document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=515421>Question 515421</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If the length and width of a rectangle are doubled, how do the perimeters of the original and new rectangles compare? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #343960 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=343960\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The perimeter of the new rectangle is double. To show this algebraically, we let L and W be the length, width of the original rectangle. The perimeter is 2L + 2W. The new rectangle would have dimensions 2L and 2W, and the perimeter would be 2(2L ) + 2(2W) = 4L + 4W, or double the original perimeter. </SPAN>\n" );
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