we know that the regular hexagon can be divided into 6 equilateral triangles
area of one equilateral triangle=s²*√3/4 for s=3 in area of one equilateral triangle=9*√3/4 in²
area of a circle=pi*r²
in this problem the radius is equal to the side of a regular hexagon r=3 in area of the circle=pi*3²-----> 9*pi in² we divide that area into 6 equal parts------> 9*pi/6----> 3*pi/2 in²
the area of a segment formed by a side of the hexagon and the circle is equal to 1/6 of the area of the circle minus the area of 1 equilateral triangle so [ (3/2)*pi in²-(9/4)*√3 in²]
