document.write( "Question 1066322: the median of a right triangle that are drawn from vertices of the acute angles have lengths of 2square root of 13 and square root of 73. what is the length of the hypotenuse \n" ); document.write( "
Algebra.Com's Answer #681502 by ikleyn(52905)\"\" \"About 
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document.write( "1.  Make a sketch.\r\n" );
document.write( "    Let ABC be your right-angled triangle with the right angle vertex A.\r\n" );
document.write( "    Let BD and CE be these two medians (|CE| = 2*sqrt(13) and |BD| = sqrt(73), for certainty).\r\n" );
document.write( "    Let x and y be the lengths of the legs AB and AC respectively.\r\n" );
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document.write( "2.  From the right triangle ACE and ABD you have\r\n" );
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document.write( "    \"abs%28AC%29%5E2+%2B+abs%28AE%29%5E2\" = \"abs%28CE%29%5E2\"   (1)\r\n" );
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document.write( "    \"abs%28AB%29%5E2+%2B+abs%28AD%29%5E2\" = \"abs%28BD%29%5E2\"   (2)\r\n" );
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document.write( "or\r\n" );
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document.write( "    \"%28x%2F2%29%5E2+%2B+y%5E2\" = \"%282%2Asqrt%2813%29%29%5E2\"     (3)\r\n" );
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document.write( "    \"x%5E2+%2B+%28y%2F2%29%5E2\" = \"%28sqrt%2873%29%29%5E2\",      (4)\r\n" );
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document.write( "By adding equations (3) and (4), you get\r\n" );
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document.write( "    \"%285%2F4%29%2Ax%5E2+%2B+%285%2F4%29%2Ay%5E2\" = 4*13 + 73 = 125,   or\r\n" );
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document.write( "    \"x%5E2+%2B+y%5E2\" = 100.\r\n" );
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document.write( "3.  Thus \"x%5E2+%2B+y%5E2\" = 100.\r\n" );
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document.write( "    It is the length of the hypotenuse squared, \"abs%28BC%29%5E2\".\r\n" );
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document.write( "    Hence, the length of the hypotenuse \"abs%28BC%29\" = \"sqrt%28100%29\" = 10 cm.\r\n" );
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