document.write( "Question 196955: make a regular hexagon of sides as you wish and divide them into six equilateral triangles and find its area ? \n" ); document.write( "
Algebra.Com's Answer #147729 by jojo14344(1513)\"\" \"About 
You can put this solution on YOUR website!

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\n" ); document.write( "We draw the hexagon:
\n" ); document.write( " --->
\n" ); document.write( "The interior angles in the center measure 60 deg. => \"360%5Eo%2F6=60%5Eo\"\r
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\n" ); document.write( "\n" ); document.write( "Being Equilateral, all interior angles are 60 deg.\r
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\n" ); document.write( "\n" ); document.write( "Let us isolate 1 Triangle:
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\n" ); document.write( "\n" ); document.write( "As you can see, we make all sides equal to 2 units.\r
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\n" ); document.write( "\n" ); document.write( "Woking Eqn, \"A%5BT%5D=%281%2F2%29%28b%29%28h%29\",where\"system%28b=%281%2F2%29%282%29=red%281%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "Solving \"h\" via Pyth Theorem:
\n" ); document.write( "\"2%5E2=b%5E2%2Bh%5E2\"
\n" ); document.write( "\"h%5E2=2%5E2-b%5E2=2%5E2-1%5E1=4-1\"
\n" ); document.write( "\"red%28h=sqrt%283%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "Going Back Working Eqn:
\n" ); document.write( "\"A%5BT%5D=%281%2F2%29%281%29%28sqrt%283%29%29\"
\n" ); document.write( "\"A%5BT%5D=%281%2F2%29%28sqrt%283%29%29\", sq.units\r
\n" ); document.write( "\n" ); document.write( "And there are 2 Right Triangles for 1 Equilateral Triangle. Therefore,
\n" ); document.write( "\"2%28A%5BT%5D%29=cross%282%29%2A%281%2Fcross%282%29%29%28sqrt%283%29%29\"
\n" ); document.write( "\"red%28sqrt%283%29%29\" ----> There are 6 Equilateral Triangles:\r
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\n" ); document.write( "\n" ); document.write( "\"highlight%28red%28%286%29%28sqrt%283%29%29%29%29\" sq.units (Answer)\r
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\n" ); document.write( "\n" ); document.write( "We can also do it by Trigo function since we know the interior angles.
\n" ); document.write( " ---> \"sin60%5Eo=opp%2Fhyp=h%2F2\"
\n" ); document.write( "\"h=%28sin60%5Eo%29%282%29\"
\n" ); document.write( "\"red%28h=sqrt%283%29%29\"
\n" ); document.write( "Or, \"cos60%5Eo=adj%2Fhyp=b%2F2\"
\n" ); document.write( "\"b=%28cos60%5Eo%29%282%29=0.50%282%29\"
\n" ); document.write( "\"red%28b=1%29\"\r
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\n" ); document.write( "\n" ); document.write( "Just do the same we did for Area of Triangle above.\r
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\n" ); document.write( "\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo
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