document.write( "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? \n" ); document.write( "
Algebra.Com's Answer #76280 by ankor@dixie-net.com(22740)\"\" \"About 
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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?
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\n" ); document.write( "Let x = length of the side of the smaller square:
\n" ); document.write( "Perimeter = 4x
\n" ); document.write( "and
\n" ); document.write( "(x+5) = length of the side of the larger square
\n" ); document.write( "Perimeter = 4(x+5)
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\n" ); document.write( "The total perimeter of the two squares = 36 in, solve for x:
\n" ); document.write( "4x + 4(x+5) = 36
\n" ); document.write( "4x + 4x + 20 = 36
\n" ); document.write( "8x = 36 - 20
\n" ); document.write( "8x = 16
\n" ); document.write( "x = 16/8
\n" ); document.write( "x = 2\" length of the smaller square
\n" ); document.write( "and
\n" ); document.write( "2 + 5 = 7\" length of the larger square
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\n" ); document.write( "Sum of the Areas:
\n" ); document.write( "2^2 + 7^2 =
\n" ); document.write( " 4 + 49 = 53 sq/inches
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