Triangles say it all

$\color{#D61F06}{\text{Solution 1:}}$

First, calculate $\triangle ADB$ and $\triangle BDC$ areas, both triangles have equal areas.

$\text{base}=10$

$\text{height}=\frac{10\sqrt{3}}{2}$ (you can use trigonometry or phytagoras theorem to find it)

$|\triangle ABD|=\frac{10\times \frac{10\sqrt{3}}{2}}{2}=\boxed{25\sqrt{3}}$

Second, calculate circular sector.

Put together the three circular sectors of $\triangle ABD$ and you get a semi-circle of radii $5$ . The area of this semi-circle is $\boxed{\frac{25\pi}{2}}$

Finally, shaded area is $2\times (25\sqrt{3}-\frac{25\pi}{2})=\color{#D61F06}{\boxed{50\sqrt{3}-25\pi\approx 8.063}}$

$\color{#3D99F6}{\text{Solution 2:}}$

$ADCB$ is a parallelogram so its area is $10\times \frac{10\sqrt{3}}{2}=\boxed{50\sqrt{3}}$ .

The six circular sectors form a cirlce of radii $5$ which area is $\boxed{25\pi}$

Shaded area is $\color{#3D99F6}{\boxed{50\sqrt{3}-25\pi=\approx 8.063}}$