Painting A Hexagram

$\begin{aligned} && OA = R = 10 \text{ units} , AP = \dfrac12 AO = 5 \text{ units} , PB = \dfrac5{\sqrt3} \text{ units} \\ && \Rightarrow \text{Area of triangle } ABC = \dfrac12 \cdot 5 \cdot \dfrac5{\sqrt3} = \dfrac{25}{\sqrt3} \text{ unit}^2 \end{aligned}$

Area of the regular hexagram is $12 \cdot \dfrac{25}{\sqrt3} = 100\sqrt3$ .

Area of the circle is $\pi R^2 = \pi \cdot 10^2 = 100 \pi$ .

Thus the area of the shaded region is $\dfrac13 ( 100\pi - 100\sqrt3) = 46.984\ldots \approx 47 \text{ unit}^2$ .

Radius of circle= 10 units
Altitude of an equilateral triangle is divided in the ratio= 2:1 at the center of the circle
Hence the altitude of the equilateral triangle= 10 * (2+1)/2 = 15 units
Altitude of equilateral triangle= Sqrt 3* side/2
Hence, we get side bigger triangle.
Side of each smaller triangle= side of big triangle/3
There are 12 such small triangles (6 external and 6 within the hexagon).
Find the area of each smaller triangle and multiply it by 12.
Subtract this area from the area of the circle.
The value that you get it the value of segments.
Divide it by 3 to get the value of 2 segments (which is the shaded portion).
The area comes out to be 47 unit2.