The length of a rectangle is 3yd more than twice the width,and the area of the rectangle is 54yd^2 find dimensions of the rectangle

Question

The length of a rectangle is 3yd more than twice the width,and the area of the rectangle is 54yd^2 find dimensions of the rectangle

Answer 1

The length of a rectangle is 3yd more than twice the width:

L = 2W + 3

 

the area of the rectangle is 54yd^2 :

LW = 54

 

Replace L with 2W+3 in 2nd equation

 

(2W+3)(W) = 54

 

2W² + 3W = 54

 

2W² + 3W - 54 = 0

 

This can be factored by grouping:

2(-54) = -108

Factors of -108 that add to 3 are (12)(-9)

Replace 3W with 12W - 9W

 

2W² + 12W - 9W- 54 = 0

2W(W + 6) -9(W+6) = 0

(2W-9)(W+6) = 0

 

2W-9 = 0    OR  W+6=0

2W=9               W = -6

W = 9/2

 

Since the width will not be negative,

your width is 9/2 yards or 4.5 yards

 

L = 2W+3 = 12 yards

 

Check:  LW = 54

            4.5(12) = 54

            54 = 54

 

Answer checks as correct

The rectangle is 12 yards by 4.5 yards