document.write( "Question 621585: The length of a rectangle is four times its width.
\n" ); document.write( "If the perimeter of the rectangle is 50in, find its area.
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Algebra.Com's Answer #390830 by dragonwalker(73)\"\" \"About 
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So let us call the shorter side, i.e the width: 'x'
\n" ); document.write( "So each length is four times the width = 4x\r
\n" ); document.write( "\n" ); document.write( "The perimeter consists of two widths and two lengths and equals 50 inches.\r
\n" ); document.write( "\n" ); document.write( "So:
\n" ); document.write( "x+x+4x+4x=50\r
\n" ); document.write( "\n" ); document.write( "solve for x:\r
\n" ); document.write( "\n" ); document.write( "10x = 50
\n" ); document.write( "to find x divide both sides by 10:\r
\n" ); document.write( "\n" ); document.write( "10x/10 = 50/10
\n" ); document.write( "x=5\r
\n" ); document.write( "\n" ); document.write( "So width equals 5 in and length is four times this i.e. 4x5=20\r
\n" ); document.write( "\n" ); document.write( "To check: 5+5+20+20=50 !!!!
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