document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=193049>Question 193049</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what is the area of a regular hexagon whose sides are each 12 inches long.  Round to the nearest squre inch.  I drew the figure.  used this formula:  A=4 + 2, A=6. Or should it be 360  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #144897 by </B><A HREF=/tutors/aboutme.mpl?userid=RAY100><B>RAY100(1637)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RAY100><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=144897\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> good  start,,somewhat  complex  solution,,your  geometry  book  probably  has  a  good  ref\r<BR>\n" );
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document.write( "lets  start  with  the  sketch,,,regular  hexagon  has  6  equal  sides,12  for  this  problem.<BR>\n" );
document.write( "If  we  find  the  center  of  the  hexagon,  and  then  draw  lines  to  each  vertex,  we  have  6 identical  triangles.\r<BR>\n" );
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document.write( "Area  of  a  triangle  =1/2 base  * height<BR>\n" );
document.write( "ok,  but  we  only  know  base,  we  need  to  find  height\r<BR>\n" );
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document.write( "We  also  know  the  central  angle,,,,,remember  sum  of  angles  of  polygon  =(n-2)(180)<BR>\n" );
document.write( "in  our  case  (6-2)(180)=720,,,,,with  6  angles,,,720/6=120,,,(central  angle  of  triangle).\r<BR>\n" );
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document.write( "note  the  120  deg  on  one  triangle.<BR>\n" );
document.write( "also ,  please,  draw  an  altitude on  that triangle,  from  center  to  base.<BR>\n" );
document.write( "this  is  the  height  of  that  triangle.<BR>\n" );
document.write( "Note  the  altitude  bisects  the  120 angle  into  2  equal  60  deg  angles.<BR>\n" );
document.write( "we  are  almost  there,<BR>\n" );
document.write( "in  our  triangle  we  have  the  60  degree  angle  at  the  center  vertex,  90 deg  where  altitude  intersects  the  base,  therefore  the  other  vertex angle  is  30 degrees,,,(sum  of  triangle =180 degrees)\r<BR>\n" );
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document.write( "lets  mark  the  30  degree  vertex,  and  the  90 degree  angle.\r<BR>\n" );
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document.write( "we  now  have  a  30-60-90  degree  triangle,  with  one  side (opposite  the  60 angle) equal  to  6<BR>\n" );
document.write( "which  is  half  of  the  full  side  12.\r<BR>\n" );
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document.write( "we  know  from  typical  triangles  that  a  30-60-90  has  sides  of  1,2,sqrt3.<BR>\n" );
document.write( "in  our  case  we  need  height,h,  which  is  opposite  the  30 deg  angle.<BR>\n" );
document.write( "if  the  height  is  proportional  to (1),  hyp  is  (2),  and  the  base  side (sqrt3),,,but  base  side  is  6\r<BR>\n" );
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document.write( "sqrt3/6=1/h<BR>\n" );
document.write( "cross  multiply,,<BR>\n" );
document.write( "(sqrt3)(h)=6<BR>\n" );
document.write( "h=6/sqrt3=3.464\r<BR>\n" );
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document.write( "now,,,area  of  triangle  is  1/2 (base) (height) = 1/2 (12)(3.464)=20.78,,,,(full base=12)\r<BR>\n" );
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document.write( "and  remember  we  have  6  triangles  in  hexagon<BR>\n" );
document.write( "area  if  hexagon=6(20.78)=124.7=125\r<BR>\n" );
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document.write( "this  is  generalized  to  Area  of  any  polygon  =1/2 (perimeter)(apothem)<BR>\n" );
document.write( "in  our  case,,,per  =6*size  of  side,  apothem  is  height<BR>\n" );
document.write( "Area=1/2 *(12*6)*3.464=  125  ok\r<BR>\n" );
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document.write( "note,  apothem  is  denoted  (a),  and  is  usually  given,,just  remember  1/2 * per*apothem </SPAN>\n" );
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