Question

The length of a rectangle is 5yd less than twice the width, and the area of the rectangle is 33yd^. Find the dimensions of the rectangle.

Answer 1

Let l be the length of the rectangle and w its width.

From this, we have:

I) w - l = 5

II) l*w = 33

From I, we have w = 5 + l

Applying this to equation II, we have: l(5+l) = 33

l^2 + 5l - 33 = 0

The positive root of this equation is l = [sqrt(157) - 5]/2 = 3.8 yd (rounded to the nearest tenth)

Applying this to equation I, we have: w - 3.8 = 5, which implies w = 5 + 3.8 = 8.8 yd