Question

⊙O是△ABC的内切圆,D、E、F为切点,设⊙O的半径为r,BC=a,CA=b,AB=c

Answer 1

(1)连接OA、OB、OC

∵⊙O为△ABC的内切圆

∴OE⊥AB,OD⊥AC,OF⊥BC

∴S△AOB=AB×OE÷2=rc/2

S△AOC=AC×OD÷2=rb/2

S△BOC=BC×OF÷2=ra/2

∴S△ABC=S△AOB+S△AOC+S△BOC

=r(a+b+c)/2

(2)由①得r=2S△ABC/(a+b+c)

=2×5÷10

=1(cm)

答：…………