初三数学题求解🙏

初三数学题求解🙏
初三数学题求解🙏
 

初三数学题求解🙏
连接OA、OB,过O作半径OD⊥AB,交AB于点E,
∵OD=0.5m,DE=0.2m,
∴OE=OD-DE=0.5-0.2=0.3m,
∴cos∠AOE=OE/OA=0.3/0.5=0.6
∴∠AOE≈53.13°
∴AE=OA•sin∠AOE=0.5×sin∠AOE=0.5×sin53.13°≈0.5×0.8=0.4
AB=2AE=0.8
∴∠AOB=2∠AOE=2×53.13°=106.26°,
∴S阴影=S扇形OAB-S△OAB=(106.26×π×0.5^2)/360-1/2×0.8×0.3=0.1118m^2．

简单