设弧BB'与AC交于点D,S∆ABC=2根号3,∠ACB=∠ACB'=60°S扇形BCD=S扇形B'CD=(πr²)/6=(2π)/3S扇形ACA'=(πr²)/3=(16π)/3∴S阴影=S∆ABC-S扇形BCD+S扇形ACA'-S扇形B'CD-S∆A'B'C=4π=12.56