在△ABC中,AB=AC=根号5,BC=2,求△ABC的内切圆半径

从A向BC作垂线,垂足为DAB=AC,∴BD=CD=1.RT△ABD中,根据勾股定理,AD=2S△ABC=1/2×2×2=2连接内切圆心和ABC顶点,设内切圆半径为XS△ABC=S△AOB+S△AOC+S△BOC=1/2×AB×X+1/2×BC×X+1/2×AC×X=1/2×(AB+BC+AC)X=21/...