有道几何的问题不懂,关于圆形性质的.
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/26 01:07:54
有道几何的问题不懂,关于圆形性质的.
![](http://img.wesiedu.com/upload/1/37/137262121f596195d99e1646b96285eb.jpg)
![](http://img.wesiedu.com/upload/1/37/137262121f596195d99e1646b96285eb.jpg)
![有道几何的问题不懂,关于圆形性质的.](/uploads/image/z/5676174-54-4.jpg?t=%E6%9C%89%E9%81%93%E5%87%A0%E4%BD%95%E7%9A%84%E9%97%AE%E9%A2%98%E4%B8%8D%E6%87%82%2C%E5%85%B3%E4%BA%8E%E5%9C%86%E5%BD%A2%E6%80%A7%E8%B4%A8%E7%9A%84.)
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
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