Find the length of a rectangle whose diagonal and area is 25 cm and 168 cm² respectively :

Question

Accepted Answer

To find the length of the rectangle, we need to use the given information: the diagonal is 25 cm, and the area is 168 cm². Let's denote the length of the rectangle as $ l $ and the width as $ w $.

1. **Using the Pythagorean Theorem:**
   The diagonal of a rectangle forms a right triangle with the length and width as the two perpendicular sides. According to the Pythagorean theorem:
   $$
   l^2 + w^2 = 25^2
   $$
   $$
   l^2 + w^2 = 625
   $$

2. **Using the Area:**
   The area of the rectangle is given by:
   $$
   l \times w = 168
   $$

3. **Solving the Equations:**
   We have two equations:
   $$
   l^2 + w^2 = 625
   $$
   $$
   l \times w = 168
   $$

To solve these, we can express $ w $ in terms of $ l $ from the area equation:
   $$
   w = \frac{168}{l}
   $$

Substitute $ w $ in the Pythagorean equation:
   $$
   l^2 + \left(\frac{168}{l}\right)^2 = 625
   $$

Simplifying:
   $$
   l^2 + \frac{28224}{l^2} = 625
   $$

Multiply through by $ l^2 $ to clear the fraction:
   $$
   l^4 - 625l^2 + 28224 = 0
   $$

Let $ x = l^2 $, then:
   $$
   x^2 - 625x + 28224 = 0
   $$

Solving this quadratic equation using the quadratic formula:
   $$
   x = \frac{625 \pm \sqrt{625^2 - 4 \times 28224}}{2}
   $$

Calculate the discriminant:
   $$
   625^2 - 4 \times 28224 = 390625 - 112896 = 277729
   $$

$$
   \sqrt{277729} = 527
   $$

So, the solutions for $ x $ are:
   $$
   x = \frac{625 \pm 527}{2}
   $$

$$
   x = \frac{1152}{2} = 576 \quad \text{or} \quad x = \frac{98}{2} = 49
   $$

Therefore, $ l^2 = 576 $ or $ l^2 = 49 $.

If $ l^2 = 576 $, then $ l = 24 $ cm.
   If $ l^2 = 49 $, then $ l = 7 $ cm.

Since the area $ l \times w = 168 $, and $ l = 24 $ satisfies this condition with $ w = \frac{168}{24} = 7 $, the correct length is 24 cm.

4. **Conclusion:**
   - **Option 2 (24 cm)** is correct because it satisfies both the area and the diagonal conditions.
   - **Option 1 (31 cm)**, **Option 3 (17 cm)**, and **Option 4 (9 cm)** do not satisfy the given conditions when calculated with the corresponding width to maintain the area of 168 cm² and the diagonal of 25 cm.

Thus, the correct answer is Option 2: 24 cm.