document.write( "Question 352012: The perimeter of a rectangle is 82 yards.The width is 15 yards less than the length.What are the dimensions of the rectangle? \n" ); document.write( "
Algebra.Com's Answer #251548 by mananth(16949)\"\" \"About 
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let length = l
\n" ); document.write( "width = l-15
\n" ); document.write( "...
\n" ); document.write( "Perimeter = 2*(L+W)
\n" ); document.write( "2*(L+L-15)=82
\n" ); document.write( "2*(2l-15)=82
\n" ); document.write( "4l-30=82
\n" ); document.write( "4l=82+30
\n" ); document.write( "4l=112
\n" ); document.write( "l= 28 yards the length
\n" ); document.write( "width = l-15 = 13 yards
\n" ); document.write( "....
\n" ); document.write( "C H E C K
\n" ); document.write( "13+13+28+28=82
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca
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