The length of a rectangle exceeds its width by 4 inches. How do you find the dimensions of the rectangle it its area is 96 square inches?

Oct 27, 2016

The dimesions of the rectangle are: Length $= 12$ inches; Width $= 8$ inches.

Explanation:

Let the width of the rectangle be $x$ inches.

Then, the length of the rectangle be $x + 4$ inches.

Hence area of the rectangle is as follows. $x \left(x + 4\right) = 96 \mathmr{and} {x}^{2} + 4 x - 96 = 0 \mathmr{and} {x}^{2} + 12 x - 8 x - 96 = 0 \mathmr{and} x \left(x + 12\right) - 8 \left(x + 12\right) = 0 \mathmr{and} \left(x - 8\right) \left(x + 12\right) = 0$

So either (x-8)=0 ;:.x=8 or (x+12)=0 ;:.x=-12. Width can not be negative. So x=8 ; x+4=12

Hence the dimesions of rectangle are as Length $= 12$ inches , Width $= 8$ inches.[Ans]