Let the width of the rectangle be denoted by $w$ (in meters), and the length is given as $l = w - 6$ meters.

The area of a rectangle is calculated by the formula:

$\text{Area} = \text{Length} \times \text{Width}$

Given that the area is 112 square meters, we can set up the equation:

$(w - 6) \times w = 112$

Expanding the equation:

$w^2 - 6w = 112$

Rearranging it into a standard quadratic form:

$w^2 - 6w - 112 = 0$

Now, solving this quadratic equation using the quadratic formula:

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

Here, $a = 1$, $b = -6$, and $c = -112$. Substituting these values:

$w = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-112)}}{2(1)}$

$w = \frac{6 \pm \sqrt{36 + 448}}{2}$

$w = \frac{6 \pm \sqrt{484}}{2}$

$w = \frac{6 \pm 22}{2}$

So, the two possible solutions for $w$ are:

$w = \frac{6 + 22}{2} = \frac{28}{2} = 14$ $w = \frac{6 - 22}{2} = \frac{-16}{2} = -8 \quad (\text{which is not a valid solution since width cannot be negative})$

Thus, the width is $w = 14$ meters, and the length is:

$l = w - 6 = 14 - 6 = 8 \text{ meters}$

Now, the perimeter $P$ of the rectangle is given by the formula:

$P = 2 \times (\text{Length} + \text{Width})$

Substituting the values:

$P = 2 \times (14 + 8) = 2 \times 22 = 44 \text{ meters}$

Therefore, the perimeter of the rectangle is 44 meters.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the problem change if the length was 6 cm more than the width?
2. How do you solve quadratic equations graphically?
3. What are other methods to solve quadratic equations besides the quadratic formula?
4. How do you derive the formula for the area of different shapes?
5. How does the perimeter change if both the length and width are doubled?

Tip: When solving word problems, always define variables clearly and check the units carefully to avoid confusion.