document.write( "Question 157086: Find the dimensions of a rectangle whose perimeter is 26 meter and whose area is 40 square meters. \n" ); document.write( "
Algebra.Com's Answer #115815 by atique.ah(27)\"\" \"About 
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Find the dimensions of a rectangle whose perimeter is 26 meter and whose area is 40 square meters.\r
\n" ); document.write( "\n" ); document.write( "Solution:\r
\n" ); document.write( "\n" ); document.write( "let length=L
\n" ); document.write( "let width=w\r
\n" ); document.write( "\n" ); document.write( "Area A=LxW=40 .....EQ 1
\n" ); document.write( "Perimeter P=2(L+W)=26
\n" ); document.write( "L+W=13.............EQ 2\r
\n" ); document.write( "\n" ); document.write( "substituting the value of W from EQ 2 into EQ 1 we have
\n" ); document.write( "LX(13-L)=40
\n" ); document.write( "13L-L^2=40
\n" ); document.write( "REARRANGING
\n" ); document.write( "L^2-13L+40=0
\n" ); document.write( "BREAKING THE MIDDLE TERM
\n" ); document.write( "L^2-5L-8L+40=0
\n" ); document.write( "L(L-5)-8(L-5)=0
\n" ); document.write( "(L-5)(L-8)=0
\n" ); document.write( "THIS IMPLIES
\n" ); document.write( "L=5 0R L=8
\n" ); document.write( "W=8 0R W=5
\n" ); document.write( " \n" ); document.write( "
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