document.write( "Question 433948: THe length of a rectangle is 2ft longer than its width. If the perimeter of a rectangle is 44ft, find its area? \n" ); document.write( "

Algebra.Com's Answer #300742 by Sarpi(32) $\"\"$ $\"About$

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Length(L), Width(W), Perimeter(P) and Area(A)\r
\n" ); document.write( "\n" ); document.write( " $\"L=W%2B2\"$ , therefore $\"P=2%28L%2BW%29\"$ where $\"P=44\"$
\n" ); document.write( "Therefore; $\"44=2%28L%2BW%29\"$
\n" ); document.write( " opening the brackets
\n" ); document.write( " $\"44=2L%2B2W\"$
\n" ); document.write( " substituting $\"L=W%2B2\"$
\n" ); document.write( " $\"44=2%28W%2B2%29%2B2W\"$
\n" ); document.write( " $\"44=2W%2B4%2B2W\"$
\n" ); document.write( " $\"44-4=2W%2B2W\"$
\n" ); document.write( " $\"40=4W\"$
\n" ); document.write( "Thus, $\"W=10\"$ and $\"L=W%2B2\"$ will be $\"L=10%2B2\"$ , $\"L=12\"$ \r
\n" ); document.write( "\n" ); document.write( "Now the Area; $\"A=L%2AW\"$
\n" ); document.write( " $\"A=12%2A10\"$ which is equal to 120
\n" ); document.write( "Hence, the area of the rectangle is $\"120ft%5E2\"$ \n" ); document.write( "

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