document.write( "Question 1149735: In the diagram, AC is the median of triangle ABD. Find the area of the triangle ABD.
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Algebra.Com's Answer #771219 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Use the law of cosines on triangles ABC and ABD\r\n" );
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document.write( "\"cos%28B%29=%28a%5E2%2Bc%5E2-b%5E2%29%2F%282ac%29\"\r\n" );
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document.write( "Using it on ABC\r\n" );
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document.write( "\"cos%28B%29=%2817%5E2%2Bx%5E2-17%5E2%29%2F%282%2A17%2Ax%29\"\r\n" );
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document.write( "which simplifies to\r\n" );
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document.write( "\"cos%28B%29=x%2F34\"\r\n" );
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document.write( "Using it on ABD\r\n" );
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document.write( "\"cos%28B%29=%2817%5E2%2B%282x%29%5E2-27%5E2%29%2F%282%2A17%2A2x%29\"\r\n" );
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document.write( "which simplifies to\r\n" );
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document.write( "\"%28x%5E2-110%29%2F17x\"\r\n" );
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document.write( "Equating the two expressions for cos(B),\r\n" );
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document.write( "\"x%2F34=%28x%5E2-110%29%2F17x\"\r\n" );
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document.write( "Solve that and get\r\n" );
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document.write( "\"x=2sqrt%2855%29\"\r\n" );
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document.write( "The height of both triangles is the green line h = AE:\r\n" );
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document.write( "\"AE%5E2=AB%5E2-BE%5E2\"\r\n" );
document.write( "\"AE%5E2=17%5E2-%28sqrt%2855%29%2F4%29%5E2\"\r\n" );
document.write( "\"AE%5E2=234\"\r\n" );
document.write( "\"AE=3sqrt%2826%29\"\r\n" );
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document.write( "\"Area=expr%281%2F2%29base%2Aheight\"\r\n" );
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document.write( "\"Area=expr%281%2F2%294sqrt%2855%29%2A3sqrt%2826%29\"\r\n" );
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document.write( "\"Area=6sqrt%281430%29\"\r\n" );
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document.write( "approx. 226.8920448\r\n" );
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document.write( "Edwin
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