document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1138788>Question 1138788</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do I find the largest possible area of an isosceles triangle if the length of each of the two equal sides is 10 m? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #756609 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=756609\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "You don't even have to get into trigonometry with the sine of an angle to solve this problem.<br><BR>\n" );
document.write( "Given the lengths of two sides of a triangle, the maximum area of a triangle with those two side lengths is when those two sides form a right angle.<br><BR>\n" );
document.write( "That should be easy to see by considering one of the two given side lengths as the base.  The area of the triangle is one-half base times height; and clearly the maximum height of the triangle is when the second given side is at right angles to the first.<br><BR>\n" );
document.write( "And, to answer the question, the maximum area of a right triangle is one-half the product of the two legs; so the maximum possible area with two legs of length 10m is (1/2)(10m)(10m) = 50m^2. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );