SOLUTION: The perimeter of a rectangle is 54 ft , and the area of the rectangle is 72 ft^2 . Find the dimensions of the rectangle. I need the length and width

Algebra ->  Rectangles -> SOLUTION: The perimeter of a rectangle is 54 ft , and the area of the rectangle is 72 ft^2 . Find the dimensions of the rectangle. I need the length and width      Log On


   

Question 677301: The perimeter of a rectangle is 54 ft , and the area of the rectangle is 72 ft^2 . Find the dimensions of the rectangle. I need the length and width
Found 3 solutions by partha_ban, aaronwiz, lynnlo:
Answer by partha_ban(41) About Me  (Show Source):
You can put this solution on YOUR website!
2*(l+w) = 54
or, l+w = 54/2 = 27
or, l = 27 - w
Now, lw=72
By substituting l, (27-w)*w = 72
27w+-+w%5E2+=+72
0+=+72+-+27w+%2B+w%5E2
w%5E2+-+27w+%2B+72+=+0
By middle term factor,
w%5E2+-+24w+-+3w+%2B+72+=+0
w%2A%28w-24%29+-+3%28w-24%29+=+0
%28w-24%29%2A%28w-3%29=0
Either, w-24=0 => w=24
OR, w-3=0 => w=3
If w=24, l = 27-24 = 3
If w=3, l = 27-3 = 24
Therefore length = 24 ft and width = 3 ft

Answer by aaronwiz(69) About Me  (Show Source):
You can put this solution on YOUR website!
set up a system of equations
x=width
y=length

x+y=54
xy=72

x+y=54
x=72/y
substitute...
72/y+y=54
multiply both sides by y to remove denominator
72+y^2=54y
y^2-54y+72=0
now use any method of your choice to solve for y (length)... factor, quadratic equation, complete the square. Plug your answer back into the original equation to solve for x

Answer by lynnlo(4176) About Me  (Show Source):
You can put this solution on YOUR website!
length is=24ft.
width is=3ft,