Question

如图,在△ABC中,∠C=90°,AB=10,AC=8,P为AC上一点,点O在BP上,⊙O与AB,AC相切,CP=4,求

Answer 1

设 O 到 AP、AB 距离,即圆半径为 r

连 AO,则有：

S△APO = AP * r / 2 = (8 - 4) * r / 2 = 2r

S△ABO = AB * r / 2 = 5r

S△ABP = S△APO + S△ABO = 7r

又勾股定理得 BC = 6

则 S△ABP = AP * BC / 2 = (8 - 4) * 6 / 2 = 12

所以 7r = 12,r = 12/7