document.write( "Question 384067: A rectangle has a length twice as long as the width and an area of 50 square feet. How long is the diagonal of the rectangle? \n" ); document.write( "

Algebra.Com's Answer #271842 by mananth(16946) $\"\"$ $\"About$

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width = x
\n" ); document.write( "lebgth =2x
\n" ); document.write( "area =50 ft^2
\n" ); document.write( "..
\n" ); document.write( "L*W= Area
\n" ); document.write( "x*2x=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 feet the width
\n" ); document.write( "length = 2x= 10 feet
\n" ); document.write( "..
\n" ); document.write( "Diagonal^2= L^2+W^2
\n" ); document.write( "D^2=5^2+10^2
\n" ); document.write( "D^2=125
\n" ); document.write( " $\"sqrt%28d%5E2%29=sqrt%28125%29\"$
\n" ); document.write( "d=11.2 feet the diagonal
\n" ); document.write( "..
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( " \n" ); document.write( "

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