document.write( "Question 1181238: What is the difference in area between the inscribed
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Algebra.Com's Answer #811242 by MathLover1(20850)\"\" \"About 
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\r
\n" ); document.write( "\n" ); document.write( "we need a diagonals across 6 sides (\"d%5B6%5D\"), \"5\" sides (\"d%5B5%5D\"), \"4\" sides (\"d%5B4%5D\"), \"3\" sides(\"d%5B3%5D\") , and \"2\" sides (\"d%5B2%5D\")\r
\n" ); document.write( "\n" ); document.write( "\"d%5B6%5D=%28sqrt%286%29%2Bsqrt%282%29%29%2Aa\" ........given \"a=2cm\"
\n" ); document.write( "\"d%5B6%5D=%283.863703305156273%29%2A2cm\"
\n" ); document.write( "\"d%5B6%5D=7.727cm\"
\n" ); document.write( "\"d%5B5%5D+=+%28+2+%2B+sqrt%283%29+%29+%2A+2=7.464cm\"
\n" ); document.write( "\"d%5B4%5D+=+%28%28+3%2Asqrt%282%29+%2B+sqrt%286%29+%29+%2F+2%29+%2A+2=6.692cm\"
\n" ); document.write( "\"d%5B3%5D+=+%28+sqrt%283%29+%2B+1+%29+%2A+2=5.464cm\"
\n" ); document.write( "\"d%5B2+%5D=+%28%28+sqrt%286%29%2Bsqrt%282%29+%29+%2F+2%29%2A+2=+d%5B6%5D+%2F+2=3.864cm\"\r
\n" ); document.write( "\n" ); document.write( "then radius of circumcircle is:\r
\n" ); document.write( "\n" ); document.write( "\"r%5Bc%5D+=+d%5B6%5D+%2F+2+=+d%5B2%5D=3.864cm\"
\n" ); document.write( "\"r%5Bc%5D+=+3.864cm\"\r
\n" ); document.write( "\n" ); document.write( "\"A%5Bcc%5D=%283.864cm%29%5E2%2Api\" ...........where \"cc\" stands for circumcircle
\n" ); document.write( "\"A%5Bcc%5D=14.93%2Api%2Acm%5E2\"\r
\n" ); document.write( "\n" ); document.write( "radius of incircle is:
\n" ); document.write( "\"r%5Bi%5D=+d%5B5%5D+%2F+2=7.464%2F2=3.732cm\"\r
\n" ); document.write( "\n" ); document.write( "\"r%5Bi%5D=3.732cm\"\r
\n" ); document.write( "\n" ); document.write( "\"A%5Bic%5D=%283.732cm%29%5E2%2Api\"...........where \"ic\" stands for incircle
\n" ); document.write( "\"A%5Bic%5D=13.93%2Api%2Acm%5E2\"\r
\n" ); document.write( "\n" ); document.write( " the difference in area between the inscribed circle and the circumscribed circle in a 12-gon is\r
\n" ); document.write( "\n" ); document.write( "\"A%5Bcc%5D-A%5Bic%5D=14.93%2Api%2Acm%5E2-13.93%2Api%2Acm%5E2=pi\"\r
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