<PRE><B>Question 26340</B>
LET ABC BE THE TRIANGLE. AND LET AD BE THE HEIGHT SHOWN BY A PERPENDICULAR FRO A TO BC.D IS ON BC. IN TRIANGLES ABD AND ACD WE HAVE 
AB=AC=6
AD=AD
ANGLE ADB = 90 = ANGLE ADC 
HENCE THE 2 TRIANGLES ARE IDENTICAL HENCE BD = DC 
BUT BC=BD+DC=6 ...SO BD=DC=3
IN RIGHT ANGLED TRIANGLE ABD WE HAVE ,ANGLE ADB =90 DEGREES
HYPOTENUSE = AB =6 
BD =3 
SO 
AD^2+BD^2=AB^2
AD^2+3^2=6^2
AD^2=36-9=27
AD=SQ.RT.27 = 3*SQRT3</PRE>