Question 10Equilateral triangle ABC has side 2. D is the midpoint of AC, and the circle with diameter BD is tangent to AC at D. What is the shaded area? $\frac{4\pi -3\sqrt{3}}{8}$$\frac{\pi -3\sqrt{3}}{4}$$\frac{7\sqrt{3}-4\pi }{16}$$\frac{5\sqrt{3}-2\pi }{8}$Cannot be determined