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Answer 1 · 2017-11-08T04:16:31

The dimensions are 5 in. and 13 in.

Let length of rectangle be #(x-2)# in

and width of rectangle be # (x+6)# in
Area of rectangle = length x width

Given : Area # = 65 # in#^2#

#therefore 65= (x-2)(x+6)#

#=> x^2 +6x -2x - 12 = 65 #

#=> x^2 +4x -65-12= 0 #

# => x^2 +4x -77= 0 #

Solve :

#=> x^2 -7x + 11x -77= 0 #

# => x(x-7) + 11(x-7) =0 #

#=>(x-7)(x+11)=0#

#=> x-7 = 0 => x=7#

or

#x+11=0 => x= -11#

As the dimensions cannot be negative , we take #x= 7#

So the dimensions will be :

#x-2= 5#in and #x+6= 13#in

The dimensions are 5 in. and 13 in.