document.write( "Question 140784: The diagonal of a rectangle platform is square root of 20 cm. If both dimensions are increased by 2 cm, then the resulting perimeter would be 5/3 of the original perimeter. Find the dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #102530 by solver91311(24713)\"\" \"About 
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If the diagonal is \"sqrt%2820%29\", then \"L%5E2%2BW%5E2=20\", thank you Mr. Pythagoras.\r
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\n" ); document.write( "\n" ); document.write( "The original perimeter is \"2L+%2B+2W\" and the larger perimeter is \"2%28L%2B2%29%2B2%28W%2B2%29\", but the larger is 5/3 of the smaller so:\r
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\n" ); document.write( "\n" ); document.write( "\"%285%2F3%29%282L+%2B+2W%29=2%28L%2B2%29%2B2%28W%2B2%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"5%282L+%2B+2W%29=6%28L%2B2%29%2B6%28W%2B2%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"10L%2B10W=6L%2B6W%2B24\"\r
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\n" ); document.write( "\n" ); document.write( "\"4L%2B4W=24\"\r
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\n" ); document.write( "\n" ); document.write( "\"L%2BW=6\" or better for our purposes \"L=6-W\"\r
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\n" ); document.write( "\n" ); document.write( "\"L%5E2%2BW%5E2=20\"
\n" ); document.write( "\"%286-W%29%5E2%2BW%5E2=20\"
\n" ); document.write( "\"36-12W%2BW%5E2%2BW%5E2-20=0\"
\n" ); document.write( "\"2W%5E2-12W%2B16=0\"
\n" ); document.write( "\"W%5E2-6W%2B8=0\"\r
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\n" ); document.write( "\n" ); document.write( "Factors:
\n" ); document.write( "\"%28W-4%29%28W-2%29=0\"\r
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\n" ); document.write( "\n" ); document.write( "So W = 4 or W = 2, meaning L = 2 or L = 4.\r
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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "\"sqrt%284%5E2%2B2%5E2%29=sqrt%2816%2B4%29=sqrt%2820%29\"
\n" ); document.write( "\"2%284%29%2B2%282%29=8%2B4=12\"\r
\n" ); document.write( "\n" ); document.write( "\"12%2A%285%2F3%29+=+20\"\r
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\n" ); document.write( "\n" ); document.write( "\"2%286%29%2B2%284%29=12%2B8=20\" Checks.
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