document.write( "Question 374746: I am haveing problems solving this problem I have tried numerous times but i just do not understand it. \r
\n" ); document.write( "\n" ); document.write( "A rectangle is 5 times as long as it is wide. If the area is 125 square feet, find its perimiter.
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Algebra.Com's Answer #266502 by andsrox(1)\"\" \"About 
You can put this solution on YOUR website!
the area of a rectangle = L x W
\n" ); document.write( "L = 5W
\n" ); document.write( "area = w x 5w
\n" ); document.write( "area = \"5w%5E2\"
\n" ); document.write( "125 = \"5w%5E2\"
\n" ); document.write( "\"sqrt%28125%29\" = \"sqrt%285w%5E2%29\"
\n" ); document.write( "\"sqrt%2825%2A5%29\" = 5w
\n" ); document.write( "\"5%2Asqrt%285%29\" = 5w
\n" ); document.write( "\"%5B5%2Asqrt%285%29+%2F+5\" = 5L / 5
\n" ); document.write( "\"sqrt%285%29\" = w\r
\n" ); document.write( "\n" ); document.write( "L/5 = w
\n" ); document.write( "L = \"5%2Asqrt%285%29\"\r
\n" ); document.write( "\n" ); document.write( "perimeter = 2W + 2L
\n" ); document.write( "p = \"2%2Asqrt%285%29\" + 2*5*sqrt(5)}}}
\n" ); document.write( "p = \"2%2Asqrt%285%29\" + 10*sqrt(5)}}}
\n" ); document.write( "p = \"12%2Asqrt%285%29\"\r
\n" ); document.write( "\n" ); document.write( "This was my best guess. Sorry if its wrong.
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