How do you find the dimensions of a rectangle if the length is 1 more than twice the the width and the area is 55 m?

Dec 4, 2016

The wide is 5m and the length is 11m.

Explanation:

First, let's have the width represented by $w$ and the length represented by $l$

We know the area of a rectangle is:

$A = l \cdot w$

We are also told the length is 1 more than twice the width so we can write"

$l = 2 w + 1$

Substituting this into the formula for area along with the given area of 55 gives:

55 = w*(2w + 1)#

Solving this gives:

$55 = 2 {w}^{2} + w$

$55 - 55 = 2 {w}^{2} + w - 55$

$0 = 2 {w}^{2} + w - 55$

$0 = \left(2 w + 11\right) \left(w - 5\right)$

We can now solve each term for 0:

$w - 5 = 0$

$w - 5 + 5 = 0 + 5$

$w = 5$

and

$2 w + 11 = 0$

$2 w + 11 - 11 = 0 - 11$

$2 w = - 11$

$\frac{2 w}{2} = - \frac{11}{2}$

$w = - \frac{11}{2}$

Because the width of a rectangle must be positive the width of this rectangle is $5$m.

Substituting $5$ for $w$ in the formula for length gives:

$l = 2 \cdot 5 + 1$

$l = 10 + 1$

$l = 11$