Expert: Sherman D. - 12/17/2004

Question
The base of a rectangle is twice the side of a square, and the height of the rectangle is two more than the side of the square. The area of the rectangle is 32 square inch units more than the area of the square. Find the dimensions of the rectangle

Answer
Ar = Area of Rectangle
As = Area of Square

Br = 2(Bs)
Hr = Bs + 2

Ar = BH
As = B^2

Ar = (2(Bs))*(Bs + 2)
Ar = As + 32

Ar = 2(Bs)^2 + 4(Bs)
Ar = (Bs)^2 + 32

(Bs)^2 + 32 = 2(Bs)^2 + 4(Bs)
(Bs)^2 + 4(Bs) - 32 = 0

or you can think of it like this

x^2 + 4x - 32 = 0
(x + 8)(x - 4) = 0
x = -8 or 4
since we can't use negative lengths
x = 4
or in this case
Bs = 4

Br = 2(Bs)
Hr = Bs + 2

Br = 2(4) = 8
Hr = 4 + 2 = 6

Check:
Ar = 8 * 10 = 48
As = 4^2 = 16, which would make the Ar 32 more than the As.