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 #124274 by </B><A HREF=/tutors/aboutme.mpl?userid=Dr.E><B>Dr.E(1)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Dr.E><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=124274\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> --------------------------------------------------------------------------------<BR>\n" );
document.write( "Dr.E's Solution:\r<BR>\n" );
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document.write( "Hello!\r<BR>\n" );
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document.write( "I would stray away from the calculus proofs by integration and other such mathematical inductions until you reach those classes or are told to use such methods in calculus to solve this problem. Since this is an algerbra tutoring website, I doubt the need to do so. Let me know if you need it.\r<BR>\n" );
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document.write( "Remember from your earlier studies that the Area of a Circle is = <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( "The area of the sector of a circle is proportional its central sector.\r<BR>\n" );
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document.write( "In simpler terms, if I said that I wanted you to find <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3%2F4 ALIGN=MIDDLE  ALT=\"3%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" >ths of the circle, you would simply: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%283%2F4%29%2Api%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A=%283%2F4%29%2Api%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Same if I asked you to find <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F2 ALIGN=MIDDLE  ALT=\"1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >,  circle: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%281%2F2%29%2Api%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A=%281%2F2%29%2Api%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "In other terms, we could have said that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F2 ALIGN=MIDDLE  ALT=\"1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > of the circle, in terms of the central angle, is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=180%5Eo%2F360%5Eo ALIGN=MIDDLE  ALT=\"180%5Eo%2F360%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi%2F360%5Eo ALIGN=MIDDLE  ALT=\"pi%2F360%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" > of the (full <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=360%5Eo ALIGN=MIDDLE  ALT=\"360%5Eo\"   DISABLED_style= \"max-width:90%; height:auto;\" >) <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%28180%5Eo%2F360%5Eo%29%2Api%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A=%28180%5Eo%2F360%5Eo%29%2Api%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%28pi%2F360%5Eo%29%2Api%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A=%28pi%2F360%5Eo%29%2Api%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Thus, we can infer that given a central angle labeled <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=BETA ALIGN=MIDDLE  ALT=\"BETA\"   DISABLED_style= \"max-width:90%; height:auto;\" > , would simply be <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%28BETA%2F360%5Eo%29pi%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A=%28BETA%2F360%5Eo%29pi%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Put everything in terms of radians (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=360%5Eo=pi ALIGN=MIDDLE  ALT=\"360%5Eo=pi\"   DISABLED_style= \"max-width:90%; height:auto;\" >), <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%28BETA%2F2pi%29pi%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A=%28BETA%2F2pi%29pi%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=+%281%2F2%29%2A%28BETA%2Fpi%29pi%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A=+%281%2F2%29%2A%28BETA%2Fpi%29pi%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > (then the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=pi ALIGN=MIDDLE  ALT=\"pi\"   DISABLED_style= \"max-width:90%; height:auto;\" >'s cancel out)<BR>\n" );
document.write( "= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=+%281%2F2%29%2A%28BETA%29r%5E2 ALIGN=MIDDLE  ALT=\"A=+%281%2F2%29%2A%28BETA%29r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Thus, by using the sector area formula of a circle, which you can derive from the area of a circle, unless your teacher said you have to derive the area of a circle, which would require your knowledge in integral calculus (which I highly doubt) then this is a valid proof for algebra and geometry.\r<BR>\n" );
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document.write( "If you have any more questions, don't hesitate to ask!\r<BR>\n" );
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document.write( "Have fun with your algebra!<BR>\n" );
document.write( "Dr.E </SPAN>\n" );
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