Tempting Circle

The shaded area is

$\textbf{Area\_center O - Area\_center E - Area of segment of circle}$

Area of segment is

$\textbf{Area of minor sector OAB} -A(\triangle OAB)$

See that $\dfrac{OD}{OA}= \dfrac{1}{2}$

Thus radius of circle with center $E$ is $10\times \dfrac{3}{4} = \dfrac{15}{2}$

Also,

area of sector OAB is $\dfrac{1}{3} \times$ area of whole circle = $\dfrac{100 \pi}{3}$

And area of $\triangle OAB = \dfrac{OD \times AB}{2} = \dfrac{5\times 10\sqrt{3}}{2} = 25\sqrt{3}$

Hence asked area is

$100 \pi - \bigl( \dfrac{15}{2} \bigr) ^2- \dfrac{100\pi}{3} + 25\sqrt{3} = \dfrac{125 \pi+ 300 \sqrt{3}}{12}$

Giving answer $125+300+12 = \boxed{437}$

Consider the diagram above. The area of the shaded part is area of the large sector plus area of the triangle minus area of the small circle. The large sector has a central angle of $240^\circ$ . So the area is

$A_{SECTOR}=\dfrac{240}{360}\pi (10^2)=\dfrac{200}{3}\pi$ .

The area of the triangle is

$A_{TRIANGLE}=\dfrac{1}{2}(10^2)(\sin 120)=50\left(\dfrac{\sqrt{3}}{2}\right)=25\sqrt{3}$

Now we need to find $h$ (see my figure) to know the diameter of the small circle. We have

$\cos 60=\dfrac{h}{10}$

$\dfrac{1}{2}=\dfrac{h}{10}$

$h=5$

The diameter of the small circle therefore is

$d=10+5=15$

So the radius is

$r=\dfrac{15}{2}$

So the area is

$A_{CIRCLE}=\pi \left(\dfrac{15}{2}\right)^2=\dfrac{225}{4}\pi$

Thus, the area of the shaded part is

$A_{SHADED}=\dfrac{200}{3}\pi+25\sqrt{3}-\dfrac{225}{4}\pi=\dfrac{125}{12}+25\sqrt{3}=\dfrac{125\pi+300\sqrt{3}}{12}$

Hence, the desired answer is

$answer=125+300+12=$ $\boxed{437}$