document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1157990>Question 1157990</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of segment AB is 20 cm. Find the distance from C to AB, given that C is a point on the circle that has AB as a diameter, and that<BR>\n" );
document.write( "(a) AC = CB; (b) AC = 10 cm; (c) AC = 12 cm. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #780914 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=780914\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "For each case, let P be the point on diameter AB such that CP is perpendicular to AB.  Then in each case the length of CP is what we need to find.<br><BR>\n" );
document.write( "(a) AC = CB<br><BR>\n" );
document.write( "Triangle ABC is isosceles; CP is a radius of the circle -- length 10.<br><BR>\n" );
document.write( "(b) AC = 10<br><BR>\n" );
document.write( "AC is half of AB, so triangles ABC, APC, and BPC are all 30-60-90 right triangles.  AC=10 means CP = 5*sqrt(3).<br><BR>\n" );
document.write( "(c) AC = 12<br><BR>\n" );
document.write( "Triangle ABC has side lengths 12, 16, and 20 and so is similar to a 3-4-5 right triangle.  Triangles APC and BPC are also similar to a 3-4-5 right triangle.  CP is the long leg in the triangle with hypotenuse 12 -- length 4/5 of 12, which is 48/5, or 9.6.<br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );