document.write( "Question 427041: How does a 4th grader find the dimensions of a rectangle when the perimeter is 58 and the area is 54. \n" ); document.write( "
Algebra.Com's Answer #297000 by Alan3354(69443)\"\" \"About 
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How does a 4th grader find the dimensions of a rectangle when the perimeter is 58 and the area is 54.
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\n" ); document.write( "Perimeter = 2L + 2W = 58 --> L + W = 29
\n" ); document.write( "Area = L*W = 54
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\n" ); document.write( "L = 54/W
\n" ); document.write( "(54/W) + W = 29
\n" ); document.write( "54 + W^2 = 29W
\n" ); document.write( "\"W%5E2+-+29W+%2B+54+=+0\"
\n" ); document.write( "(W - 2)*(W - 27) = 0
\n" ); document.write( "W = 2
\n" ); document.write( "L = 27
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\n" ); document.write( "I'm not sure that'll work in the 4th grade, it's been awhile since I was there.
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\n" ); document.write( "Try this:
\n" ); document.write( "Factor 54 and look for a pair that adds to 29
\n" ); document.write( "54 =
\n" ); document.write( "1 x 54
\n" ); document.write( "2 x 27
\n" ); document.write( "3 x 18
\n" ); document.write( "6 x 9\r
\n" ); document.write( "\n" ); document.write( "Only 2 by 27 fits.
\n" ); document.write( "If the solution is not integers, that won't work.
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