document.write( "Question 667067: the lengths of three medians of a triangle are 9,12 & 15cm. The area of the triangle is \n" ); document.write( "
Algebra.Com's Answer #414799 by MathLover1(20850)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "A median in a triangle is a line segment drawn from an angle to the midpoint of the opposite side. The medians of a triangle all intersect in one point. This point divides each median in the ratio \"1%3A+2\".\r
\n" ); document.write( "\n" ); document.write( "let \"ABC\" be a triangle \r
\n" ); document.write( "\n" ); document.write( "if CM[1]=15...it is equal to the hight of triangle \"h\", \"AM%5B2%5D=12\", and \"BM%5B3%5D=9\"\r
\n" ); document.write( "\n" ); document.write( "if intersection point of all three medians is \"T\", and \"AM%5B2%5D\" is median then \r
\n" ); document.write( "\n" ); document.write( "\"AT%3ATM%5B2%5D=2%3A1\" .......=> \"AT=%2812%2F3%292=8\" and \"TM%5B2%5D=4\"\r
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\n" ); document.write( "\n" ); document.write( "and \"CT%3ATM%5B1%5D=2%3A1\" .......=>\"CT%3ATM%5B1%5D=%2815%2F3%292=10\" and\"TM%5B1%5D=%2815%2F3%291=5\"\r
\n" ); document.write( "\n" ); document.write( "now we can find \"AM%5B1%5D\" which is half of base \"AB\"\r
\n" ); document.write( "\n" ); document.write( "\"AM%5B1%5D%5E2=%28AT%29%5E2-%28TM%5B1%5D%29%5E2\"\r
\n" ); document.write( "\n" ); document.write( "\"AM%5B1%5D%5E2=8%5E2-5%5E2\"\r
\n" ); document.write( "\n" ); document.write( "\"AM%5B1%5D%5E2=64-25\"\r
\n" ); document.write( "\n" ); document.write( "\"AM%5B1%5D%5E2=39\"\r
\n" ); document.write( "\n" ); document.write( "\"AM%5B1%5D=sqrt%2839%29\"\r
\n" ); document.write( "\n" ); document.write( "\"AM%5B1%5D=6.25\"...........so, the base \"AB=6.25%2A2=12.5\"\r
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\n" ); document.write( "\n" ); document.write( "then the area will be:\r
\n" ); document.write( "\n" ); document.write( "\"A=%281%2F2%2912.5%2A15\"\r
\n" ); document.write( "\n" ); document.write( "\"A=6.2.5%2A15\"\r
\n" ); document.write( "\n" ); document.write( "\"A=93.75\"\r
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