document.write( "Question 805623: Two squares, each of side length 1-sqrt( 2 ) unit, overlap. The overlapping region is a regular octagon. What is the area (in square units) of the octagon? \n" ); document.write( "
Algebra.Com's Answer #485387 by AnlytcPhil(1806)\"\" \"About 
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Question 805623
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document.write( "The side of a square cannot be 1-√2 because that's a negative number, 1-1.414 = -0.414,\r\n" );
document.write( "so I will assume you meant √2-1 instead, which is positive.\r\n" );
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document.write( "All 8 little triangles around the points of the \"star\" are congruent,\r\n" );
document.write( "so if we add the areas of the two squares we will have twice the\r\n" );
document.write( "area of the octagon + 8 times the area of one of those triangles. So\r\n" );
document.write( "we will seek to find the area of one of those little triangles.\r\n" );
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document.write( "The area of one of those squares is (√2-1)² = 2-2√2+1 = 3-2√2\r\n" );
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document.write( "So if we add both of them we get 2(3-2√2)\r\n" );
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document.write( "Let the area of the octagon be x, and let the area of each triangle be t.\r\n" );
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document.write( "Then \r\n" );
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document.write( "2(3-2√2) = 2x-8t\r\n" );
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document.write( "Divide through by 2:\r\n" );
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document.write( "3-2√2 = x-4t\r\n" );
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document.write( "Solve for x:\r\n" );
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document.write( "(1)    x = 3-2√2+4t\r\n" );
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document.write( "So we will now seek to find t, the area of each of those little triangles.\r\n" );
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document.write( "Draw a diagonal AB of the red square cutting the green square\r\n" );
document.write( "at D and E.  We label C,F, and G \r\n" );
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document.write( "The area of the octagon is the area of a square minus 4 times\r\n" );
document.write( "the area of the ΔAFG\r\n" );
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document.write( "We will calculate diagonal AB by using the Pythagorean theorem on\r\n" );
document.write( "isosceles right triangle ΔABC:\r\n" );
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document.write( "AB² = AC² + BC²\r\n" );
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document.write( "AB² = (√2-1)² + (√2-1)²\r\n" );
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document.write( "AB² = 2(√2-1)² \r\n" );
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document.write( "AB  = √2(√2-1)\r\n" );
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document.write( "AB = 2-√2\r\n" );
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document.write( "DE = √2-1, the same\r\n" );
document.write( "as a side of a square.\r\n" );
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document.write( "AD+DE+BE = AB and since AD=BE\r\n" );
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document.write( "2AD+DE = AB\r\n" );
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document.write( "2AD = AB-DE\r\n" );
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document.write( "2AD = (2-√2)-(√2-1)\r\n" );
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document.write( "2AD = 2-√2-√2+1\r\n" );
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document.write( "2AD = 3-2√2\r\n" );
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document.write( "AD = \"%283-2sqrt%282%29%29%2F2\"\r\n" );
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document.write( "ΔADF is an isosceles right triangle so DF=AD\r\n" );
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document.write( "Area of ΔADF = \"expr%281%2F2%29AD%2ADF%29\" = \"expr%281%2F2%29AD%2AAD%29\" = \"expr%281%2F2%29AD%5E2\" = \"expr%281%2F2%29%28%283-2sqrt%282%29%29%2F2%29%5E2\" = \"expr%281%2F2%29%28%289-12sqrt%282%29%2B4%2A2%29%2F4%29\" = \"expr%281%2F2%29%28%289-12sqrt%282%29%2B8%29%2F4%29\" = \"expr%281%2F2%29%28%2817-12sqrt%282%29%29%2F4%29\" \r\n" );
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document.write( "The area of ΔAFG = t = twice the area of ΔADF = \"%28%2817-12sqrt%282%29%29%2F4%29\"\r\n" );
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document.write( "Now we go back to equation (1) back near the top:\r\n" );
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document.write( "(1)    x = 3-2√2+4t\r\n" );
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document.write( "       x = 3-2√2+4\"%28%2817-12sqrt%282%29%29%2F4%29\"\r\n" );
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document.write( "Cancel the 4's\r\n" );
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document.write( "       x = 3-2√2-17+12√2\r\n" );
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document.write( "       x = -14+10√2\r\n" );
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document.write( "       x = 10√2-14\r\n" );
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document.write( "Answer: Area = 10√2-14  \r\n" );
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document.write( "Edwin

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