document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=156649>Question 156649</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the perimeter of a rectangle is 60 inches and the length is 4 inches longer than the width. Find the dimensions of the triangle.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #115457 by </B><A HREF=/tutors/aboutme.mpl?userid=mizjoice><B>mizjoice(22)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mizjoice><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=115457\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Once you know the rules you can break them!  In algebra, start with the formula.  the formula for a perimeter is P=2(w + l).  I don't have a conventional way of solving for the perimeter but this does work.  The perimeter of the rectangle is 60 inches.  The first thing is do is divide by four.60/4=15.  now, if this perimeter is 15 on all sides, that's not a rectangle, thats a square!  But remember, the length is 4 inches longer than the width.  Add 2 inches to two sides 15 + 2 =17.  Now take two from the other side15-2=13.  Check your answer 17-4=13. And double check by substituting values:  P = 2(17+13) P=34+26 P=60 </SPAN>\n" );
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