document.write( "Question 1210520: A regular hexagon has a perimeter of p (in length units) and an area of A (in square units). If A=3/2 then find the side length of the hexagon. \n" ); document.write( "

Algebra.Com's Answer #853456 by ikleyn(53646) $\"\"$ $\"About$

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\n" ); document.write( "A regular hexagon has a perimeter of p (in length units) and an area of A (in square units).
\n" ); document.write( "If A=3/2 then find the side length of the hexagon.
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\n" ); document.write( "\n" ); document.write( "As I read this post by @CPhill, it saddens me to see how clumsily the problem is formulated
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\n" ); document.write( "\n" ); document.write( "The part \" hexagon has a perimeter of p \" is totally irrelevant to the problem and should be omitted.\r
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\n" ); document.write( "\n" ); document.write( "The normal formulation to this problem is as follow\r
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document.write( "    A regular hexagon has the area of 3/2 square units. \r\n" );
document.write( "    Find the side length of the hexagon.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Below is a normal mathematical solution in a form as it should be.\r
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document.write( "Let 'a' be the side length of the regular hexagon.\r\n" );
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document.write( "This hexagon is the union of 6 equilateral triangles with the side length of 'a'.\r\n" );
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document.write( "So, the area of each such a triangle is   =  of the square unit.\r\n" );
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document.write( "The area of each such a triangle is  .\r\n" );
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document.write( "So, for 'a' we have this equation\r\n" );
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document.write( "     = ,\r\n" );
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document.write( "which implies\r\n" );
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document.write( "     = ,\r\n" );
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document.write( "    a =  = 3^(-1/4) = 0.7598  (rounded).\r\n" );
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document.write( "ANSWER.  The side of the regular hexagon is  a =  = 3^(-1/4) = 0.7598  (rounded).\r\n" );
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document.write( "CHECK.   = 1.49986  for the full area, which is a good approximation.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved.\r
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\n" ); document.write( "\n" ); document.write( "Compare this solution with the mess of words in the post by @CPhill.\r
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