If perimeter of an isoceles `Delta ` is 64 cm and height is 8 cm find area .

Answer

To solve the problem of finding the area of an isosceles triangle with a perimeter of 64 cm and a height of 8 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Isosceles Triangle**:
- An isosceles triangle has two equal sides. Let's denote the length of the equal sides as \( A \) and the base as \( B \).

2. **Using the Perimeter**:
- The perimeter of the triangle is given by the formula:
\[ \text{Perimeter} = 2A + B \]
- According to the problem, the perimeter is 64 cm:
\[ 2A + B = 64 \]

3. **Rearranging the Equation**:
- We can rearrange the equation to express \( B \) in terms of \( A \):
\[ B = 64 - 2A \]

4. **Understanding the Height**:
- The height of the triangle divides the base \( B \) into two equal parts. Therefore, each part is:
\[ \frac{B}{2} \]

5. **Using the Pythagorean Theorem**:
- In the right triangle formed by the height, we can apply the Pythagorean theorem:
\[ A^2 = \left(\frac{B}{2}\right)^2 + \text{Height}^2 \]
- Substituting the height (8 cm):
\[ A^2 = \left(\frac{B}{2}\right)^2 + 8^2 \]

6. **Substituting for \( B \)**:
- Substitute \( B = 64 - 2A \) into the equation:
\[ A^2 = \left(\frac{64 - 2A}{2}\right)^2 + 8^2 \]
- Simplifying:
\[ A^2 = \left(32 - A\right)^2 + 64 \]

7. **Expanding and Rearranging**:
- Expanding \( (32 - A)^2 \):
\[ A^2 = 1024 - 64A + A^2 + 64 \]
- Cancel \( A^2 \) from both sides:
\[ 0 = 1088 - 64A \]
- Rearranging gives:
\[ 64A = 1088 \]
- Solving for \( A \):
\[ A = \frac{1088}{64} = 17 \]

8. **Finding \( B \)**:
- Substitute \( A \) back into the equation for \( B \):
\[ B = 64 - 2 \times 17 = 30 \]

9. **Calculating the Area**:
- The area \( A \) of the triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
- Substituting the values:
\[ \text{Area} = \frac{1}{2} \times 30 \times 8 = 120 \text{ cm}^2 \]

### Final Answer:
The area of the isosceles triangle is \( 120 \text{ cm}^2 \).

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