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?

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Answer 1 · 2018-05-03T12:05:45

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

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 $A=l*w= x(2 x+1) = 66$ sq.yd .

$:. 2 x^2+x=66 or 2 x^2+x-66=0$ or

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

$2 x(x+6) -11( x +6)=0 or (x+6)(2 x-11)=0 :.$ 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]