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?

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Answer 1 · 2016-12-04T16:13:31

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

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

We know the area of a rectangle is:

$A = l*w$

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

$l = 2w + 1$

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

55 = w*(2w + 1)#

Solving this gives:

$55 = 2w^2 + w$

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

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

$0 = (2w + 11)(w - 5)$

We can now solve each term for 0:

$w - 5 = 0$

$w - 5 + 5 = 0 + 5$

$w = 5$

and

$2w + 11 = 0$

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

$2w = -11$

$(2w)/2 = -11/2$

$w = -11/2$

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:

$l = 2*5 + 1$

$l = 10 + 1$

$l = 11$