document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1204108>Question 1204108</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the radius of the circle inscribed in an equilateral triangle whose perimeter is 10.8 units </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #840149 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=840149\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "This is what the diagram could look like<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%0D%0Adrawing%28400%2C400%2C-4%2C4%2C-4%2C4%2C%0D%0Aline%28-2%2C-1%2C2%2C-1%29%2C%0D%0Aline%282%2C-1%2C0%2C2.4641%29%2C%0D%0Aline%280%2C2.4641%2C-2%2C-1%29%2C%0D%0Aline%280%2C0.1547%2C-2%2C-1%29%2C%0D%0Aline%280%2C0.1547%2C0%2C-1%29%2C%0D%0A%0D%0Aline%280%2C-1%2B0.2%2C0-0.2%2C-1%2B0.2%29%2C%0D%0Aline%280-0.2%2C-1%2B0.2%2C0-0.2%2C-1%29%2C%0D%0A%0D%0Acircle%280%2C0.1547%2C1.1547%29%2C%0D%0A%0D%0Alocate%28-2-0.2%2C-1-0.1%2C%22A%22%29%2C%0D%0Alocate%282%2B0.1%2C-1-0.1%2C%22B%22%29%2C%0D%0Alocate%280%2B0.1%2C2.4641%2B0.4%2C%22C%22%29%2C%0D%0A%0D%0Alocate%280%2B0.1%2C0.1547%2B0.4%2C%22D%22%29%2C%0D%0Alocate%280%2B0.1%2C-1-0.1%2C%22E%22%29%2C%0D%0A%0D%0Acircle%28-2%2C-1%2C0.05%29%2Ccircle%28-2%2C-1%2C0.07%29%2Ccircle%28-2%2C-1%2C0.09%29%2Ccircle%28-2%2C-1%2C0.11%29%2Ccircle%282%2C-1%2C0.05%29%2Ccircle%282%2C-1%2C0.07%29%2Ccircle%282%2C-1%2C0.09%29%2Ccircle%282%2C-1%2C0.11%29%2Ccircle%280%2C2.4641%2C0.05%29%2Ccircle%280%2C2.4641%2C0.07%29%2Ccircle%280%2C2.4641%2C0.09%29%2Ccircle%280%2C2.4641%2C0.11%29%2Ccircle%280%2C0.1547%2C0.05%29%2Ccircle%280%2C0.1547%2C0.07%29%2Ccircle%280%2C0.1547%2C0.09%29%2Ccircle%280%2C0.1547%2C0.11%29%2Ccircle%280%2C-1%2C0.05%29%2Ccircle%280%2C-1%2C0.07%29%2Ccircle%280%2C-1%2C0.09%29%2Ccircle%280%2C-1%2C0.11%29%2C%0D%0A%0D%0Alocate%28-3.8%2C-2.8%2Cmatrix%281%2C4%2C%22Diagram%22%2C%22is%22%2C%22to%22%2C%22scale%22%29%29%0D%0A%0D%0A%29%0D%0A ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "A,B,C = vertices of the equilateral triangle<BR>\n" );
document.write( "D = center of the inscribed circle<BR>\n" );
document.write( "E = midpoint of AB\r<BR>\n" );
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document.write( "Triangle ADE is a right triangle. More specifically, it is a 30-60-90 triangle.<BR>\n" );
document.write( "This is because angle CAB = 60 is bisected to help form angle DAE = 30 degrees.<BR>\n" );
document.write( "Also, angle ADE = 60 degrees.\r<BR>\n" );
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document.write( "The perimeter of the equilateral triangle is 10.8 units.<BR>\n" );
document.write( "Each side must be (10.8)/3 = 3.6 units<BR>\n" );
document.write( "AB = 3.6<BR>\n" );
document.write( "BC = 3.6<BR>\n" );
document.write( "AC = 3.6\r<BR>\n" );
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document.write( "Because E is the midpoint of AB, we then know<BR>\n" );
document.write( "AE = AB/2 = (3.6)/2 = 1.8\r<BR>\n" );
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document.write( "Segment AE is the longer leg of the 30-60-90 triangle ADE (notice it's opposite the 60 degree angle).\r<BR>\n" );
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document.write( "For any 30-60-90 triangle we have this template<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=longLeg+=+shortLeg%2Asqrt%283%29 ALIGN=MIDDLE  ALT=\"longLeg+=+shortLeg%2Asqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "In this particular case it means<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AE+=+DE%2Asqrt%283%29 ALIGN=MIDDLE  ALT=\"AE+=+DE%2Asqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Isolating DE gets us <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DE+=+%28AE%29%2F%28sqrt%283%29%29+=+%281.8%29%2Fsqrt%283%29+=+1.8%2A%28sqrt%283%29%2F3%29+=+0.6%2Asqrt%283%29+=+1.03923 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Which is the approximate radius of the inscribed circle.\r<BR>\n" );
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document.write( "Round this value however needed. Or you can stick to the exact value <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0.6%2Asqrt%283%29 ALIGN=MIDDLE  ALT=\"0.6%2Asqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > to avoid worrying about rounding.\r<BR>\n" );
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document.write( "Side note: 0.6 = 3/5<BR>\n" );
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