If the area of rectangle is 24 sq. cm and the length and breadth are integers in cm, then find the maximum possible perimeter of the rectangle.

Explanation

To solve the problem, we need to find the maximum possible perimeter of a rectangle given that its area is 24 square centimeters and both the length (L) and breadth (B) are integers.

### Step-by-Step Solution:

1. **Understand the Area Formula**:
The area of a rectangle is given by the formula:
\[ \text{Area} = \text{Length} \times \text{Breadth} \quad (A = L \times B) \]
Given that the area is 24 sq. cm, we have:
\[ L \times B = 24 \]

2. **Find Integer Pairs**:
We need to find pairs of integers (L, B) such that their product is 24. The pairs of integers that satisfy this equation are:
- (1, 24)
- (2, 12)
- (3, 8)
- (4, 6)
- (6, 4)
- (8, 3)
- (12, 2)
- (24, 1)

3. **Calculate the Perimeter**:
The perimeter (P) of a rectangle is given by the formula:
\[ P = 2 \times (L + B) \]
We will calculate the perimeter for each pair:
- For (1, 24):
\[ P = 2 \times (1 + 24) = 2 \times 25 = 50 \text{ cm} \]
- For (2, 12):
\[ P = 2 \times (2 + 12) = 2 \times 14 = 28 \text{ cm} \]
- For (3, 8):
\[ P = 2 \times (3 + 8) = 2 \times 11 = 22 \text{ cm} \]
- For (4, 6):
\[ P = 2 \times (4 + 6) = 2 \times 10 = 20 \text{ cm} \]
- For (6, 4):
\[ P = 2 \times (6 + 4) = 2 \times 10 = 20 \text{ cm} \]
- For (8, 3):
\[ P = 2 \times (8 + 3) = 2 \times 11 = 22 \text{ cm} \]
- For (12, 2):
\[ P = 2 \times (12 + 2) = 2 \times 14 = 28 \text{ cm} \]
- For (24, 1):
\[ P = 2 \times (24 + 1) = 2 \times 25 = 50 \text{ cm} \]

4. **Identify the Maximum Perimeter**:
Now, we compare all the calculated perimeters:
- 50 cm
- 28 cm
- 22 cm
- 20 cm
- 20 cm
- 22 cm
- 28 cm
- 50 cm

The maximum perimeter is **50 cm**.

### Final Answer:
The maximum possible perimeter of the rectangle is **50 cm**.

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