document.write( "Question 384282: A rectangle that the length is twice the with with an area of 50yd^2 \n" ); document.write( "
Algebra.Com's Answer #272039 by mananth(16949)\"\" \"About 
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width = x
\n" ); document.write( "length = 2x
\n" ); document.write( "..
\n" ); document.write( "L*W=Area
\n" ); document.write( "2x*x=50
\n" ); document.write( "2x^2=50
\n" ); document.write( "/2
\n" ); document.write( "x^2=25
\n" ); document.write( "\"sqrt%28x%5E2%29=sqrt%2825%29\"
\n" ); document.write( "x= 5 yard the width
\n" ); document.write( "..
\n" ); document.write( "length = 2x = 10 yard the length
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( "
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