The width of a rectangle is 3 less than twice its length. If the area of the rectangle is 131 cm^2, what is the length of the diagonal?
I set up the basic equation to solve for l and I know I need to use Pythagorean therm to find the actual value but I keep getting the answer incorrect.
Best Answer
width of rectangle = $w$
length of rectangle = $2w-3$
$w(2w-3) = 131$
$2w^2 - 3w - 131 = 0$
Solve for $w$ using the quadratic formula
$w = -7.3779 $
$w = 8.8779$
We obviously only want the second answer.
$w^2+l^2=d^2$
where d is the diagonal.
$\sqrt{w^2+l^2}=\sqrt{d^2}$
$d =\sqrt{w^2+(2w-3)^2}$
Plug in w to solve for d.