document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1008994>Question 1008994</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the area of the shaded region in the figure. (Round your answer to one decimal place.)\r<BR>\n" );
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document.write( "USE LINK FOR PICTURE <BR>\n" );
document.write( "http://www.webassign.net/waplots/7/7/84eaf04fe7ca06731479a31a56bf2e.gif<BR>\n" );
document.write( "PLEASE HELP!!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #624511 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=624511\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Okay, the area of the shaded region is of course the area of the circle minus the area of the segment, as shown.<BR>\n" );
document.write( "We can find the area of the circle as A = (pi)r^2 = 324(pi)<BR>\n" );
document.write( "The segment can be found by subtracting the equilateral triangle of side 18 from 1/6 the area of the circle, so<BR>\n" );
document.write( "A(segment) = (1/6)(pi*r^2) - s^2*sqrt(3) / 4<BR>\n" );
document.write( "A(segment) = (1/6)(324pi) - 18^2*sqrt(3) / 4 <BR>\n" );
document.write( "A(segment) = 54(pi) - 81*sqrt(3)<BR>\n" );
document.write( "And then the final shaded area is<BR>\n" );
document.write( "A(shaded) = 324(pi) - [54(pi) - 81*sqrt(3)] = 270(pi) + 81*sqrt(3) </SPAN>\n" );
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