Question 798439: A 40-inch length of wire is cut in two. One of the pieces is bent to form a square and the other is bent to form a rectangle three times as long as wide. If the combined area of the square and the rectangle is 55 3/4 inches, where was the wire cut? There are two places where the wire can be cut. One of them is a fraction answer.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A 40-inch length of wire is cut in two.
One of the pieces is bent to form a square and the other is bent to form a rectangle three times as long as wide.
If the combined area of the square and the rectangle is 55 3/4 inches, where was the wire cut?
There are two places where the wire can be cut.
One of them is a fraction answer.
:
Let x = length of the one side of the square
then
x^2 = the area of the square
and
(40-4x) = the perimeter of the rectangle
Let W = the width of the rectangle
then
3W = the length of the rectangle
therefore
2(3W) + 2W = 40 - 4x
Simplify, divide by 2
3W + W = 20 - 2x
4W = 20-2x
Simplify, divide by 2
2W = 10 - x
x = -2W + 10
:
"the combined area of the square and the rectangle is 55 3/4 inches"
x^2 + 3W^2 = 55.75
Replace x with (-2W+10)
(-2w+10)^2 + 3W^2 = 55.75
Foil
4w^2 -40w + 100 + 3w^2 = 55.75
7W^2 - 40w + 100 - 55.75 = 0
7w^2 - 40w + 44.25 = 0
Using the quadratic formula, two solutions
w = 1.5, we will use this one
w = 4.2143
find x
x = -2(1.5) + 10
x = 7 in, the side of square
then
4(7) = 28 inches cut the wire
:
Check this, 40 - 28 = 12 in is perimeter of the rectangle
see if this checks out: w= 1.5, L = 4.5
2(4.5) + 2(1.5) = 12
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