document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=164005>Question 164005</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Prove that the area of a circular sector of radius r with central angle theta is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%281%2F2%29%28theta%29r%5E2 ALIGN=MIDDLE  ALT=\"A=%281%2F2%29%28theta%29r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, where theta is measured in radians.\r<BR>\n" );
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document.write( "The following is what I've come up with so far, am I on the right track?\r<BR>\n" );
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document.write( "I can tell that a circular sector looks similar to a triangle with an arc replacing its flat base.  The area of a triangle is computed using A=1/2bh.  Therefore:\r<BR>\n" );
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document.write( "h=r<BR>\n" );
document.write( "b=s=r(theta)\r<BR>\n" );
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document.write( "A=(1/2)rr(theta) which is simplified to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%281%2F2%29r%5E2 ALIGN=MIDDLE  ALT=\"A=%281%2F2%29r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #120836 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=120836\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Prove that the area of a circular sector of radius r with central angle theta is , where theta is measured in radians. <BR>\n" );
document.write( "The following is what I've come up with so far, am I on the right track? <BR>\n" );
document.write( "I can tell that a circular sector looks similar to a triangle with an arc replacing its flat base. The area of a triangle is computed using A=1/2bh. Therefore: <BR>\n" );
document.write( "h=r<BR>\n" );
document.write( "b=s=r(theta) <BR>\n" );
document.write( "A=(1/2)rr(theta) which is simplified to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%281%2F2%29r%5E2 ALIGN=MIDDLE  ALT=\"A=%281%2F2%29r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%281%2F2%29r%5E2%2Atheta ALIGN=MIDDLE  ALT=\"A=%281%2F2%29r%5E2%2Atheta\"   DISABLED_style= \"max-width:90%; height:auto;\" >  (You left out the theta)<BR>\n" );
document.write( "What you've done is the correct approach.  This can be done as a limit, or by integral calculus, which is essentially the same thing.  I don't know if your instructor would accept an integration, tho, it depends on the level of your class. </SPAN>\n" );
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