# The length of a rectangle is 1 more than twice its width, and the area of the rectangle is 66 yd^2, how do you find the dimensions of the rectangle?

May 3, 2018

Dimensions of the rectangle are $12$ yards long and $5.5$ yards wide.

#### Explanation:

Let the width of the rectangle is $w = x$ yd , , then the

length of the rectangle is $l = 2 x + 1$ yd , therefore, the area of the

rectangle is $A = l \cdot w = x \left(2 x + 1\right) = 66$ sq.yd .

$\therefore 2 {x}^{2} + x = 66 \mathmr{and} 2 {x}^{2} + x - 66 = 0$ or

$2 {x}^{2} + 12 x - 11 x - 66 = 0$ or

$2 x \left(x + 6\right) - 11 \left(x + 6\right) = 0 \mathmr{and} \left(x + 6\right) \left(2 x - 11\right) = 0 \therefore$ either,

x+6=0:. x =-6 or 2 x-11= 0 :. x= 5.5 ; x  cannot be

negative.  :. x= 5.5 ;2 x+1= 2*5.5+1=12 . Dimensions

of the rectangle are $12$ yards long and $5.5$ yards wide.[Ans]