A triangle with integral sides has perimeter 8 cm. Then find the area of the triangle

Explanation

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

### Step 1: Define the sides of the triangle
Let the sides of the triangle be \( A \), \( B \), and \( C \). According to the problem, the perimeter is given by:
\[ A + B + C = 8 \]

### Step 2: Determine possible integral side lengths
Since \( A \), \( B \), and \( C \) are integral sides, we need to find combinations of \( A \), \( B \), and \( C \) that satisfy the triangle inequality, which states that the sum of the lengths of any two sides must be greater than the length of the third side.

### Step 3: List possible combinations
We can start by listing down possible combinations of \( A \), \( B \), and \( C \) that add up to 8:
1. \( (1, 1, 6) \)
2. \( (1, 2, 5) \)
3. \( (1, 3, 4) \)
4. \( (2, 2, 4) \)
5. \( (2, 3, 3) \)

### Step 4: Check which combinations satisfy the triangle inequality
Now we will check which of these combinations satisfy the triangle inequality:
- For \( (1, 1, 6) \): \( 1 + 1 \not> 6 \) (not valid)
- For \( (1, 2, 5) \): \( 1 + 2 \not> 5 \) (not valid)
- For \( (1, 3, 4) \): \( 1 + 3 \not> 4 \) (not valid)
- For \( (2, 2, 4) \): \( 2 + 2 \not> 4 \) (not valid)
- For \( (2, 3, 3) \): \( 2 + 3 > 3 \) and \( 3 + 3 > 2 \) and \( 2 + 3 > 3 \) (valid)

Thus, the only valid combination of sides is \( (2, 3, 3) \).

### Step 5: Calculate the semi-perimeter
Now we can calculate the semi-perimeter \( s \):
\[ s = \frac{A + B + C}{2} = \frac{8}{2} = 4 \]

### Step 6: Use Heron's formula to find the area
Heron's formula for the area \( A \) of a triangle is given by:
\[ A = \sqrt{s(s - A)(s - B)(s - C)} \]
Substituting the values:
\[ A = \sqrt{4(4 - 2)(4 - 3)(4 - 3)} = \sqrt{4 \times 2 \times 1 \times 1} = \sqrt{8} = 2\sqrt{2} \]

### Step 7: State the final area
Thus, the area of the triangle is:
\[ \text{Area} = 2\sqrt{2} \text{ cm}^2 \]

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