The area of an isosceles triangle each of whose equal sides is `13 cm` and whose base is `24 cm` is:

Options

A. `45cm^(2)`
B. `48cm^(2)`
C. `60cm^(2)`
D. `75cm^(2)`

Answer

To find the area of the isosceles triangle with equal sides of 13 cm and a base of 24 cm, we can follow these steps:

### Step 1: Identify the components of the triangle
- Let the triangle be ABC, where AB = AC = 13 cm (the equal sides) and BC = 24 cm (the base).

### Step 2: Find the height of the triangle
- To find the height (h) of the triangle, we can drop a perpendicular from point A to the base BC at point K. This creates two right triangles, AKB and AKC.
- Since K is the midpoint of BC, we have BK = KC = 24 cm / 2 = 12 cm.

### Step 3: Apply the Pythagorean theorem
- In triangle AKB, we can apply the Pythagorean theorem:
\[ AB^2 = AK^2 + BK^2 \]
Substituting the known values:
\[ 13^2 = h^2 + 12^2 \]
This simplifies to:
\[ 169 = h^2 + 144 \]

### Step 4: Solve for the height
- Rearranging the equation gives:
\[ h^2 = 169 - 144 \]
\[ h^2 = 25 \]
Taking the square root:
\[ h = \sqrt{25} = 5 \text{ cm} \]

### Step 5: Calculate the area of the triangle
- The area (A) of the triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting the values we have:
\[ A = \frac{1}{2} \times 24 \times 5 \]
\[ A = 12 \times 5 = 60 \text{ cm}^2 \]

### Final Answer
The area of the isosceles triangle is **60 cm²**.