document.write( "Question 1199182: Find the area of region ABCD in cm^2 if the radius of the circle is 2 cm and both AB and CD are perpendicular to EF.
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Algebra.Com's Answer #832915 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Draw the two diameters AC and BD (in red) \r\n" );
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document.write( "OA is a radius so OA = 2. And by the marks along the horizontal diameter,\r\n" );
document.write( "OE is half a radius or half of 2 which is 1.  So OE = 1\r\n" );
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document.write( "So triangle AOE is a 30-60-90 right triangle, so AE=\"sqrt%283%29\"\r\n" );
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document.write( "Triangle AOE has area = \"expr%281%2F2%29%2Abase%2Aheight=expr%281%2F2%29%281%29sqrt%283%29\"\"%22%22=%22%22\"\"sqrt%283%29%2F2%29\"\r\n" );
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document.write( "The four triangles AOE, BOE, COE, DOE are congruent. \r\n" );
document.write( "So the four of them have area \"4%28expr%28sqrt%283%29%2F2%29%29=2sqrt%283%29\" \r\n" );
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document.write( "Next we find the area of the sector AOD.\r\n" );
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document.write( "Angle AOD is 60o because angles AOE and DOE are both 60o,\r\n" );
document.write( "so angle AOD = 180o-60o-60o = 60o.\r\n" );
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document.write( "The area of a sector is \"expr%281%2F2%29%28theta%2F360%29pi%2Aradius%5E2\" \r\n" );
document.write( "and substituting the values, the area of sector AOD is\r\n" );
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document.write( "\"%281%2F2%29%2860%2F360%29%28pi%29%282%5E2%29\"\"%22%22=%22%22\"\"pi%2F3\"\r\n" );
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document.write( "The sector BOC is congruent to the sector AOD\r\n" );
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document.write( "Adding the 4 triangles and the two sectors,\r\n" );
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document.write( "Area of ABCD = \"%282sqrt%283%29%2Bpi%2F3%29\"\"cm%5E2\"\r\n" );
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document.write( "Edwin
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