(2005•佛山)一座拱型桥,桥下水面宽度AB是20米,拱高CD是4米.若水面上升3米至EF,则水面宽度EF是多少?
(1)若把它看作是抛物线的一部分,在坐标系中(如图1)可设抛物线的表达式为y=ax2+c.请你填空:
a=______,c=______,EF=______米.
(2)若把它看作是圆的一部分,则可构造图形(如图2)计算如下:
设圆的半径是r米,在Rt△OCB中,易知r2=(r-4)2+102,r=14.5
同理,当水面上升3米至EF,在Rt△OGF中可计算出GF=______
【答案】
分析:求a、c的值可以利用待定系数法,求出A,D的坐标就可以.计算EF的差的近似值,可以利用函数解析式求出准确值,然后利用垂径定理求出近似值,两者求差.
解答:解:(1)AB是20米,则AC=10米,拱高CD是4米.则A,D的坐标分别是(-10,0),(0,4)
把这两点的坐标代入解析式得到:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/0.png)
解得:a=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/1.png)
,c=4,
则解析式是y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/2.png)
x
2+4.
把y=3代入解析式解得x=±5,则EF=10米.
(2)在Rt△OGF中,由题可知,OF=14.5,OG=14.5-1=13.5,
根据勾股定理知:GF
2=OF
2-OG
2,
即GF
2=14.5
2-13.5
2=28,
所以GF=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/3.png)
,此时水面宽度EF=4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/4.png)
米.
(3)误差估计如下:
解法一:∵2.6<
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/5.png)
<2.7,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/6.png)
≈2.65,4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/7.png)
≈10.6
∴4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/8.png)
-10≈0.6.(8分)
∴差的近似值约为0.6米.(9分)
解法二:∵4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/9.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/10.png)
在10到11之间,
∴可得10.5<4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/11.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/12.png)
<10.6,
∴0.5<4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470130795/SYS201310191058594701307023_DA/13.png)
-10<0.6,(8分)
∴差的近似值约为0.5或0.6米.(9分)
点评:求函数的解析式,常用的方法是待定系数法,涉及圆中求半径的问题,此类在圆中涉及弦长、半径、圆心角的计算的问题,常把半弦长,半圆心角,圆心到弦距离转换到同一直角三角形中,然后通过直角三角形求解.