document.write( "Question 1205024: In a right angle triangle ABC side AC is 4cm shorter than the hypotenuse and side BC is also 4cm shorter than the hypotenuse. Find the dimensions of the triangle \n" ); document.write( "
Algebra.Com's Answer #841607 by ikleyn(52782)\"\" \"About 
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\n" ); document.write( "In a right angle triangle ABC side AC is 4cm shorter than the hypotenuse and side BC
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document.write( "x^2 + x^2 = (x+4)^2    <<<---=== Pythagorea equation;  x is the common length of any of the two its legs.\r\n" );
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document.write( "2x^2 = x^2 + 8x + 16\r\n" );
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document.write( "x^2 - 8x - 16 = 0\r\n" );
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document.write( "x = \"%288+%2B-+sqrt%28%28-8%29%5E2+-+4%2A1%2A%28-16%29%29%29%2F2\" = \"%288+%2B-+sqrt%28128%29%29%2F2\" = \"%288+%2B-+8%2Asqrt%282%29%29%2F2\" = \"4+%2B-+4%2Asqrt%282%29\".\r\n" );
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document.write( "Only positive root is the solution to the problem.\r\n" );
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document.write( "ANSWER.  The triangle is isosceles right angled triangle.\r\n" );
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document.write( "         Its legs are \"4%2Asqrt%282%29%2B4\"  cm long.  Its hypotenuse is  \"%284%2Asqrt%282%29%2B4%29%2Asqrt%282%29\" = \"8%2B4%2Asqrt%282%29\" cm long.\r\n" );
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Warning to a reader

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\n" ); document.write( "\n" ); document.write( "        The solution by @mananth, giving two possible solutions for the hypotenuse, is NOT precisely correct.\r
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\n" ); document.write( "\n" ); document.write( "        Only one (greater) of the two values of the hypotenuse length is admittable; \r
\n" ); document.write( "\n" ); document.write( "        the second (lesser) value is NOT admittable, since it leads to negative value of the leg length.\r
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