document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=197934>Question 197934</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There is a right triangle. The short leg is 7 and the long leg is 10. X is across from the right angle.  How do I find X? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #148423 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley77><B>checkley77(12844)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley77><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=148423\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The sides of a right triangle are defined by the following formula:<BR>\n" );
document.write( "a^2+b^2=C^2 where a & b are the 2 short legs & c is the hypotenuse ( the side opposite the right angle)<BR>\n" );
document.write( "7^2+10^2=c^2<BR>\n" );
document.write( "49=100=c^2<BR>\n" );
document.write( "149=c^2<BR>\n" );
document.write( "sqrt149=c<BR>\n" );
document.write( "12.2=c ans.<BR>\n" );
document.write( "Proof:<BR>\n" );
document.write( "49+100=12.2^2<BR>\n" );
document.write( "149~149 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );