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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)![]() ![]() ![]() 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 \n" ); document.write( " \n" ); document.write( " |