document.write( "Question 1007115: A regular hexagon is cut from a square of side 6 inches, a.) what is the
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Algebra.Com's Answer #623205 by Edwin McCravy(20055)\"\" \"About 
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A regular hexagon is cut from a square of side 6 inches, a.) what is
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document.write( "The red line below is the apothem.\r\n" );
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document.write( "In the figure \r\n" );
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document.write( "AB is given 6 inches.  It is divided into 4 parts,\r\n" );
document.write( "AD = DC = CE = EB = 6/4 = 3/2 inches.\r\n" );
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document.write( "Triangle DOE is equilateral, so DO = DE = DC+CE = 3/2+3/2 = 3 in.\r\n" );
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document.write( "Using the Pythagorean theorem on right triangle DOC,\r\n" );
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document.write( "     DOē = DCē + OCē\r\n" );
document.write( "      3ē = (3/2)ē + OCē\r\n" );
document.write( "       9 = 9/4 + OCē\r\n" );
document.write( "   9-9/4 = OCē\r\n" );
document.write( "36/4-9/4 = OCē\r\n" );
document.write( "    27/4 = OCē\r\n" );
document.write( "   \"sqrt%2827%2F4%29\" = OC\r\n" );
document.write( "   \"sqrt%2827%29%2F2\" = OC\r\n" );
document.write( "   \"sqrt%289%2A3%29%2F2\" = OC\r\n" );
document.write( "   \"3sqrt%283%29%2F2\" = OC = the length of the apothem in inches.\r\n" );
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document.write( "For the area of the hexagon, first we find the area of the \r\n" );
document.write( "equilateral triangle DOE:\r\n" );
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document.write( "\"A\"\"%22%22=%22%22\"\"expr%281%2F2%29DE%2AOC\"\"%22%22=%22%22\"\"expr%281%2F2%29%283%29%283sqrt%283%29%2F2%29\"\"%22%22=%22%22\"\"9sqrt%283%29%2F4\"inē\r\n" );
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document.write( "And as we see below, the hexagon is made up of 6 congruent equilateral\r\n" );
document.write( "triangles all congruent to DOE, its area is 6 times the area of triangle\r\n" );
document.write( "DOE.\r\n" );
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document.write( "So the area of the hexagon is \"6%2Aexpr%289sqrt%283%29%2F4%29\" or \"27sqrt%283%29%2F2\"inē.\r\n" );
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document.write( "Edwin
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