document.write( "Question 742527: \r
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\n" ); document.write( "\n" ); document.write( "the perimeter of a rectangle is 42 ft and the area of the rectangle is 54 ft^2. find the dimensions of the rectangle ?
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Algebra.Com's Answer #452515 by checkley79(3341)\"\" \"About 
You can put this solution on YOUR website!
P=2L+2W
\n" ); document.write( "42=2L+2W
\n" ); document.write( "AREA=LW
\n" ); document.write( "54=LW
\n" ); document.write( "L=54/W
\n" ); document.write( "42=2(54/W)+2W
\n" ); document.write( "42=108/W+2W
\n" ); document.write( "42=(108+2W*W)/W
\n" ); document.write( "42=(108+2W^2)/W CROSS MULTIPLY.
\n" ); document.write( "42*W=108+2W^2
\n" ); document.write( "42W=108+2W^2
\n" ); document.write( "2W^2-42W+108=0
\n" ); document.write( "2(W^2-21W+54)=0
\n" ); document.write( "2(W-18)(W-3)=0
\n" ); document.write( "W-18=0
\n" ); document.write( "W=18 ANS.
\n" ); document.write( "W-3=0
\n" ); document.write( "W=3 ANS.
\n" ); document.write( "42=2L+2*18
\n" ); document.write( "42=2L+36
\n" ); document.write( "2L=42-36
\n" ); document.write( "2L=6
\n" ); document.write( "L=6/2
\n" ); document.write( "L=3 WHEN W=18
\n" ); document.write( "42=2L+2*3
\n" ); document.write( "42=2L+6
\n" ); document.write( "2L=42-6
\n" ); document.write( "2L=36
\n" ); document.write( "L=36/2
\n" ); document.write( "L=18 WHEN W=3
\n" ); document.write( "PROOF:
\n" ); document.write( "54=18*3
\n" ); document.write( "54=54
\n" ); document.write( "54=3*18
\n" ); document.write( "54=54
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