Answer
For this rectangle, the length is $11$ meters and the width is $5$ meters.
Work Step by Step
We know that the area of a rectangle can be given by the following formula:
$$A = lw$$
where $A$ is the area, $l$ is the length, and $w$ is the width.
We have the area for this rectangle as $55$ square meters, and the length is $6$ meters more than the width, so if we have $w$ as the width, then the length $l$ is $w + 6$. We can plug these values into the equation for area:
$$55 = (w + 6)(w)$$
We use the distributive property to simplify the right side of the equation:
$$55 = w(w) + 6(w)$$
We multiply to simplify:
$$55 = w^2 + 6w$$
We can turn this into a quadratic equation and set it to $0$ to solve for $w$:
$$w^2 + 6w - 55 = 0$$
To factor this equation, we need to see what combination of factors for the constant $-55$ when added together will give us $6$, the coefficient of the second term. We will want to find one negative factor and one positive factor where the positive factor is a greater number than the negative factor. Let's look at the factors:
$-55$ and $1$ or $55$ and $-1$
$-11$ and $5$ or $11$ and $-5$
It seems like the factors $11$ and $-5$ will work, so we plug these into the equation:
$$(w + 11)(w - 5) = 0$$
According to the Zero Product Property, either factor can equal zero, so we set each factor equal to zero to solve for $w$:
$$w + 11 = 0$$
Subtract $11$ from each side to solve for $w$:
$$w = -11$$
We set the other factor equal to $0$:
$$w - 5 = 0$$
Add $5$ to each side to solve for $w$:
$$w = 5$$
Since measurements cannot be negative, we discard the $-11$ answer, so we know that the width is $5$ meters.
If the width is $5$ meters, we can plug this value into the equation for area and solve for length:
$$55 = (l)(5)$$
Divide both sides by $5$ to solve for $l$:
$$l = 11$$
For this rectangle, the length is $11$ meters and the width is $5$ meters.
