题目内容
![精英家教网](http://thumb.1010pic.com/pic3/upload/images/201106/32/3db56546.png)
A、![]() | B、![]() | C、![]() | D、![]() |
分析:△AMN的面积=
AP×MN,通过题干已知条件,用x分别表示出AP、MN,根据所得的函数,利用其图象,可分两种情况解答:(1)0<x≤1;(2)1<x<2;
1 |
2 |
解答:解:(1)当0<x≤1时,如图,![精英家教网](http://thumb.1010pic.com/pic3/upload/images/201106/32/2413bef4.png)
在菱形ABCD中,AC=2,BD=1,AO=1,且AC⊥BD;
∵MN⊥AC,∴MN∥BD;
∴△AMN∽△ABD,
∴
=
,
即,
=
,MN=x;
∴y=
AP×MN=
x2(0<x≤1),
∵
>0,∴函数图象开口向上;
(2)当1<x<2,如图,
同理证得,△CDB∽△CNM,
=
,
即,
=
,MN=2-x;![精英家教网](http://thumb.1010pic.com/pic3/upload/images/201201/48/df0c01ba.png)
∴y=
AP×MN=
x×(2-x),
y=-
x2+x;
∵-
<0,∴函数图象开口向下;
综上,答案C的图象大致符合;
故选C.
![精英家教网](http://thumb.1010pic.com/pic3/upload/images/201106/32/2413bef4.png)
在菱形ABCD中,AC=2,BD=1,AO=1,且AC⊥BD;
∵MN⊥AC,∴MN∥BD;
∴△AMN∽△ABD,
∴
AP |
AO |
MN |
BD |
即,
x |
1 |
MN |
1 |
∴y=
1 |
2 |
1 |
2 |
∵
1 |
2 |
(2)当1<x<2,如图,
同理证得,△CDB∽△CNM,
CP |
OC |
MN |
BD |
即,
2-x |
1 |
MN |
1 |
![精英家教网](http://thumb.1010pic.com/pic3/upload/images/201201/48/df0c01ba.png)
∴y=
1 |
2 |
1 |
2 |
y=-
1 |
2 |
∵-
1 |
2 |
综上,答案C的图象大致符合;
故选C.
点评:本题考查了二次函数的图象,考查了学生从图象中读取信息的数形结合能力,体现了分类讨论的思想.
![](http://thumb2018.1010pic.com/images/loading.gif)
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