The area of a rectangle is 65 yd^2, and the length of the rectangle is 3 yd less than twice the width. How do you find the dimensions of the rectangle?

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Answer 1 · 2018-07-10T08:36:10

$\text{Length}=10$ , $\text{width}=13/2$

Let $L$ & $B$ be the length and width of rectangle then

as per given condition

$L=2B-3\ ..........(1)$

And the area of rectangle

$LB=65$

setting value of $L=2B-3$ from (1) in above equation, we get

$(2B-3)B=65$

$2B^2-3B-65=0$

$2B^2-13B+10B-65=0$

$B(2B-13)+5(2B-13)=0$

$(2B-13)(B+5)=0$

$2B-13=0 \ \ or\ \ B+5=0$

$B=13/2\ \ or \ \ B=-5$

But the width of rectangle can't be negative hence

$B=13/2$

setting $B=13/2$ in (1), we get

$L=2B-3$

$=2(13/2)-3$

$=10$