SOLUTION: A wire 36 inches long is cut into two pieces and then bent into two square frames. The two frames have sides differing by five inches. What is the sum of the two areas?

Question 104786: A wire 36 inches long is cut into two pieces and then bent into two square frames. The two frames have sides differing by five inches. What is the sum of the two areas?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A wire 36 inches long is cut into two pieces and then bent into two square frames. The two frames have sides differing by five inches. What is the sum of the two areas?
:
Let x = length of the side of the smaller square:
Perimeter = 4x
and
(x+5) = length of the side of the larger square
Perimeter = 4(x+5)
:
The total perimeter of the two squares = 36 in, solve for x:
4x + 4(x+5) = 36
4x + 4x + 20 = 36
8x = 36 - 20
8x = 16
x = 16/8
x = 2" length of the smaller square
and
2 + 5 = 7" length of the larger square
:
Sum of the Areas:
2^2 + 7^2 =
4 + 49 = 53 sq/inches

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