The area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm is `k sqrt30 cm^(2)`. Find the value of k.

Options

A. 8
B. 6
C. 7
D. 9

Answer

To find the value of \( k \) in the area of the triangle given by \( k \sqrt{30} \, \text{cm}^2 \), we will follow these steps:

### Step 1: Identify the sides of the triangle
We know two sides of the triangle are \( a = 8 \, \text{cm} \) and \( b = 11 \, \text{cm} \). The perimeter of the triangle is given as \( 32 \, \text{cm} \).

### Step 2: Calculate the third side
Let the third side be \( c \). The perimeter of the triangle is the sum of all its sides:
\[ a + b + c = 32 \]
Substituting the known values:
\[ 8 + 11 + c = 32 \]
\[ c = 32 - 19 = 13 \, \text{cm} \]
So, the sides of the triangle are \( 8 \, \text{cm}, 11 \, \text{cm}, \) and \( 13 \, \text{cm} \).

### Step 3: Calculate the semi-perimeter
The semi-perimeter \( s \) of the triangle is given by:
\[ s = \frac{a + b + c}{2} = \frac{32}{2} = 16 \, \text{cm} \]

### Step 4: Apply Heron's Formula
Heron's formula for the area \( A \) of a triangle is:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values:
\[ A = \sqrt{16(16-8)(16-11)(16-13)} \]
Calculating each term:
\[ A = \sqrt{16 \times 8 \times 5 \times 3} \]

### Step 5: Simplify the expression
Calculating the product inside the square root:
\[ A = \sqrt{16 \times 8 \times 5 \times 3} = \sqrt{16} \times \sqrt{8} \times \sqrt{5} \times \sqrt{3} \]
\[ = 4 \times \sqrt{8} \times \sqrt{15} \]
Now simplify \( \sqrt{8} \):
\[ \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \]
Thus,
\[ A = 4 \times 2\sqrt{2} \times \sqrt{15} = 8\sqrt{30} \]

### Step 6: Compare with the given area
We are given that the area can be expressed as \( k \sqrt{30} \). From our calculation:
\[ A = 8\sqrt{30} \]
Thus, we can equate:
\[ k \sqrt{30} = 8 \sqrt{30} \]
This implies:
\[ k = 8 \]

### Final Answer
The value of \( k \) is \( \boxed{8} \).