Two equilateral triangles of side 12 units are drawn one upon the other to form a star and a common circum circle is drawn to them then finds the area enclosed in circle but not in star.

• According to the question,
  area of the equilateral triangles are sqrt(3)*a*a/4
  here a is 12 units
  so the area of the triangle is 36*sqrt(3) sq.units
  now the radius of the circle is s/sqrt(3)
  where s is the side of the equilateral triangle
  which is 12 units.
  so radius is 12/sqrt(3) units.
  so area of the circle is pi*(12/sqrt(3))^2
  which is 48*pi sq.units.
  now in between those two triangle a hexagon is formed,
  whose side is 12/3 = 4 units,
  so the area of the hexagon is 3*sqrt(3)*4*4/2 = 24*sqrt(3) sq.units....according to the formula
  now the area of the circle leaving the area of the star is,
  48*pi - (24*sqrt(3)+24*sqrt(3)) sq.units
  which is 48*(pi - sqrt(3)) sq.units.
  I am not sure about the solution.
  Correct me if i am wrong.
• 9 years agoHelpfull: Yes(3) No(1)
• i think ans will be 12pi unit^2.
• 9 years agoHelpfull: Yes(0) No(1)
• 
  length of median of equi. triangle=a(root[3]/2)
  pi(6*root[3])^2-(root[3])/4(12)^2 ans
• 9 years agoHelpfull: Yes(0) No(2)