document.write( "Question 1210156: An envelope is pictured below. Solve for the area of the shaded region.
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Algebra.Com's Answer #851413 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "An envelope is pictured below. Solve for the area of the regions\r\n" );
document.write( "marked XXX.\r\n" );
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document.write( "DE is half of 12 or 6. So we find AD with the Pythagorean theorem.\r\n" );
document.write( "(Or you may recognize it as a 6-8-10 right triangle, a 3-4-5 right triangle\r\n" );
document.write( "with all sides doubled):\r\n" );
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document.write( "\"AD+=+sqrt%28AE%5E2-DE%5E2%29+=sqrt%2810%5E2-6%5E2%29+=+sqrt%28100-36%29+=+sqrt%2864%29+=+8\" \r\n" );
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document.write( "So tha area of right triangle ADE is \"expr%281%2F2%29%2Abase%2Aheight\"\"%22%22=%22%22\"\r\n" );
document.write( "\"expr%281%2F2%29%2ADE%2AAD\"\"%22%22=%22%22\"\"expr%281%2F2%29%2A6%2A8%29\"\"%22%22=%22%22\"\"24\"in2.\r\n" );
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document.write( "Triangle BCE is congruent to triangle ADE so it also has area \"24\"in2. \r\n" );
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document.write( "Total shaded area (or area marked XXX) = (2)(24) = 48 in2.\r\n" );
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document.write( "Edwin
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