document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=91982>Question 91982</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Terence&#8217;s dog has a dog house that is 3&#8217; wide and 5&#8217; long. He is building a fence so that way the dog can stay within 4&#8217; of the doghouse on each side. What would be the area inside the fence? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #66893 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=66893\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> adding 4' to BOTH ends of a 3' width and 5' length gives an area that is 11' by 13' ... or a=11(13) ... a=143sq' </SPAN>\n" );
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