document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=29982>Question 29982</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I really need help with this problem. I know it will be a system of equtions but I don't know what those equations would be! Here it is: \r<BR>\n" );
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document.write( "A right triangle has a perimeter of 40 units and one leg has a length of 8 units. What are the lengths of the other legs? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #16742 by </B><A HREF=/tutors/aboutme.mpl?userid=Fermat><B>Fermat(136)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Fermat><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=Fermat><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=16742\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Label the three sides as a,b,c, and let a be the hypotenuse.\r<BR>\n" );
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document.write( "Suppose a = 8.<BR>\n" );
document.write( "Since a is the hypotenuse, then the shortest the other two sides could be is 16 units each, and there is no way that 16² + 16² = 8².<BR>\n" );
document.write( "In other words, the side of length 8 cannot be a hypotenuse so it must be one of the other sides.<BR>\n" );
document.write( "let c = 8, say.\r<BR>\n" );
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document.write( "The perimeter: a + b + c = 40<BR>\n" );
document.write( "i.e. a + b = 32<BR>\n" );
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document.write( "Pathagoras' theorem: a² = b² + c²<BR>\n" );
document.write( "i.e. a² = b² + 64<BR>\n" );
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document.write( "Now substitute for a = 32 - b into the quadratic eqn and solve for a. </SPAN>\n" );
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