document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=891679>Question 891679</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Three identical circles are tangent to each other externally. If the area of the curvilinear triangle enclosed between the points of tangency of the 3 circles is 16.13, compute the radius of each circle. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #539942 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=539942\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Note that the figure formed is an equilateral, equiangular, triangle with side 2r and interior angles of 60 degrees.  For each circle, the area of the intercepted sector is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%5Bs%5D+=+%28pi%2F6%29%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A%5Bs%5D+=+%28pi%2F6%29%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "==> the combined area of the sectors is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3%2AA%5Bs%5D+=+%28pi%2F2%29%2Ar%5E2 ALIGN=MIDDLE  ALT=\"3%2AA%5Bs%5D+=+%28pi%2F2%29%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "\n" );
document.write( "But the area of the equilateral triangle with side 2r is <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%283%29%2Ar%5E2 ALIGN=MIDDLE  ALT=\"sqrt%283%29%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, hence\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=16.13+=+sqrt%283%29%2Ar%5E2+-+%28pi%2F2%29%2Ar%5E2 ALIGN=MIDDLE  ALT=\"16.13+=+sqrt%283%29%2Ar%5E2+-+%28pi%2F2%29%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<==> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=16.13%2F%28sqrt%283%29+-+pi%2F2%29+=+r%5E2 ALIGN=MIDDLE  ALT=\"16.13%2F%28sqrt%283%29+-+pi%2F2%29+=+r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "==> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%2816.13%2F%28sqrt%283%29+-+pi%2F2%29%29+=+r ALIGN=MIDDLE  ALT=\"sqrt%2816.13%2F%28sqrt%283%29+-+pi%2F2%29%29+=+r\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Then r = 10.00141131, or practically, r = 10 units. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );