposite these sides are also equal (45° each). Let's denote the equal sides as \( A \) and the hypotenuse as \( AC \).

2. **Using the Pythagorean Theorem**:
- For an isosceles right triangle, the relationship between the sides is given by:
\[ AC = \sqrt{A^2 + A^2} = \sqrt{2A^2} = A\sqrt{2} \]

3. **Calculating the Perimeter**:
- The perimeter \( P \) of the triangle can be expressed as:
\[ P = A + A + AC = 2A + A\sqrt{2} \]
- We know from the problem that the perimeter is \( 16\sqrt{2} + 16 \).

4. **Setting Up the Equation**:
- We can set up the equation:
\[ 2A + A\sqrt{2} = 16\sqrt{2} + 16 \]

5. **Factoring Out \( A \)**:
- Rearranging gives:
\[ A(2 + \sqrt{2}) = 16\sqrt{2} + 16 \]

6. **Solving for \( A \)**:
- To isolate \( A \), we divide both sides by \( (2 + \sqrt{2}) \):
\[ A = \frac{16\sqrt{2} + 16}{2 + \sqrt{2}} \]

7. **Rationalizing the Denominator**:
- Multiply the numerator and denominator by \( (2 - \sqrt{2}) \):
\[ A = \frac{(16\sqrt{2} + 16)(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} \]
- The denominator simplifies to:
\[ 2^2 - (\sqrt{2})^2 = 4 - 2 = 2 \]
- The numerator expands to:
\[ 32\sqrt{2} - 16 + 32 - 16\sqrt{2} = 16\sqrt{2} + 16 \]
- Thus:
\[ A = \frac{16\sqrt{2} + 16}{2} = 8\sqrt{2} + 8 \]

8. **Finding the Area**:
- The area \( A_{triangle} \) of the triangle is given by:
\[ A_{triangle} = \frac{1}{2} \times A \times A \]
- Substituting \( A \):
\[ A_{triangle} = \frac{1}{2} \times (8\sqrt{2}) \times (8\sqrt{2}) = \frac{1}{2} \times 64 \times 2 = 64 \text{ cm}^2 \]

### Final Answer:
The area of the triangle is \( 64 \text{ cm}^2 \).