document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=288845>Question 288845</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An isosceles right triangle has an area of 98 square units, what is the perimeter? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #209312 by </B><A HREF=/tutors/aboutme.mpl?userid=dabanfield><B>dabanfield(803)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=dabanfield><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=209312\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the length of the common leg. Then both the base and the height of the triangle are x so the area is:\r<BR>\n" );
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document.write( "1/2*base*height = 98<BR>\n" );
document.write( "(1/2)*x*x = 98<BR>\n" );
document.write( "x^2/2 = 98<BR>\n" );
document.write( "x^2 = 2*98<BR>\n" );
document.write( "x^2 = 196<BR>\n" );
document.write( "x = 14\r<BR>\n" );
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document.write( "Let h be the hypotenuse. Then, by the Pythagorean Theorem, we have;\r<BR>\n" );
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document.write( "h^2 = x^2 + x^2<BR>\n" );
document.write( "h^2 = 2*(x^2)<BR>\n" );
document.write( "h = x*sqrt(2)<BR>\n" );
document.write( "h = 14*sqrt(2)\r<BR>\n" );
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document.write( "The perimeter then is x + x + h = 14 + 14 + 14*sqrt(2) = 14*(2 + sqrt(2)) </SPAN>\n" );
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