document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=31363>Question 31363</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An equilateral triangle has an area of 300square root of 3.  Find the apothem. \r<BR>\n" );
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document.write( "My teacher said that there is a formula that can be used to find this. I have looked and looked but have not been successful yet. She said it has to do with the connection between the area and perimeter of equilateral triangles. I tried to find the perimeter of the triangle by using the area formula A=1/2bh. I was unable to get anywhere close to the area. I don't know what else to try.  Could you please help me? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #18053 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><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=18053\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Well, first, let's see the definition of an apothem of a regular polygon, of which, an equilateral triangle is certainly an example:<BR>\n" );
document.write( "\"An apothem of a regular polygon is a line drawn from its centre perpendicular to one of its sides\" This is also the radius of the inscribed circle.\r<BR>\n" );
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document.write( "The formula for finding  an apothem of a regular polygon is:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r+=+%281%2F2%29%28s%29cot%28180%2Fn%29 ALIGN=MIDDLE  ALT=\"r+=+%281%2F2%29%28s%29cot%28180%2Fn%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Where:<BR>\n" );
document.write( "r = is the length of the apothem.<BR>\n" );
document.write( "s = the length of one side of the regular polygon (equilateral triangle).<BR>\n" );
document.write( "n = the number of sides in the regular polygon (3).\r<BR>\n" );
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document.write( "One minor problem is...you don't know the length of one side (s) of the equilateral tringle!\r<BR>\n" );
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document.write( "Not to worry however because you do know the area and you can use Heron's formula for finding the length of the side of the equilateral triangle.\r<BR>\n" );
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document.write( "Heron's formula, which gives the area of a triangle as a function of the length of the sides is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+sqrt%28s%28s-a%29%28s-b%29%28s-c%29%29 ALIGN=MIDDLE  ALT=\"A+=+sqrt%28s%28s-a%29%28s-b%29%28s-c%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Where:<BR>\n" );
document.write( "s = the semi-perimeter of the triangle.<BR>\n" );
document.write( "a, b, c, are the lengths of the sides of the triangle.\r<BR>\n" );
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document.write( "But, in an equilateral triangle, a = b = c and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s+=+%28a%2Bb%2Bc%29%2F2 ALIGN=MIDDLE  ALT=\"s+=+%28a%2Bb%2Bc%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3a%2F2 ALIGN=MIDDLE  ALT=\"3a%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "So, let's find the length (a) of one side of the triangle using Heron's formula <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+sqrt%28s%28s-a%29%28s-b%29%28s-c%29%29 ALIGN=MIDDLE  ALT=\"A+=+sqrt%28s%28s-a%29%28s-b%29%28s-c%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >and the known area of the triangle<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=300sqrt%283%29 ALIGN=MIDDLE  ALT=\"300sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Rewrite Heron's formula for the case of an equilateral triangle where <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s+=+3a%2F2 ALIGN=MIDDLE  ALT=\"s+=+3a%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+sqrt%28%283a%2F2%29%28%283a%2F2%29-a%29%5E3%29 ALIGN=MIDDLE  ALT=\"A+=+sqrt%28%283a%2F2%29%28%283a%2F2%29-a%29%5E3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > Simplify.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+sqrt%28%283a%2F2%29%28a%2F2%29%5E3%29 ALIGN=MIDDLE  ALT=\"A+=+sqrt%28%283a%2F2%29%28a%2F2%29%5E3%29\"   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+=+sqrt%28%283a%2F2%29%28a%5E3%2F8%29%29 ALIGN=MIDDLE  ALT=\"A+=+sqrt%28%283a%2F2%29%28a%5E3%2F8%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=A+=+sqrt%283a%5E4%2F16%29 ALIGN=MIDDLE  ALT=\"A+=+sqrt%283a%5E4%2F16%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+%28a%5E2%2F4%29sqrt%283%29 ALIGN=MIDDLE  ALT=\"A+=+%28a%5E2%2F4%29sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > But the area of the triangle is given as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+300sqrt%283%29 ALIGN=MIDDLE  ALT=\"A+=+300sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=300sqrt%283%29+=+%28a%5E2%2F4%29sqrt%283%29 ALIGN=MIDDLE  ALT=\"300sqrt%283%29+=+%28a%5E2%2F4%29sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > Simplifying, we get:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=300+=+a%5E2%2F4 ALIGN=MIDDLE  ALT=\"300+=+a%5E2%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" > Solving for a, the length of the side of the triangle:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2+=+4%28300%29 ALIGN=MIDDLE  ALT=\"a%5E2+=+4%28300%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2+=+1200 ALIGN=MIDDLE  ALT=\"a%5E2+=+1200\"   DISABLED_style= \"max-width:90%; height:auto;\" > Taking the square root of both sides.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a+=+20sqrt%283%29 ALIGN=MIDDLE  ALT=\"a+=+20sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Now we can substitute this for s in the formula for the apothem.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r+=+%281%2F2%2920sqrt%283%29cot%28180%2F3%29 ALIGN=MIDDLE  ALT=\"r+=+%281%2F2%2920sqrt%283%29cot%28180%2F3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > Simplifying.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r+=+10sqrt%283%29cot%2860%29 ALIGN=MIDDLE  ALT=\"r+=+10sqrt%283%29cot%2860%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r+=+10sqrt%283%29%280.577%29 ALIGN=MIDDLE  ALT=\"r+=+10sqrt%283%29%280.577%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r+=+5.77sqrt%283%29 ALIGN=MIDDLE  ALT=\"r+=+5.77sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > This is the length of the apothem. </SPAN>\n" );
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