Two equilateral triangles of side 12 cm are placed one on top of another, such that a 6 pointed star is formed. If the six vertices lie on a circle, what is the area of the circle not enclosed by the star? (Give answers to the nearest sq cm.)
a)61
b)57
c)68
d)83

• area of 1 equilateral triangle= [sqrt(3)/4]*a^2=62.28
  area of three small equilateral triangle (sides=4)= 20.76
  total area covered by triangle= 62.28+20.76=83.04
  radius of crcle = 4*sqrt(3)
  area of radius=3.14* 4*sqrt(3)*4*sqrt(3)=150.72
  so, area of the circle not enclosed by the star= 150.72-83.04= 67.68(aprox 68)
  ansr is 68...
• 11 years agoHelpfull: Yes(13) No(4)
• ans is 68...
• 11 years agoHelpfull: Yes(11) No(1)
• why side is 4 cm for 3 small equilateral triangle?
  
• 9 years agoHelpfull: Yes(2) No(0)
• firstly we find the radius of circle that will be 20.784 cm and area of circle will be 150.72..then find area enclosed by traingle is 83.136..left area is(150.72-83.136=67.59).ans=68 sqcm
• 11 years agoHelpfull: Yes(1) No(2)
• ans is
  68
  
• 11 years agoHelpfull: Yes(0) No(1)
• 68 will be the answer !!!
• 11 years agoHelpfull: Yes(0) No(0)
• how to find side of equilateral triangle
• 9 years agoHelpfull: Yes(0) No(0)
• Sol: Given that two equilateral triangles of length 12 has inscribed in a circle.
  
  
  
  Altitude of the triangle = 3√2a = 3√2(12) = 63√
  We know that centroid divides the altitude in the ratio 2 : 1 and 23(Altitude) = Circum radius
  Circum radius = 23(63√)=43√
  Area of the circle = πr2=3.14×(43√)2
  Now the two triangles in the circle forms 12 small equilateral triangles with side 4. So their total area = 12×3√4a2= 12×3√442
  Area which is not covered by the equilateral triangles = 3.14×(43√)2 - 12×3√442 = 67.65 ≃68
  
• 9 years agoHelpfull: Yes(0) No(0)