document.write( "Question 1155796: In triangle ABC, point P is on side BC such that PA = 13, PB = 14, PC = 4, and the circumcircle of triangles APB and APC have the same radius. Find the area of triangle ABC. \n" ); document.write( "
Algebra.Com's Answer #778481 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
\r
\n" ); document.write( "\n" ); document.write( "In triangle \"ABC\", point \"P\" is on side \"BC\" such that\r
\n" ); document.write( "\n" ); document.write( "\"+PA+=+13\", \r
\n" ); document.write( "\n" ); document.write( "\"PB+=+14\", \r
\n" ); document.write( "\n" ); document.write( "\"PC+=+4\", \r
\n" ); document.write( "\n" ); document.write( "and the circumcircle of triangles \"APB\" and \"APC\" have the same radius. Find the area of triangle \"ABC\".\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"triangle-ABC.png\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Note that \"AP\" is a \"chord\" of both circles. Since both circles have the \"same\" \"radius\", chord \"AP\" must subtend the \"same\" angle, i.e. < \"ABP\"= < \"ACP\". \r
\n" ); document.write( "\n" ); document.write( "Thus, triangle \"ABC\" is \"isosceles\" with \"AB=AC\".\r
\n" ); document.write( "\n" ); document.write( "Let\"+M\" be the \"midpoint\" of \"BC\". We see that \r
\n" ); document.write( "\n" ); document.write( "\"BC=BP%2B+PC=14%2B4=18\"\r
\n" ); document.write( "\n" ); document.write( "so\"+BM=CM=BC%2F2=9\" \r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "Then \"PM=MC-PC=9-4=5\". \r
\n" ); document.write( "\n" ); document.write( "Since \"AM\" is also an \"altitude\", we can apply Pythagoras theorem to right triangle \"AMP\" to get \"AM\":\r
\n" ); document.write( "\n" ); document.write( "\"AM=sqrt%2813%5E2-5%5E2%29=sqrt%28169-25%29=sqrt%28144%29=12\"\r
\n" ); document.write( "\n" ); document.write( "Hence, the area of triangle \"ABC\" is:\r
\n" ); document.write( "\n" ); document.write( "\"area=%281%2F2%29AM%2ABC\"\r
\n" ); document.write( "\n" ); document.write( "\"area=%281%2F2%2912%2A18\"\r
\n" ); document.write( "\n" ); document.write( "\"area=6%2A18\"\r
\n" ); document.write( "\n" ); document.write( "\"area=108\"\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );