document.write( "Question 291941: a rectangle has a perimeter of 38 centimeters and a area of 84 centimeters.what are the dimension of the rectangle \n" ); document.write( "
Algebra.Com's Answer #210992 by oberobic(2304)\"\" \"About 
You can put this solution on YOUR website!
P = 2(L+W) = 38 cm
\n" ); document.write( "A = L*W = 84 cm^2
\n" ); document.write( "L = 84/W
\n" ); document.write( "2(84/W+W)=38
\n" ); document.write( "Divide both sides by 2.
\n" ); document.write( "84/W + W = 19
\n" ); document.write( "Multiply through by W to remove denominator.
\n" ); document.write( "84 + W^2 = 19W
\n" ); document.write( "Subtract 19W from both sides.
\n" ); document.write( "W^2 - 19W + 84 = 0
\n" ); document.write( "Factor.
\n" ); document.write( "(W-7)(W-12) = 0
\n" ); document.write( "So,
\n" ); document.write( "W = 7 or W = 12
\n" ); document.write( "It is a rectangle, which means logically, if
\n" ); document.write( "W = 7, then L = 12
\n" ); document.write( "and if
\n" ); document.write( "W = 12, then L = 7.
\n" ); document.write( "Checking the perimeter...
\n" ); document.write( "2(7+12) = 38??
\n" ); document.write( "2(19) = 38
\n" ); document.write( "Correct.
\n" ); document.write( ".
\n" ); document.write( "Answer
\n" ); document.write( "L = 7 and W = 12
\n" ); document.write( "or
\n" ); document.write( "L = 12 and W = 7
\n" ); document.write( ".
\n" ); document.write( "Done \n" ); document.write( "
\n" );