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#