document.write( "Question 1009441: An equilateral triangle, each side of which measures 12, is
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Algebra.Com's Answer #624931 by AnlytcPhil(1806)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "We will need: \r\n" );
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document.write( "1. the height of the triangle so that we can find its area, and \r\n" );
document.write( "2. the radius of the circle so that we can find its area.\r\n" );
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document.write( "Then we can subtract to find the other part of the circle. \r\n" );
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document.write( "First we find the height of the triangle. \r\n" );
document.write( "We draw the altitude h.\r\n" );
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document.write( "By the Pythagorean theorem,\r\n" );
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document.write( "\"6%5E2%2Bh%5E2=12%5E2\"\r\n" );
document.write( "\"36%2Bh%5E2=144\"\r\n" );
document.write( "\"h%5E2=108\"\r\n" );
document.write( "\"h=sqrt%28108%29\"\r\n" );
document.write( "\"h=sqrt%2836%2A3%29\"\r\n" );
document.write( "\"h=6sqrt%283%29\"\r\n" );
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document.write( "The area of the equilateral triangle\r\n" );
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document.write( "\"A=expr%281%2F2%29base%2Aheight\"\r\n" );
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document.write( "\"A=expr%281%2F2%29%2812%29%286sqrt%283%29%29\"\r\n" );
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document.write( "\"A=36sqrt%283%29\"\r\n" );
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document.write( "Next we draw the radii of the circle to the vertices\r\n" );
document.write( "of the triangle.  Notice that this divides the height of\r\n" );
document.write( "the triangle into the radius r and the apothem a which is\r\n" );
document.write( "the height of the lower triangle:\r\n" );
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document.write( "We know that \"r%2Ba+=+h\", so \"a=h-r\"\r\n" );
document.write( "We know that \"h=6sqrt%283%29\"\r\n" );
document.write( "So \"a=6sqrt%283%29-r\"\r\n" );
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document.write( "By the Pythagorean theorem, \r\n" );
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document.write( "\"6%5E2%2Ba%5E2=r%5E2\"\r\n" );
document.write( "\"36%2B%286sqrt%283%29-r%29%5E2=r%5E2\"\r\n" );
document.write( "\"36%2B%2836%2A3-12sqrt%283%29%2Ar%2Br%5E2%29=r%5E2\"\r\n" );
document.write( "\"36%2B108-12sqrt%283%29%2Ar%2Br%5E2=r%5E2\"\r\n" );
document.write( "\"144-12sqrt%283%29%2Ar%2Br%5E2=r%5E2\"\r\n" );
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document.write( "Subtract \"r%5E2\" from both sides:\r\n" );
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document.write( "\"144-12sqrt%283%29%2Ar=0\"\r\n" );
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document.write( "Solve for r:\r\n" );
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document.write( "\"144=12sqrt%283%29%2Ar\"\r\n" );
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document.write( "\"144%2F%2812sqrt%283%29%29=r\"\r\n" );
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document.write( "\"12%2Fsqrt%283%29=r\"\r\n" );
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document.write( "The area of the circle is given by\r\n" );
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document.write( "\"A=pi%2Ar%5E2\"\r\n" );
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document.write( "\"A=pi%2A%2812%2Fsqrt%283%29%29%5E2\"\r\n" );
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document.write( "\"A=pi%2A%28144%2F3%29\"\r\n" );
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document.write( "\"A=pi%2A48\"\r\n" );
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document.write( "\"A=48pi\"\r\n" );
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document.write( "So the area of the other part of the circle besides the\r\n" );
document.write( "equilateral triangle is found by subtracting the area of \r\n" );
document.write( "the equilateral triangle from the area of the circle:\r\n" );
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document.write( "Answer: \"48pi-36sqrt%283%29\", or about 88.4\r\n" );
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document.write( "If we like, we can factor out 12\r\n" );
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document.write( "Answer:  \"12%284pi-3sqrt%283%29%29\" but that isn't necessary.\r\n" );
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document.write( "Edwin

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