?$解:連接BP,CP,則BP=CP=BC=2$?
?$則△BCP是等邊三角形$?
?$∴∠BPC=60°,∴∠ABP=90°-60°=30°$?
?$S_{陰影}=2×[S_{扇形ABP}-(S_{扇形BCP}-S_{BCP})]$?
?$=2×[\frac {30°}{360°}×π×22-(\frac {60°}{360°}×π×22-\frac {1}{2}×2×\sqrt{3})]$?
?$=2×[\frac {1}{3}π-\frac {2}{3}π+\sqrt{3}]$?
?$=2\sqrt{3}-\frac {2π}{3}$?