Two equilateral triangles of sides 12cms are placed one on top of another,such that a six pointed star is formed. If the six vertices lie on the circle,what is the area of the circle not enclosed by the star?

• ans 48(pi-root3)
  area of 1 triangle= ((root3)/4)*12^2
  area of small triangle=((root3)/4)*4^2
  radius of circle = 12/ root3
  area not covered= pi(12/ root3)-(((root3)/4)*12^2+3*((root3)/4)*4^2)
• 10 years agoHelpfull: Yes(12) No(1)
• Eq. tri Area= side2*√3/4
  Given Side=12 cm
  One eq.triangle placed over other they cut the line in ratio 1:3
  Side of smaller triangle=12/3=4 cm
  Total area covered by two triangles= ( (12*12)+3(4*4))*√3/4
  =48√3
  Radius of circle=side/√3=12/√3
  Area of Circle=A=πr^2=48π
  Uncovered area=48(π-√3)
  
• 10 years agoHelpfull: Yes(9) No(1)
• 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
• 10 years agoHelpfull: Yes(7) No(0)
• 48(pi-root3)
• 10 years agoHelpfull: Yes(5) No(4)
• 12(4pi-root3)
  
• 10 years agoHelpfull: Yes(2) No(5)
• 72pi-120(sqrt3)=18.348 cm sqr.
• 10 years agoHelpfull: Yes(1) No(5)
• area of 1 triangle= ((root3)/4)*12^2=62.352
  area of small triangle=((root3)/4)*4^2=6.928
  radius of circle= ((root3)/2)*12*3/4=9*(root3)/2=7.794
  area of circle= pi*(7.794)^2=216.86
  ans will be=216.86-(62.352+3*6.928)=133.724
• 10 years agoHelpfull: Yes(1) No(2)
• answer is 152.064cmsq
• 10 years agoHelpfull: Yes(0) No(2)