Area of a Segment of a Circle.

Using the above diagram $2(180 - 2\theta) + \lambda = 180 \implies \lambda = 4\theta - 180 \implies$

$m = 180 - 2\theta$ .

For $\triangle{AOB}$ using the law of cosines we have:

$36 = 2r^2(1 - \cos(180 - 2\theta)) \implies 18 = r^2(1 + \cos(2\theta)) \implies r^2 = \dfrac{18}{1 + \cos(2\theta)}$

and

for $\triangle{COD}$ using the law of cosines we have:

$196 - 2r^2(1 + \cos(4\theta)) \implies 49 = r^2(\cos^2(2\theta)) \implies$

$49 + 49\cos(2\theta) = 18\cos^2(2\theta) \implies 18\cos^2(\theta) - 49\cos(2\theta) - 49 = 0 \implies$

$\cos(2\theta) = \dfrac{49 \pm 77}{2}$ and $|\cos(u)| \leq 1 \implies \cos(2\theta) = -\dfrac{7}{9}$

$\implies r^2 = \dfrac{18}{1 - \dfrac{7}{9}} = 81 \implies r = 9.$

Let $h$ be the height of $\triangle{COD} \implies h = \sqrt{81 -49} = \sqrt{32} = 4\sqrt{2} \implies$ $A_{\triangle{COD}} = 28\sqrt{2}$

From above $\cos(2\theta) = -\dfrac{7}{9} \implies$

The area of sector $COD$ is $A^{*} = \dfrac{1}{2}(4\theta - \pi)81 =$ $\dfrac{81}{2}(2\arccos(-\dfrac{7}{9}) - \pi) = 81(\arccos(-\dfrac{7}{9}) - \dfrac{\pi}{2})$

$\implies$

The desired area is $A = A^{*} - A_{\triangle{COD}} = 81(\arccos(-\dfrac{7}{9}) - \dfrac{\pi}{2}) - 28\sqrt{2} =$

$9^2(\arccos(-\dfrac{7}{9}) - \dfrac{\pi}{2}) - 28\sqrt{2} = a^b(\arccos(-\dfrac{c}{a}) - \dfrac{\pi}{b}) - d\sqrt{b} \implies$

$a + b + c + d = \boxed{46}$ .