The area of a rectangle is 52 yd2 , and the length of the rectangle is 5 yd less than twice the width. Find the dimensions of the rectangle.

Question

The area of a rectangle is
52 yd2
, and the length of the rectangle is
5 yd
less than twice the width. Find the dimensions of the rectangle.

Answer 1

Let L = length

Let W = width

 

Length is 5 yd less than twice the width  ==>  L = 2*W - 5

 

Area is 52 yd^2  ==>  52 = L*W

 

Substitute the value for L from the 1st equation into the 2nd equation 

 

52 = (2*W - 5) * W

52 = 2*W^2 - 5*W

0 = 2*W^2 - 5*W - 52 

0 = (2W - 13)*(W + 4)

 

2W - 13 = 0  or  W + 4 = 0

W = 13/2      or  W = -4

 

Width cannot be negative so discard W = -4 

 

W = 13/2  ==>  L = 2*(13/2) - 5 = 13 - 5 = 8 

 

So the width is 13/2 yards and the length is 8 yards 

 

Check:  2*(13/2) - 5 = 13 - 5 = 8 

            (13/2)*8 = 4*13 = 52 yd^2