The base of an isosceles triangle is 12 cm and its perimeter is 32 cm. Find its area.

Answer

To find the area of the isosceles triangle with a base of 12 cm and a perimeter of 32 cm, we can follow these steps:

### Step 1: Identify the sides of the triangle
Let the lengths of the two equal sides be \( x \) cm. The base is given as 12 cm. Therefore, the sides of the triangle can be represented as:
- Side 1: \( x \) cm
- Side 2: \( x \) cm
- Base: 12 cm

### Step 2: Set up the equation for the perimeter
The perimeter of the triangle is the sum of all its sides:
\[ x + x + 12 = 32 \]
This simplifies to:
\[ 2x + 12 = 32 \]

### Step 3: Solve for \( x \)
Subtract 12 from both sides:
\[ 2x = 32 - 12 \]
\[ 2x = 20 \]
Now, divide by 2:
\[ x = 10 \text{ cm} \]
So, the lengths of the two equal sides are 10 cm each.

### Step 4: Calculate the semi-perimeter (s)
The semi-perimeter \( s \) is half of the perimeter:
\[ s = \frac{32}{2} = 16 \text{ cm} \]

### Step 5: Use Heron's formula to find the area
Heron's formula for the area \( A \) of a triangle is given by:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Where \( a, b, c \) are the lengths of the sides of the triangle. In our case:
- \( a = 10 \) cm
- \( b = 10 \) cm
- \( c = 12 \) cm

Substituting the values into Heron's formula:
\[ A = \sqrt{16(16-10)(16-10)(16-12)} \]
\[ A = \sqrt{16 \times 6 \times 6 \times 4} \]

### Step 6: Simplify the expression
Calculating the product inside the square root:
\[ A = \sqrt{16 \times 6 \times 6 \times 4} = \sqrt{16 \times 144} = \sqrt{2304} \]
Calculating the square root:
\[ A = 48 \text{ cm}^2 \]

### Final Answer
The area of the isosceles triangle is \( 48 \text{ cm}^2 \).

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