The question in the image is:

"The length of the rectangle is greater than its breadth by 2 cm. The area of the rectangle is 24 sq. cm. Find its length and breadth."

To solve this, let’s break down the problem:

1. Let the breadth of the rectangle be $x$ cm.

2. Then, the length would be $x + 2$ cm (since the length is 2 cm greater than the breadth).

3. The area of a rectangle is given by $\text{length} \times \text{breadth}$.

4. According to the question, the area is 24 sq. cm. So,

   $x \times (x + 2) = 24$

Expanding this equation:

$x^2 + 2x = 24$

Rearrange to form a standard quadratic equation:

$x^2 + 2x - 24 = 0$

Now, we can solve for $x$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For this equation:

• $a = 1$
• $b = 2$
• $c = -24$

Plugging these values in:

$x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1}$

$x = \frac{-2 \pm \sqrt{4 + 96}}{2}$

$x = \frac{-2 \pm \sqrt{100}}{2}$

$x = \frac{-2 \pm 10}{2}$

This gives two solutions:

1. $x = \frac{-2 + 10}{2} = \frac{8}{2} = 4$
2. $x = \frac{-2 - 10}{2} = \frac{-12}{2} = -6$

Since the breadth $x$ cannot be negative, we have $x = 4$ cm.

Thus:

• The breadth is $4$ cm.
• The length is $4 + 2 = 6$ cm.

Answer

The length of the rectangle is 6 cm, and the breadth is 4 cm.

Would you like further details on this solution, or have any questions?

Related Questions:

1. How would the solution change if the area of the rectangle was 48 sq. cm instead?
2. What if the length was 3 cm greater than the breadth?
3. Can we solve this using a different method, such as factoring?
4. What are the properties of quadratic equations used here?
5. How would this problem change if we dealt with a different shape, like a triangle?

Tip

When solving quadratic equations, always check if you can factor them easily before using the quadratic formula. It can save time and simplify the process.