document.write( "Question 294013: The width of a rectangle is 5 meters less than its length. The area is 84 square meters. Find the dimensions of the rectangle and the perimeter of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #212160 by peter_mccarter(5)\"\" \"About 
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The width of a rectangle is 5 meters less than its length. The area is 84 square meters. Find the dimensions of the rectangle and the perimeter of the rectangle.
\n" ); document.write( "First it helps to draw a picture so draw a rectangle \r
\n" ); document.write( "\n" ); document.write( "Remember that area of a rectangle =Length times width and the perimeter is P= 2 times the length plus 2 times the width
\n" ); document.write( "So A=LW and P=2L+2W
\n" ); document.write( "84= area
\n" ); document.write( "L=length
\n" ); document.write( "W= length -5
\n" ); document.write( "L(L-5)=84 distribute the L and you have L2 -5L=84 now subtract 84 from both sides and get L2 –5L-84 next factor to get (L+7)(L-12) next set both to =0 (L+7)=0 (L+12)=0 solve for L
\n" ); document.write( "L=-7 and L=12 you can’t have a negative length so -7 is out. So L=12 now plug in to the original area equation
\n" ); document.write( "84=LW 84= 12w now divide both sides by 12 to isolate W, this gives you W=7
\n" ); document.write( "L=12 and W= 7
\n" ); document.write( "For perimeter use 12 for the length and 7 for width Remember that P= 2L+2W
\n" ); document.write( "P=2(12) +2(7) =38\r
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