球O球面上有三点A、B、C，已知AB=18，BC=24，AC=30，且球半径是球心O到平面ABC的距离的2倍，求球O的表面 球面上有三个点A、B、C组成球的一个内接三角形，若AB=18...

\u5df2\u77e5\u7403\u9762\u4e0a\u6709\u4e09\u70b9A\u3001B\u3001C\uff0cAB=6cm\uff0cBC=8cm\uff0cAC=10cm\uff0c\u4e14\u7403\u5fc3O\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb\u4e3a12\uff0c\u5219\u7403\u7684\u534a\u5f84\u4e3a\uff08\u3000\u3000\uff09

\u5982\u56fe\u6240\u793a\uff1a\u2235AB=6 cm\uff0cBC=8cm\uff0cCA=10cm\uff0c\u2234\u2220CBA=90\u00b0\u2234\u53d6AC\u7684\u4e2d\u70b9M\uff0c\u5219\u7403\u9762\u4e0aA\u3001B\u3001C\u4e09\u70b9\u6240\u5728\u7684\u5706\u5373\u4e3a\u2299M\uff0c\u8fde\u63a5OM\uff0c\u5219OM\u5373\u4e3a\u7403\u5fc3\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb\uff0c\u5728Rt\u25b3OMA\u4e2d\uff0cOM=13cm\uff0cMA=5cm\uff0c\u2234OA=13cm\uff0c\u5373\u7403\u7403\u7684\u534a\u5f84\u4e3a13cm\uff0e\u6545\u9009A\uff0e

\u89e3\uff1a\u7403\u9762\u4e0a\u4e09\u70b9A\u3001B\u3001C\uff0c\u5e73\u9762ABC\u4e0e\u7403\u9762\u4ea4\u4e8e\u4e00\u4e2a\u5706\uff0c\u4e09\u70b9A\u3001B\u3001C\u5728\u8fd9\u4e2a\u5706\u4e0a\u2235AB=18\uff0cBC=24\uff0cAC=30\uff0cAC2=AB2+BC2\uff0c\u2234AC\u4e3a\u8fd9\u4e2a\u5706\u7684\u76f4\u5f84\uff0cAC\u4e2d\u70b9M\u5706\u5fc3\u7403\u5fc3O\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb\u5373OM=\u7403\u534a\u5f84\u7684\u4e00\u534a=12R\u25b3OMA\u4e2d\uff0c\u2220OMA=90\u00b0\uff0cOM=12R\uff0cAM=12AC=30\u00d712=15\uff0cOA=R\u7531\u52fe\u80a1\u5b9a\u7406\uff08 12R\uff092+152=R2\uff0c34R2=225\u89e3\u5f97R=10 3\u7403\u7684\u8868\u9762\u79efS=4\u03c0R2=1200\u03c0\u6545\u7b54\u6848\u4e3a\uff1a1200\u03c0\uff0e

解答：解：球面上三点A、B、C，平面ABC与球面交于一个圆，三点A、B、C在这个圆上
∵AB=18，BC=24，AC=30，
AC²=AB²+BC²，∴AC为这个圆的直径，AC中点M圆心
球心O到平面ABC的距离即OM=球半径的一半=

1

2

R
△OMA中，∠OMA=90°，OM=

1

2

R，AM=

1

2

AC=30×

1

2

=15，OA=R
由勾股定理（

1

2

R）²+15²=R²，

3

4

R²=225
解得R=10

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