document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=467255>Question 467255</A>: <I><SPAN STYLE=\"word-break: break-all;\"> QR is an arc of a circle with a center P. If the length of the arc is 6pi and the degree of the angle formed at center P is 30 degrees, what is the area of the sector PQR?<BR>\n" );
document.write( " The answer is 108pi. <BR>\n" );
document.write( "Please help me understand how they got this answer. <BR>\n" );
document.write( "Thank you!<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #364461 by </B><A HREF=/tutors/aboutme.mpl?userid=TutorDelphia><B>TutorDelphia(193)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=TutorDelphia><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=TutorDelphia><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=364461\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 30 degrees is 30/360 which reduces to 1/12 of the whole circle (there are 360 degrees in a circle).\r<BR>\n" );
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document.write( "if 6pi is 1/12 of the circumference than the whole circumference can be found from 6pi*12 which gives us 72pi as my circumference. <BR>\n" );
document.write( "Using the formula <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=c=2pi%2Ar ALIGN=MIDDLE  ALT=\"c=2pi%2Ar\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=72pi=2%2Api%2Ar ALIGN=MIDDLE  ALT=\"72pi=2%2Api%2Ar\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "36=r<BR>\n" );
document.write( "I now know that my radius of the circle is 36.\r<BR>\n" );
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document.write( "Therefore my area of the circle is 1296 pi from the formula <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=pi%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A=pi%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Now that we have the area of the whole circle, I can just divide it by 12 to get the area of the sector. So my area is 108 pi as you said. </SPAN>\n" );
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