B2. $6 \pi$

Point $F$ is the midpoint of $A B$ (and the center of the semicircle). Points $D$ and $E$ are the midpoints of $B C$ and $A C$. They lie on the semicircle because triangles $B D F$ and $A E F$ are equilateral. Draw the arc between $D$ and $E$ of the circle with center $C$ and radius $|C E|=6$. Since $\angle A F E=\angle E F D=60^{\circ}$, the circular segment corresponding to line segment $A E$ has exactly the same dimensions as the circular segment above line segment $D E$. The circular segment below $D E$ also has these dimensions because the two circles around $C$ and $F$ have equal radii. The circular segment corresponding to

![](https://cdn.mathpix.com/cropped/2024_04_17_ecd98442595314acd6b1g-2.jpg?height=337&width=325&top_left_y=1353&top_left_x=1531)
line segment $B D$ can thus be exactly moved to the circular segment below $D E$. The three gray areas together have exactly the same area as circular sector $C E D$, and that has an area of $\frac{1}{6}\left(\pi \cdot 6^{2}\right)=6 \pi$.
B2. An equilateral triangle $A B C$ has a side length of 12. A semicircle with diameter $A B$ intersects the sides $A C$ and $B C$.

Determine the total area of the gray region consisting of the two circular segments outside the triangle and the part of the triangle outside the circle.

![](https://cdn.mathpix.com/cropped/2024_04_17_dd45dfb811a24abc1595g-2.jpg?height=352&width=355&top_left_y=1823&top_left_x=1519)