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Find the radius of an inscribed circle in a triangle with legs 12, 12, 8.
\n" ); document.write( "LET O BE ORIGIN AND ABC THE TRIANGLE SUCH THAT
\n" ); document.write( "1.AB=c=AC=b=12...
\n" ); document.write( "2.BC=8=a
\n" ); document.write( "3.MID POINT OF BC IS ORIGIN AND BC IS ALONG X AXIS.HENCE B IS (-4,0) AND C IS (4,0)
\n" ); document.write( "4.SINCE ABC IS AN ISOCELLES TRIANGLE AO IS PERPENDICULAR TO BC THAT IS THE ALTITUDE,MEDIAN AND INTERNAL ANGLE BISECTOR OF A.HENCE
\n" ); document.write( "AO^2+OB^2=AB^2
\n" ); document.write( "AO^2=12^2-4^2=144-16=128
\n" ); document.write( "AO=SQRT(128)=8*SQRT(2)
\n" ); document.write( "HENCE A IS (0,8SQRT(2))
\n" ); document.write( "INCENTRE ,I ...IS ON AO AND IS GIVEN BY THE FORMULA
\n" ); document.write( "(a*Xa+b*Xb+c*Xc)/(a+b+c).....AND (a*Ya+b*Yb+c*Yc)/(a+b+c)
\n" ); document.write( "(8*0+12*4+12*(-4))/((8+12+12)=0
\n" ); document.write( "AND (8*8SQRT(2)+12*0+12*0)/(8+12+12)=64SQRT(2)/32=2SQRT(2)
\n" ); document.write( "THAT IS......... I IS(0,2SQRT(2)
\n" ); document.write( "IN RADIUS =IO=2SQRT(2) \n" ); document.write( "