I just asked one of my friend who has given tcs test he said me that there is a question give:
A circle is given inside it there are two equilateral triangle which forms a start and said to find the area of the circle where diameter of circle is given my test is on 15 if anybody knows this please tell me how to solve it !!!

• in which at the center there common part of two triangle is hexagon so from he center of the circle draw 6 triangle in hexagon.
  so total there is equilateral 12 triangle
  after that their find altitude of one of the main triangle that is
  (sqrt(3)/4)*(a*a)=1/2 * 12(base) * altitude
  
  so altitude is=6sqrt(3)
  
  now the small triangle below that altitude having the altitude is
  (sqrt(3)/4)*(a*a) where a=4 because all 12 triangle is equilateral = 1/2 * 4(base) * altitude
  
  sp altitude=2sqrt(3)
  
  so diameter of circle=6sqrt(3) + 2sqrt(3) = 8sqrt(3)
  so radius is = 4sqrt(3)
  
  remaining area (ans) = pi * r * r – 12* (sqrt(3)/4) * a * a where a=4
  
  so ans is=67.58
• 12 years agoHelpfull: Yes(12) No(0)
• Hi i too have a tcs campus on 20th...please send me some ques...i am just tired searching the net for ques..
  
  send me the ques pls at : harmohitgaba2000@gmail.com
  
  i`ll surely help u find the solution to ur ques...
• 12 years agoHelpfull: Yes(4) No(1)
• Actually it was asked to find out the area of the circle that was not covered by star.its not the diameter it is side of the equilateral triangle given as 12.
  and now to find the radius u need to find the height of equilateral triangle..
  (sqrt3)/2a. n the triangle is surrouded by crcle that is circum circle ...and the circumcircle divides height in the ratio of 2:1 soo finally radius is 2parts of{(sqrt3)/2a}[here given 'a=12' is substituted].now area of circle is(pi)*r*r. now to find area of two equilateral triangles it is 12*area of equilateral traingle that is 12*((sqrt3)/4)*a*a[here a/3 value is substituted in place of 'a'].for suppose given a=12 then 12/3=4 is to be substituted in finding the equilateral triangle area and finally subtratct (area of circle -area of triangle).
• 12 years agoHelpfull: Yes(4) No(0)
• area of circle(having r=12/sqrt(3))-(area of 2 equilateral triangles-area of hexagon having edge of 4cm) is the answer i.e. 67.59cm
• 12 years agoHelpfull: Yes(3) No(0)
• sorry its STAR not START
• 12 years agoHelpfull: Yes(1) No(1)
• area of which part??
  is the vertices of the two triangles touching the circle boundry?
  please clear you question first...
• 12 years agoHelpfull: Yes(0) No(0)
• i can tel one thing to all of them is pls dont fully believe on m4maths.. aps is really very tough!! one 2 to 3 questions coming dats alll..!!! concentrate on ur basicss!!!!
• 12 years agoHelpfull: Yes(0) No(0)
• JAHNAVIMANGINA i really don't know the question my friend who has given the test sent me message with the problem so i posted it here if it comes possibly to me on 15 oct i can attempt thnx for you giving the detailed description, can you tell me the exact answer to this question !!!
• 12 years agoHelpfull: Yes(0) No(0)
• ans is 48(3.14-3)
• 12 years agoHelpfull: Yes(0) No(0)
• 22/7*(15/2)^2
• 12 years agoHelpfull: Yes(0) No(0)
• SASIKUMAR explanation please !!!!
• 12 years agoHelpfull: Yes(0) No(0)
• HIMADRI please elaborate how you did that ????
• 12 years agoHelpfull: Yes(0) No(0)