Find the altitude and area of an isosceles triangle whose perimeter is 64 cm and where base is 2 cm.

Answer

To find the altitude and area of an isosceles triangle with a perimeter of 64 cm and a base of 2 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Triangle**:
- Start by sketching an isosceles triangle with a base of 2 cm. Label the vertices as A, B, and C, where AB = AC (the equal sides) and BC = 2 cm (the base).

2. **Identify the Perimeter**:
- The perimeter of the triangle is given as 64 cm. The perimeter (P) of a triangle is the sum of the lengths of its sides. Therefore, we can express this as:
\[ P = AB + AC + BC = x + x + 2 = 2x + 2 \]
- Here, let the length of the equal sides (AB and AC) be \(x\).

3. **Set up the Equation**:
- Since the perimeter is 64 cm, we can set up the equation:
\[ 2x + 2 = 64 \]

4. **Solve for x**:
- Subtract 2 from both sides:
\[ 2x = 62 \]
- Divide by 2:
\[ x = 31 \text{ cm} \]

5. **Determine the Height**:
- Draw a perpendicular line from vertex A to the base BC, meeting at point D. This line represents the height (h) of the triangle.
- Since the triangle is isosceles, the perpendicular bisects the base. Thus, BD = DC = 1 cm (half of the base).
- Now we can use the Pythagorean theorem in triangle ABD:
\[ AB^2 = AD^2 + BD^2 \]
- Substituting the known values:
\[ 31^2 = h^2 + 1^2 \]
\[ 961 = h^2 + 1 \]
- Rearranging gives:
\[ h^2 = 961 - 1 = 960 \]
- Taking the square root:
\[ h = \sqrt{960} = \sqrt{64 \times 15} = 8\sqrt{15} \text{ cm} \]

6. **Calculate the Area**:
- The area (A) of a triangle is given by the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
- Substituting the values we have:
\[ A = \frac{1}{2} \times 2 \times 8\sqrt{15} \]
\[ A = 8\sqrt{15} \text{ cm}^2 \]

### Final Answers:
- **Altitude (Height)**: \( h = 8\sqrt{15} \text{ cm} \)
- **Area**: \( A = 8\sqrt{15} \text{ cm}^2 \)