SOLUTION: a square and rectangle have the same area. the length of the rectangle is 8 less than twice the side of the square. the width of the rectangle is 3 less than the side of the square

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Question 280482: a square and rectangle have the same area. the length of the rectangle is 8 less than twice the side of the square. the width of the rectangle is 3 less than the side of the square. let x represent the length of the side of the square. a)write expressions for the dimensions of the rectangle. b) write an equation that represents the situation. c) find the dimension of the square and rectangle.
Found 2 solutions by richwmiller, mananth:
Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website!
l=2s-8
w=s-3
(2s-8)*(s-3)=s^2
2s^2-8s-6s+24=s^2
s^2-14s+24=0
(s-12)*(s-2)=0
s=12
s=2
l=2s-8
l=24-8=16
w=9
16*9=144
12^2=144
s=2 can't be since s-3=-1 we can'tr have a negative number for a side

Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
a square and rectangle have the same area. the length of the rectangle is 8 less than twice the side of the square. the width of the rectangle is 3 less than the side of the square. let x represent the length of the side of the square. a)write expressions for the dimensions of the rectangle. b) write an equation that represents the situation. c) find the dimension of the square and rectangle.
Let the side of the square be x
Length of rectangle = 2x-8
width of rectangle= x-3
(2x-8)(x-3)= x^2 ( since area of recatangle = area of the square
2x^2-6x-8x+24=x^2
x^2-14x+24=0
x^2-12x-2x+24=0
x(x-12)-2(x-12)=0
(x-12)(x-2)=0

x=2 or 12
x=12 satisifies the condition
side of square =12------------------------- Area =144
the sides of the rectangle are 16 and 9---------------- Area = 144