ABC is an equilateral triangle.A circle incsribed in it.Another small circle is inside such that BC and AC are tangents to it.
Find the ratio of area of two circles. Option not remembered.
Where is Rakesh plz solve this. There are two questions like this asked in todas exam.

• draw a common tangent b/n two circles
  let 'R' be radius of big circle & 'r' for small circle & side of triangle = x
  
  height of triangle = h = √3*x/2
  R = h/3 = x/2√3
  
  height of smaller triangle = (h-2R)= (h- 2h/3) = h/3 = x/2√3
  r = 1/3 * (h-2R) = x/6√3
  
  ratio = A1/A2 = R^2/r^2 = 9:1
  
• 10 years agoHelpfull: Yes(37) No(11)
• Hi rakesh, height of smaller triangle = (h-2R) how it is?
• 10 years agoHelpfull: Yes(12) No(0)
• #Raka I strongly agree wid your answer.Are you dedicated to solve elitmus question?Have u ever taken the test ?
• 10 years agoHelpfull: Yes(7) No(1)
• Rakesh bhai, Zindabad, hats off 2 u boss
• 10 years agoHelpfull: Yes(6) No(1)
• 4:1 answer
• 10 years agoHelpfull: Yes(5) No(2)
• @ Rakesh,,, height of smaller triangle = (h-2R) ???
• 10 years agoHelpfull: Yes(3) No(0)
• is this formula of height of eqilateral triangle is given in question paper
• 10 years agoHelpfull: Yes(2) No(0)
• how small triangle is formed
  
• 10 years agoHelpfull: Yes(1) No(0)
• outer circle: inner circle= 4:1
• 10 years agoHelpfull: Yes(1) No(0)
• how become height of triangle = h = √3*x/2 & height of smaller triangle = (h-2R)
• 10 years agoHelpfull: Yes(0) No(0)
• radius of incircle is a/root3*2 and radius of circumcrcle is a/root3
  a= side of equilateral triangle
  ratio =pi(a/root3)^2:pi(a/root3*2)
  so answer is 4:1
  
• 10 years agoHelpfull: Yes(0) No(0)
• can any one tell me clearly how the answer 4:1 came????
• 10 years agoHelpfull: Yes(0) No(0)