in an equilateral triangle ABC D is a point on BC such that BD:DC =1:2 and AB =3 cm then find AD

• BC=3cm Let AE be a perpendicular from vertex A to side BC. Thus, BE=1.5cm and given BD=1cm, thus DE=0.5cm. AE=(3)^1/2*3/2 So, AD^2 = AE^2 + DE^2 i.e. AD^2=27/4 +0.25=7 Thus, AD=(7)^1/2=2.64cm
• 10 years agoHelpfull: Yes(2) No(0)
• as, BC=3 , so BD=1 and CD=2
  now draw a perpendicular from A to BC and let the point be O
  now applying area of triangle
  root(3)/4 * 3^2 = 1/2 *3* AO
  AO=(3^1.5)/2
  now applying pythagorus theorem
  AD^2 = AO^2 + DO^2
  DO = 1.5-1 = 0.5 (as, the perpendicular bisect the side in equilateral )
  AD = 7^0.5 = 2.645
• 10 years agoHelpfull: Yes(1) No(0)