The length and breadth of a rectangle are in the ratio $3 : 2$ respectively. When the length of the rectangle is increased by $25\%$ while the breadth remains unchanged, the area of the rectangle increases by $24 ext{ m}^2$. What is the original length (in meters) of the rectangle?

Question

The length and breadth of a rectangle are in the ratio $3 : 2$ respectively. When the length of the rectangle is increased by $25\%$ while the breadth remains unchanged, the area of the rectangle increases by $24 	ext{ m}^2$. What is the original length (in meters) of the rectangle?

Accepted Answer

**Solution:**

1. Let the original length be $3x$ and the breadth be $2x$ since the ratio of length to breadth is $3 : 2$.

2. The original area of the rectangle is:
   $$	ext{Area}_{	ext{original}} = 3x 	imes 2x = 6x^2$$

3. The length is increased by $25\%$, so the new length becomes:
   $$3x + 25\% 	ext{ of } 3x = 3x + \frac{25}{100} 	imes 3x = 3x + \frac{3x}{4} = \frac{12x + 3x}{4} = \frac{15x}{4}$$

4. The breadth remains the same, $2x$.

5. The new area is:
   $$	ext{Area}_{	ext{new}} = \frac{15x}{4} 	imes 2x = \frac{30x^2}{4} = \frac{15x^2}{2}$$

6. The increase in area is given as $24 	ext{ m}^2$, so:
   $$	ext{Area}_{	ext{new}} - 	ext{Area}_{	ext{original}} = 24$$
   $$\frac{15x^2}{2} - 6x^2 = 24$$

7. Simplify the equation:
   $$\frac{15x^2}{2} - \frac{12x^2}{2} = 24$$
   $$\frac{3x^2}{2} = 24$$

8. Multiply both sides by 2:
   $$3x^2 = 48$$

9. Divide both sides by 3:
   $$x^2 = 16$$

10. Taking the positive root (since length is positive):
    $$x = 4$$

11. The original length is:
    $$3x = 3 	imes 4 = 12 	ext{ meters}$$

**Answer:** $12$

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