题目内容
如图,⊙O的直径CD=10cm,AB是⊙O的弦,AB⊥CD,垂足为M,OM:OC=3:5,则AB的长是
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A.4cm | B.6cm | C.8cm | D.10cm |
C.
试题分析:连接OB;
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∵CD=10cm,∴OC=5cm;
∵OM:OC=3:5,∴OM=3cm;
Rt△OCP中,OC=OA=5cm,OM=3cm;
由勾股定理,得:
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所以AB=2AM=8cm,
故选C.
考点: 1.垂径定理;2.勾股定理.
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