有道几何的问题不懂,关于圆形性质的.

设E,F为AB、CD的中点.则由垂径定理,OE⊥AB,OF⊥CD
由勾股定理：OE= √(OB^2 - BE^2) = √(32.5^2 - 16.5^2) = √(32.5 + 16.5)(32.5-16.5) =√(49 * 16)= 28
OF = √(OC^2 - CF^2) = √(32.5^2 - 31.5^2) =√(32.5 + 31.5)(32.5-31.5) =√(64 * 1) = 8
因为AB⊥CD,所以OE⊥OF
再由勾股定理得OK = √(OF^2 + OE^2) = 4√53