Problema Solution
The area of a rectangle is 54yd^2, and the length of the rectangle is 3yds more than twice the width. Find the dimensions of the rectangle.
Answer provided by our tutors
let
l = the length of the rectangle, l>0
w = the width of the rectangle, w>0
the area of a rectangle is 54 yd^2
l*w = 54
the length of the rectangle is 3 yds more than twice the width
l = 3 + 2w
plug l = 3 + 2w into l*w = 54
(3 + 2w)*w = 54
by solving we find
w = 4.5 yd
click here to see the step by step solution of the quadratic equation
l = 3 + 2*4.5
l = 12 yd
the dimensions of the rectangle are: length is 12 yd and width is 4.5 yd.