Regular hexagon ABCDEF is inscribed in a circle. If the shortest diago

Regular hexagon has 9 diagonal, of which the shortest one connects A and C, and A and E.
If we take diagonal AE, then it is given that AE = 3 cm ------(1)
Angle \( AFE = (6-2)*\frac{180}{6}\) = 120 and FE = FA (isosceles triangle)
If we draw a perpendicular line from F to diagonal AE, touching at G, then angle AFG = 60.
From (1), \(AG = \frac{AE}{2}= \frac{3}{2}\).
From triangle AFG, \(AF = \sqrt{3} \) and \(GF = \sqrt{3}/2, ED = AF = \sqrt{3}\) (regular hexagon)
From right triangle AED, hypotenuse AD (dia of the circle) = \(\sqrt{(AE^2 + ED^2)} = 2*\sqrt{3} \)
So, the area of the circle = \(\pi*(\sqrt{3})^2 = 3*\pi\)
Area of the hexagon = area of rectangle ABDE + 2*(area of triangle AFE) = \(AE*ED + 2*(1/2*FG*AE) = 3*\sqrt{3} + 2*(1/2*\sqrt{3}/2*3)\)
= \(3*\sqrt{3} + \frac{3*\sqrt{3}}{2}\)
Now, the difference between area of the circle and area of the hexagon = \(3*\pi - (3*\sqrt{3}+\frac{3*\sqrt{3}}{2}) = 3*\pi - \frac{9*\sqrt{3}}{2}\)
The shaded area = 1/6*(difference in area)
= \(\frac{1}{6}(3*\pi-\frac{9*\sqrt{3}}{2})\)
Answer (C)

Please Find The Attached File

∆ AED is a right-angled triangle.
\(3^{2} +R^{2}= (2R)^{2}\)

\(3R^{2} = 9\)

\(--> R= √3\)

The shaded region = \(\frac{1}{6} (πR^{2}- 6(\frac{√3}{4} R^{2})= \frac{1}{6} (3π -\frac{9√3}{2} )\)

Answer (C).