SOLUTION: Two square wire frames are to be constructed from a piece of wire 100 inches long. If the area enclosed by one frame is to be one-half the area enclosed by the other, find the dime
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-> SOLUTION: Two square wire frames are to be constructed from a piece of wire 100 inches long. If the area enclosed by one frame is to be one-half the area enclosed by the other, find the dime
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Question 149714: Two square wire frames are to be constructed from a piece of wire 100 inches long. If the area enclosed by one frame is to be one-half the area enclosed by the other, find the dimensions of each frame. (Disregard the thickness of the wire.) Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Let x = side of smaller square
and y = side of larger square
.
x^2 is the area of smaller square
y^2 is the area of larger square
.
4x is the perimeter of smaller square
4y is the perimeter of larger square
.
Since we have two unknowns, we need two equations:
From: "Two square wire frames are to be constructed from a piece of wire 100 inches long." we get:
4x + 4y = 100
x + y = 25 (equation 1)
.
From: "If the area enclosed by one frame is to be one-half the area enclosed by the other" we get:
x^2 = (1/2)y^2 (equation 2)
.
solving equation 1 for x:
x + y = 25
x = 25 - y
.
plug the above into equation 2 and solve for y:
x^2 = (1/2)y^2
(25-y)^2 = (1/2)y^2
625 - 50y + y^2 = (1/2)y^2
1250 - 100y + 2y^2 = y^2
1250 - 100y + y^2 = 0
y^2 - 100y + 1250 = 0
.
From the "quadratic equation" you'll get:
y = 85.36
x = 14.64
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Therefore the dimensions are:
85.36 in by 85.36 in (larger square)
14.64 in by 14.64 in (smaller square)
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Details of the quadratic solution follows:
