在半径为1的圆O中,弦AB,AC的长分别为根号3和根号2,求AB`AC和BC弧围成的图形面积

1个回答

  • 要分二种情况,弦AB和AC是圆心的同侧和异侧.

    1、异侧,

    从A作直径AD,连结BD,CD,根据半圆上圆周角是直角性质,

    △ABD和△ACD都是RT△,

    AD==2,AB=√2,BD=√2,

    CD=√(2^2-3)=1,

    〈CAB=75°,

    S作OE⊥AB,OE=1/2R=1/2,AE=√3/2,

    S△AOB=√3/4,

    S扇形AOB=π*120/360=π/3,

    S弓形AB=π/3-√3/4,

    S△AOC=1/2,

    S扇形AOC=π/4,

    S弓形AC=π/4-1/2

    S△ABC+BD弓形=π-(π/3-√3/4)-(π/4-1/2)

    =5π/12+√3/4+1/2.

    第二种情况在同侧,

    S△ABC=AC*ABsin15°/2=√3*√2*(√6-√2)/4

    =(3-√3)/2.

    S△BOE=1*1*sin30°/2=1/4,

    S扇形BE=π*30/360=π/12,

    S弓形BE=π/12-1/4,

    S△AOE=1/2,

    S扇形AE=π/4,

    S弓形AE=π/4-1/2,

    S△ABC+BD弓形=圆面积-弓形BE-弓形AE=π-(π/12-1/4)-(π/4-1/2)

    =2π+3/4.