if the side of the regular hexagon above is 2 , what is the circumference of the inscribed circle ?
now ac will be 1 because circle is touching the midpoints of the circle . then we get kite acbo . we have ac = 1 and cb = 1 , now we need to find bo which is radius from triplets we get 30 : 60 : 90 = 1 : √ 3 : 2 now cbo is right angled triangle , then we get bo as √ 3 . r = √ 3 . . . circumference is 2 ∏ r is 2 ∏ √ 3 . answer : option a .
circumface(sqrt(subtract(power(2, const_2), power(divide(2, const_2), const_2))))