题目
题型:不详难度:来源:
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003137-74982.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003137-13282.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003138-43286.png)
(1)求AB长;
(2)求⊙C截AB所得弦BD的长.
答案
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003138-70310.png)
解析
试题分析:(1)在△ABC中,根据锐角三角函数定义可得BC的长,由勾股定理可得AB的长;
(2) 过点C作AB垂线CE,由等积法可求得CE的长,在Rt△BCE中,由勾股定理可得BE的长,根据垂径定理可得BD的长.
试题解析:(1)∵在△ABC中,
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003137-74982.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003137-13282.png)
∴根据勾股定理,得AB=5.
(2)如图,过点C作AB垂线,垂足为E,由等积法得CE=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003138-90911.png)
在Rt△BCE中,由勾股定理,得
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003138-43766.png)
∴BD=2BE=
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003138-70310.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003139-93520.png)
核心考点
举一反三
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003130-69010.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003130-32794.jpg)
(1)求直径BC的长;
(2)求弦AB的长.
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003126-77858.png)
A.外离; | B.外切; | C.内含; | D.内切. |
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003121-44263.png)
A.第①块; | B.第②块; | C.第③块; | D.第④块. |
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003117-73638.jpg)
A.
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003117-14955.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003117-71072.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105003110-50283.jpg)