Dimension Problem

Question

The length of a rectangle is
3 ft
less than double the width, and the area of the rectangle is
44 ft2
. Find the dimensions of the rectangle.

Answer 1

Hello, Colin

 

Let's set the problem up;

 

Let w represent the width of the rectangle

Let 2w-3 represent the length f the rectangle.. Since the problem says double the width, 2w, less 3.

 

Now, the area of a rectangle is l*w.. So, what we need to do next is substitute, use the distributive property and solve for w. And then, find y.  Here we go,

 

a=lw

a=(2w-3)w                 Substituting

44=2w^2-3w              Using the distributive property.

0= 2w^2-3w-44= 0    Factoring: Hint: Use the AC method.. if you cant see it, use the quadratic formula!!

0=(2w-11)(w+4)

2w-11=0     0r  w+4=0

2w=11             w=-4

w=5.5       0r w=-4

 

We cannot use w=-4 because dimensions( length, width, height, radius, diameter e.t.c) cannot be negative.  So, w=5.5 and l= 2w-3= 2(5.5)-3= 11-3= 8. 

 

Our final answer, Colin, is w=5.5 and l=8.  Glad to help.. Please don't forget to rate my answer. Take care. :)

 

D. Y. Taylor

Answer 2

 

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