document.write( "Question 1188279: https://imgur.com/c7zQ0QS In the diagram, AD is the median of triangle ABC. Find the area of triangle ABC. \n" ); document.write( "

Algebra.Com's Answer #819423 by Edwin McCravy(20054) $\"\"$ $\"About$

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document.write( "We drew in the altitude AE in green to the right.  The green line is the\r\n" );
document.write( "altitude of both triangles ABD and ADC.\r\n" );
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document.write( "Both triangles, ABD and ADC, have equal bases because AD is a median of triangle\r\n" );
document.write( "ABC.  So BD = DC.  Let x = BD = DC, then BC = 2x\r\n" );
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document.write( "By the law of cosines,\r\n" );
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document.write( "x=0;  x-22cos(B)=0\r\n" );
document.write( "         x = 22cos(B)\r\n" );
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document.write( "Ignore x=0, so x = 22cos(B)\r\n" );
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document.write( "Also by the law of cosines,\r\n" );
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document.write( "Divide through by 4\r\n" );
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document.write( "Substitute 22cos(B) for x\r\n" );
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document.write( "Angle B cannot be obtuse. It must be acute, so we ignore \r\n" );
document.write( "the negative sign\r\n" );
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document.write( "Using the SAS formula for the area of triangle ABC\r\n" );
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document.write( "We find sin(B) from \r\n" );
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document.write( "We find BC = 2x = (2)( 22cos(B) ) = 44cos(B) =  =  \r\n" );
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document.write( "Edwin

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