2024 AMC 10A Problem 22

Each kite is two $30$-$60$-$90$ triangles (legs $1$ and $\sqrt3,$ hypotenuse $2$); set up coordinates from these lengths

Take $A=(0,0)$ and $AB$ horizontal; $AB$ is six unit lengths, while the boundary from $A$ to $C$ uses lengths $\sqrt3,\sqrt3,1$ at angles $30^\circ,90^\circ,0^\circ$

Each half of a kite is a $30$-$60$-$90$ triangle, so its edges have the shown lengths and directions. Take $A=(0,0)$ and $AB$ horizontal. The horizontal span in the figure is six unit lengths, so $B=(6,0).$ Along the outer boundary from $A$ to $C,$ the three edges have vectors $\sqrt3(\cos30^\circ,\sin30^\circ),$ $\sqrt3(\cos90^\circ,\sin90^\circ),$ and $(1,0).$ Their sum is $(\tfrac52,\tfrac{3\sqrt3}{2}),$ so $C=(\tfrac52,\tfrac{3\sqrt3}{2}).$ Thus $AB=6$ and the altitude from $C$ is $\tfrac{3\sqrt3}{2},$ giving area $\tfrac12\cdot6\cdot\tfrac{3\sqrt3}{2}=\tfrac{9\sqrt3}{2}.$ Therefore, the answer is B.