document.write( "Question 621585: The length of a rectangle is four times its width.
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document.write( "If the perimeter of the rectangle is 50in, find its area. \n" );
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Algebra.Com's Answer #390830 by dragonwalker(73)![]() ![]() ![]() You can put this solution on YOUR website! 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 !!!! \n" ); document.write( " |