document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=847225>Question 847225</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, I have not been able to get this question as I don't know where to start.<BR>\n" );
document.write( "Thank you in advance for any help.\r<BR>\n" );
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document.write( "There is a diagram showing a sector oab of center o in which the angle aob is 120 degrees. There is a horizontal line drawn (ab) splitting the sector into a semicircle and a triangle.\r<BR>\n" );
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document.write( "i) Show that;\r<BR>\n" );
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document.write( "Area of sector oab/area of triangle oab = 4&#960;/3&#8730;3\r<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 #510292 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39620)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=510292\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Area of the sector, based on it being <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=120%2F360 ALIGN=MIDDLE  ALT=\"120%2F360\"   DISABLED_style= \"max-width:90%; height:auto;\" > of the area of the circle of radius r, is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight_green%28%2812%2F36%29pi%2Ar%5E2%29 ALIGN=MIDDLE  ALT=\"highlight_green%28%2812%2F36%29pi%2Ar%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "The triangle with bottom side AB has two equal sides of r each.  Note that this is two special triangles sharing a common side.  If you split the triangle from the center point of the circle to the middle of AB, you have TWO 30-60-90 triangles.  The bottom leg is (1/2)AB, the leg with endpoint at the circle center has length <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r%2F2 ALIGN=MIDDLE  ALT=\"r%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and hypotenuse is r.  We know all this because we have a special right triangle.  Drawing the picture of everything will help.  With all this, first find the length AB, and then find the expression for the area of the central triangle (with the central angle 120 degrees).\r<BR>\n" );
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document.write( "*<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28AB%2F2%29%5E2%2B%28r%2F2%29%5E2=r%5E2 ALIGN=MIDDLE  ALT=\"%28AB%2F2%29%5E2%2B%28r%2F2%29%5E2=r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >;<BR>\n" );
document.write( "the algebra steps, not shown here, will give <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight_green%28AB=r%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"highlight_green%28AB=r%2Asqrt%283%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >; the area for the 120 degree central triangle is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29%28AB%29%28r%2F2%29 ALIGN=MIDDLE  ALT=\"%281%2F2%29%28AB%29%28r%2F2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, half of base times height,<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29%2Ar%2Asqrt%283%29%2Ar%2F2 ALIGN=MIDDLE  ALT=\"%281%2F2%29%2Ar%2Asqrt%283%29%2Ar%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight_green%28r%5E2%2Asqrt%283%29%2F4%29 ALIGN=MIDDLE  ALT=\"highlight_green%28r%5E2%2Asqrt%283%29%2F4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, the triangle's area.\r<BR>\n" );
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document.write( "The question asks, what is the ratio of the sector area to the area of triangle AOB.  Form that ratio and simplify.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%281%2F3%29pi%2Ar%5E2%29%2F%28%281%2F4%29r%5E2%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"%28%281%2F3%29pi%2Ar%5E2%29%2F%28%281%2F4%29r%5E2%2Asqrt%283%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%284%2F3%29pi%2Fsqrt%283%29 ALIGN=MIDDLE  ALT=\"%284%2F3%29pi%2Fsqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%284%2Api%2F%283%2Asqrt%283%29%29%29 ALIGN=MIDDLE  ALT=\"highlight%284%2Api%2F%283%2Asqrt%283%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "-------------<BR>\n" );
document.write( "*  A little more help from the Pythagorean Theorem relationship.<BR>\n" );
document.write( "The \"AB\" is the geometry point notation for the measure of the segment, AB.  The two special 30-60-90 triangles which make up the one central angled triangle have a leg of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29AB ALIGN=MIDDLE  ALT=\"%281%2F2%29AB\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  The other \"leg\" is of length <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r%2F2 ALIGN=MIDDLE  ALT=\"r%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, while the hypotenuse is r.<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28AB%2F2%29%5E2%2B%28r%2F2%29%5E2=r%5E2 ALIGN=MIDDLE  ALT=\"%28AB%2F2%29%5E2%2B%28r%2F2%29%5E2=r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > and we want to solve for AB.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28AB%2F2%29%5E2=r%5E2-%28r%2F2%29%5E2 ALIGN=MIDDLE  ALT=\"%28AB%2F2%29%5E2=r%5E2-%28r%2F2%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AB%5E2=%282%5E2%29%28r%5E2-r%5E2%2F2%5E2%29 ALIGN=MIDDLE  ALT=\"AB%5E2=%282%5E2%29%28r%5E2-r%5E2%2F2%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AB%5E2=4%28r%5E2-%281%2F4%29r%5E2%29 ALIGN=MIDDLE  ALT=\"AB%5E2=4%28r%5E2-%281%2F4%29r%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AB%5E2=4r%5E2-r%5E2 ALIGN=MIDDLE  ALT=\"AB%5E2=4r%5E2-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=AB%5E2=3r%5E2 ALIGN=MIDDLE  ALT=\"AB%5E2=3r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AB=sqrt%283r%5E2%29 ALIGN=MIDDLE  ALT=\"AB=sqrt%283r%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AB=r%2Asqrt%283%29 ALIGN=MIDDLE  ALT=\"AB=r%2Asqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
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