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#