SOLUTION: the perimeter of a rectangle is 42 ft and the area of the rectangle is 54 ft^2. find the dimensions of the rectangle ?

Question 742527:

the perimeter of a rectangle is 42 ft and the area of the rectangle is 54 ft^2. find the dimensions of the rectangle ?

Answer by checkley79(3341) (Show Source): You can put this solution on YOUR website!
P=2L+2W
42=2L+2W
AREA=LW
54=LW
L=54/W
42=2(54/W)+2W
42=108/W+2W
42=(108+2W*W)/W
42=(108+2W^2)/W CROSS MULTIPLY.
42*W=108+2W^2
42W=108+2W^2
2W^2-42W+108=0
2(W^2-21W+54)=0
2(W-18)(W-3)=0
W-18=0
W=18 ANS.
W-3=0
W=3 ANS.
42=2L+2*18
42=2L+36
2L=42-36
2L=6
L=6/2
L=3 WHEN W=18
42=2L+2*3
42=2L+6
2L=42-6
2L=36
L=36/2
L=18 WHEN W=3
PROOF:
54=18*3
54=54
54=3*18
54=54

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