document.write( "Question 1103134: Find the perimeter of a right triangle whose hypotenuse is 2 and whose area is 2. Express your answer in simplest radical form. \n" ); document.write( "

Algebra.Com's Answer #717813 by ikleyn(52772) $\"\"$ $\"About$

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\n" ); document.write( "The condition describes an IMPOSSIBLE situation,\r
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\n" ); document.write( "\n" ); document.write( "and below I give the proof to this fact.\r
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document.write( "Let the legs be x and y.\r\n" );
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document.write( "Then from the condition you have THESE two equations:\r\n" );
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document.write( "x^2 + y^2 = 2^2\r\n" );
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document.write( " = 2,\r\n" );
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document.write( "or, EQUIVALENTLY,\r\n" );
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document.write( "x^2 + y^2 = 4,   (1)\r\n" );
document.write( "xy = 4.          (2)\r\n" );
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document.write( "Then  (x-y)^2 = x^2 -2xy + y^2 = (x^2+y^2) - 2xy = 4 - 2*4 = -4.\r\n" );
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document.write( "But the left side (x-y)^2 is a NON-NEGATIVE value/quantity,  while the right side is NEGATIVE.\r\n" );
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document.write( "Contradiction.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Conclusion: The condition is self-contradictory.\r
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