document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=996452>Question 996452</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of an isoceles right angled triangle is 2p unit. The area of the same triangle is ? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #614919 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=614919\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the length of the leg of that triangle x.<BR>\n" );
document.write( "Thus the perimeter is<BR>\n" );
document.write( "P = x + x + x(sqrt(2)) = 2p<BR>\n" );
document.write( "The area would be <BR>\n" );
document.write( "A = (1/2)bh = 1/2 (x^2)<BR>\n" );
document.write( "so we need to solve for x above and plug it in here.<BR>\n" );
document.write( "2x + x(sqrt(2)) = 2p<BR>\n" );
document.write( "Factor out x and divide by what's left and get<BR>\n" );
document.write( "x(2 + sqrt(2)) = 2p<BR>\n" );
document.write( "x = 2p / (2 + sqrt(2))<BR>\n" );
document.write( "Now plug in and get<BR>\n" );
document.write( "A = (1/2)[2p / (2 + sqrt(2))]^2<BR>\n" );
document.write( "A = (1/2)(4p^2 / (2 + sqrt(2))^2)<BR>\n" );
document.write( "A = 2p^2 / (2 + sqrt(2))^2<BR>\n" );
document.write( "or if you have to simplify further<BR>\n" );
document.write( "A = 2p^2 / (6 + 4sqrt(2)) = p^2 / (3 + 2sqrt(2))<BR>\n" );
document.write( "You may still have to rationalize the denominator. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );