Find the area of triangle whose two sides are 8 cm and 11 cm and perimeter is 32 cm:

Options

A. A)`42 cm^2`
B. B)`30 cm^2`
C. C)`8sqrt(30) cm^2`
D. D)`5sqrt(30) cm^2`

Answer

To find the area of the triangle with two sides measuring 8 cm and 11 cm and a perimeter of 32 cm, we can follow these steps:

### Step 1: Determine the length of the third side
We know the perimeter of the triangle is the sum of all its sides. Let the third side be \( K \).

\[ \text{Perimeter} = 8 + 11 + K = 32 \]

Now, we can solve for \( K \):

\[ K = 32 - 8 - 11 = 32 - 19 = 13 \text{ cm} \]

### Step 2: Calculate the semi-perimeter
The semi-perimeter \( s \) is half of the perimeter:

\[ s = \frac{\text{Perimeter}}{2} = \frac{32}{2} = 16 \text{ cm} \]

### Step 3: Use Heron's formula to find the area
Heron's formula states that the area \( A \) of a triangle can be calculated using the semi-perimeter and the lengths of the sides:

\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]

Where \( a = 8 \text{ cm} \), \( b = 11 \text{ cm} \), and \( c = 13 \text{ cm} \).

Substituting the values into the formula:

\[ A = \sqrt{16(16-8)(16-11)(16-13)} \]

Calculating each term:

\[ A = \sqrt{16 \times 8 \times 5 \times 3} \]

### Step 4: Simplify the expression
Now, we can simplify the expression:

\[ A = \sqrt{16} \times \sqrt{8 \times 5 \times 3} \]

Calculating \( \sqrt{16} \):

\[ \sqrt{16} = 4 \]

Now we need to calculate \( 8 \times 5 \times 3 \):

\[ 8 \times 5 = 40 \]
\[ 40 \times 3 = 120 \]

Thus, we have:

\[ A = 4 \times \sqrt{120} \]

### Step 5: Further simplify \( \sqrt{120} \)
We can simplify \( \sqrt{120} \):

\[ \sqrt{120} = \sqrt{4 \times 30} = 2\sqrt{30} \]

So now substituting back:

\[ A = 4 \times 2\sqrt{30} = 8\sqrt{30} \text{ cm}^2 \]

### Final Answer
The area of the triangle is:

\[ \boxed{8\sqrt{30} \text{ cm}^2} \]