document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=257521>Question 257521</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If the perimeter of a rectangle is 22 meters, and the perimeter of a right triangle is 12 meters (the sides of the triangle are half the length of the rectangle, the width of the rectangle, and the hypotenuse is 5 meters). How do you solve for L and W, the dimensions of the rectangle. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #189536 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=189536\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <BR>\n" );
document.write( "P=2L+2W<BR>\n" );
document.write( "22=2L+2W<BR>\n" );
document.write( "P=a+b+c<BR>\n" );
document.write( "12=L/2+W+5<BR>\n" );
document.write( "22=2L+2W<BR>\n" );
document.write( "L=8 and W=3<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );