The perimeter of a triangle is 16 cm. One ofthe sides is of length 6 cm. If the area of thetriangle is 12 sq. cm, then the triangle is

Options

A. right angled
B. isoscles
C. equilateral
D. scalene

Answer

To solve the problem step by step, we will follow the information provided and apply relevant formulas.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Perimeter of the triangle (P) = 16 cm
- One side (a) = 6 cm
- Area of the triangle (Δ) = 12 cm²

2. **Calculate the Sum of the Other Two Sides:**
The perimeter of a triangle is the sum of all its sides:
\[ P = a + b + c \]
Given \( P = 16 \) cm and \( a = 6 \) cm, we can write:
\[ 16 = 6 + b + c \]
Rearranging gives:
\[ b + c = 10 \quad \text{(Equation 1)} \]

3. **Calculate the Semi-Perimeter (s):**
The semi-perimeter \( s \) is half of the perimeter:
\[ s = \frac{P}{2} = \frac{16}{2} = 8 \text{ cm} \]

4. **Use Heron's Formula for Area:**
Heron's formula states that the area of a triangle can be calculated as:
\[ Δ = \sqrt{s(s-a)(s-b)(s-c)} \]
Plugging in the values we know:
\[ 12 = \sqrt{8(8-6)(8-b)(8-c)} \]
Simplifying gives:
\[ 12 = \sqrt{8 \cdot 2 \cdot (8-b) \cdot (8-c)} \]
Squaring both sides:
\[ 144 = 16(8-b)(8-c) \]
Dividing by 16:
\[ 9 = (8-b)(8-c) \quad \text{(Equation 2)} \]

5. **Substituting from Equation 1 into Equation 2:**
From Equation 1, we know \( c = 10 - b \). Substitute this into Equation 2:
\[ 9 = (8-b)(8-(10-b)) \]
Simplifying gives:
\[ 9 = (8-b)(b-2) \]
Expanding:
\[ 9 = 8b - b^2 - 16 + 2b \]
Rearranging leads to:
\[ b^2 - 10b + 25 = 0 \]

6. **Solving the Quadratic Equation:**
The equation can be factored as:
\[ (b-5)^2 = 0 \]
Thus, we find:
\[ b = 5 \]

7. **Finding the Length of Side c:**
Using Equation 1 again:
\[ c = 10 - b = 10 - 5 = 5 \]

8. **Final Side Lengths:**
Now we have:
- \( a = 6 \) cm
- \( b = 5 \) cm
- \( c = 5 \) cm

9. **Determine the Type of Triangle:**
Since two sides are equal (b = c), the triangle is **isosceles**.

### Conclusion:
The triangle is isosceles.