document.write( "Question 363792: The perimeter of a rectangle is 64cm. Twice the width is 4cm more than the length. Find the dimensions of the rectangle \n" ); document.write( "

Algebra.Com's Answer #259446 by vasumathi(46) $\"\"$ $\"About$

You can put this solution on YOUR website!
twice the width is 4 cm more than the length
\n" ); document.write( "so...
\n" ); document.write( "2w = l+4 (where w is the width and l is the length)
\n" ); document.write( "the perimetre = 64 cms
\n" ); document.write( "perimetre in terms of w and l is p = 2l + 2w
\n" ); document.write( "so..
\n" ); document.write( "64 = 2l + l+4
\n" ); document.write( "64 = 3l+4
\n" ); document.write( "subtract 4 on both sides
\n" ); document.write( "64-4 = 3l +4 -4
\n" ); document.write( "60=3l
\n" ); document.write( "divide both sides by 3
\n" ); document.write( "60/3 = 3l/3
\n" ); document.write( "20 = l
\n" ); document.write( "so the length is 20 cms
\n" ); document.write( "we have 2w = l+4
\n" ); document.write( "2w = 20+4
\n" ); document.write( "2w = 24
\n" ); document.write( "so w = 24/2 = 12
\n" ); document.write( "so width = 12 cms
\n" ); document.write( "so the length = 20 cms
\n" ); document.write( "and width = 12 cms \n" ); document.write( "

\n" );