Let's solve the problem step by step.

Problem Details:

• The length of the rectangle: $x + 10$ meters.
• The breadth of the rectangle: $x + 6$ meters.
• Area of the rectangle: $396 \, \text{m}^2$.
• After increasing the length by 2 meters, we are to find the perimeter of the rectangle.

Step 1: Use the area formula to find $x$.

The area of a rectangle is given by: $\text{Area} = \text{Length} \times \text{Breadth}$ Substitute the given values: $396 = (x + 10)(x + 6)$

Expand: $396 = x^2 + 6x + 10x + 60$ $396 = x^2 + 16x + 60$

Rearrange into standard quadratic form: $x^2 + 16x + 60 - 396 = 0$ $x^2 + 16x - 336 = 0$

Step 2: Solve for $x$ using the quadratic formula.

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 16$, and $c = -336$. Substituting: $x = \frac{-16 \pm \sqrt{16^2 - 4(1)(-336)}}{2(1)}$ $x = \frac{-16 \pm \sqrt{256 + 1344}}{2}$ $x = \frac{-16 \pm \sqrt{1600}}{2}$ $x = \frac{-16 \pm 40}{2}$

Calculate the two possible solutions: $x = \frac{-16 + 40}{2} = \frac{24}{2} = 12$ $x = \frac{-16 - 40}{2} = \frac{-56}{2} = -28$

Since $x$ represents a length, it must be positive: $x = 12$

Step 3: Calculate the new length and breadth.

• Original length: $x + 10 = 12 + 10 = 22 \, \text{m}$.
• Original breadth: $x + 6 = 12 + 6 = 18 \, \text{m}$.

After increasing the length by 2 meters:

• New length: $22 + 2 = 24 \, \text{m}$.
• New breadth remains $18 \, \text{m}$.

Step 4: Find the new perimeter.

The perimeter of a rectangle is: $\text{Perimeter} = 2(\text{Length} + \text{Breadth})$ Substitute the values: $\text{Perimeter} = 2(24 + 18) = 2(42) = 84 \, \text{m}$

Final Answer:

The perimeter of the rectangle is 84 meters.

Would you like further explanation or have any related questions? Here are five possible related queries to explore:

1. How do we verify the solution for $x$ using the original area?
2. What would the perimeter be if the breadth were also increased by 2 meters?
3. How does the quadratic formula work, and why do we discard negative solutions in this case?
4. Could this problem be solved by completing the square instead of using the quadratic formula?
5. What happens to the area if both the length and breadth are increased by 2 meters?

Tip: Always verify your results by substituting back into the original equations!