SOLUTION: The width of a rectangle is 2 less than twice its length. If the area of the rectangle is 53 cm squared, what is the length of the diagonal?
Algebra ->
Rectangles
-> SOLUTION: The width of a rectangle is 2 less than twice its length. If the area of the rectangle is 53 cm squared, what is the length of the diagonal?
Log On
Question 499367: The width of a rectangle is 2 less than twice its length. If the area of the rectangle is 53 cm squared, what is the length of the diagonal? Answer by chessace(471) (Show Source):
You can put this solution on YOUR website! W = 2*L-2 (cm I assume)
A = L*W = 53 cm^2
A = 2L^2 -2L
2L^2 -2L -53 = 0
L = (2 +- sqrt(4+2*53))/2 = 1 + sqrt(110)/2 = 6.244
W = 10.488
Diag^2 = L^2 + W^2 = 39 + 110 = 149
D = 12.207 cm
Looks like a typo in the problem.
