document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=245477>Question 245477</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the area of a regular hexagon with the given measurement 48inch perimeter A=sq.in. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #179303 by </B><A HREF=/tutors/aboutme.mpl?userid=JimboP1977><B>JimboP1977(311)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JimboP1977><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=JimboP1977><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=179303\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Ok the perimeter is 48 inches. It is a regular hexagon (six sides) so this must mean each of the sides is equal in length. So 48/6 = 8 inches per side.\r<BR>\n" );
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document.write( "We know from the formula <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%2A%28n-2%29%2A360%29%2Fn ALIGN=MIDDLE  ALT=\"%281%2F2%2A%28n-2%29%2A360%29%2Fn\"   DISABLED_style= \"max-width:90%; height:auto;\" > (where n is the number of sides) that each interior angle must equal 120 degrees. \r<BR>\n" );
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document.write( "Best thing here is to draw a hexagon. If you would like to see the one I drew, respond back so I email it to you. It will look like a square with two triangle on each end.\r<BR>\n" );
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document.write( "Now if we take say the left side of the hexagon, with the triangle shape, we know that the angle must be 120 degrees. If we draw a line from the left hand point down to the bottom we form a triangle.\r<BR>\n" );
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document.write( "We know that the angle top angle must be (120-90) = 30 degrees. So to find the base of length of the triangle we can use the Sine rule which is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%2FSinA+=+b%2FSinB ALIGN=MIDDLE  ALT=\"a%2FSinA+=+b%2FSinB\"   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%2FSin+120+=+8%2FSin30 ALIGN=MIDDLE  ALT=\"a%2FSin+120+=+8%2FSin30\"   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=+Sin120+%2A+8%2FSin30 ALIGN=MIDDLE  ALT=\"a=+Sin120+%2A+8%2FSin30\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "If we half this number we get one side of a right angled triangle with the hypotenuse being 8 inches. The third side can be obtained using <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r%5E2+=+x%5E2%2By%5E2 ALIGN=MIDDLE  ALT=\"r%5E2+=+x%5E2%2By%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "So <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=8%5E2-48+=+x%5E2 ALIGN=MIDDLE  ALT=\"8%5E2-48+=+x%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=x+=+4 ALIGN=MIDDLE  ALT=\"x+=+4\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "So we have a little right angled triangle with sides 4, sqrt48, and 8. The area of this triangle is given by 1/2 base times the height.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F2%2A4%2Asqrt58=sqrt192 ALIGN=MIDDLE  ALT=\"1%2F2%2A4%2Asqrt58=sqrt192\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "So we have the area of one of the small triangles. So the four triangles equals <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4%2Asqrt192 ALIGN=MIDDLE  ALT=\"4%2Asqrt192\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "The remaining area is the square in the hexagon which is simply <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=8%2Asqrt192 ALIGN=MIDDLE  ALT=\"8%2Asqrt192\"   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=Total+area+=+4%2Asqrt192%2B8%2Asqrt192=+12%2Asqrt+192+=+166.277 ALIGN=MIDDLE  ALT=\"Total+area+=+4%2Asqrt192%2B8%2Asqrt192=+12%2Asqrt+192+=+166.277\"   DISABLED_style= \"max-width:90%; height:auto;\" > inches^2 (to 3 decimal places). \r<BR>\n" );
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document.write( "This is the long method, but there is a simple formula which is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+L%5E2+%283%2Asqrt3%2F2%29 ALIGN=MIDDLE  ALT=\"A+=+L%5E2+%283%2Asqrt3%2F2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "where L equals the length of one of the sides of a regular hexagon.\r<BR>\n" );
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document.write( "It will make loads more sense with a diagram! Trust me! :-)\r<BR>\n" );
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