如图24-5-26所示,在扇形$O A B$中,半径$O A = 2$,$\angle A O B = 120 ^ {\circ}$,$C$是$\hat {A B}$的中点,连接$A C$,$B C$,则图中阴影部分的面积是( )
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如图D-24-36所示,连接$O C$,
∵$\angle A O B = 120 ^ {\circ}$,$C$为$\hat {A B}$的中点,
∴$\angle A O C = \angle B O C = 60 ^ {\circ}$.
∵$O A = O C = O B = 2$,
∴$\triangle A O C$,$\triangle B O C$是等边三角形,
∴$A C = B C = O A = 2$,
∴$\triangle A O C$的边$A C$上的高是$\sqrt {2 ^ {2} - 1 ^ {2}} = \sqrt {3}$.
∴$S _ {\text {阴影}} = S _ {\text {扇形 O A B}} - 2S _ {\triangle A O C}$$= \frac {120 \pi \times 2 ^ {2}} {360} - 2 \times \frac {1} {2} \times 2 \times \sqrt {3} = \frac {4 \pi} {3} - 2 \sqrt {3}$.