新建
上传
首页
助手
最?/div>
资料?/div>
工具

数学精品复习资料

 

规律探索

 

一、选择?/p>

 

1. 

?/p>

2014?/p>

山东威海?/p>

?/p>

12

?/p>

3

分)

如图?/p>

在平面直角坐标系

xOy

中,

Rt

?/p>

OA

1

C

1

?/p>

Rt

?/p>

OA

2

C

2

?/p>

Rt

?/p>

OA

3

C

3

?/p>

Rt

?/p>

OA

4

C

4

?/p>

的斜边都在坐标轴上,

?/p>

A

1

OC

1

=

?/p>

A

2

OC

2

=

?/p>

A

3

OC

3

=

?/p>

A

4

OC

4

=?30°

?/p>

若点

A

1

的坐标为

?/p>

3

?/p>

0

?/p>

?/p>

OA

1

=

OC

2

?/p>

OA

2

=

OC

3

?/p>

OA

3

=

OC

4

?/p>

,则依此规律,点

A

2014

的纵坐标为(

 

 

?/p>

 

 

 

 

A

?/p>

 

0

 

B

?/p>

 

?/p>

3×

?/p>

?/p>

2013

 

C

?/p>

 

?/p>

2

?

2014

 

D

?/p>

 

3×

?

?/p>

2013

 

考点

?/p>

 

规律型:点的坐标

 

专题

?/p>

 

规律型.

 

分析?/p>

 

根据?/p>

30

度的直角三角形三边的关系?/p>

OA

2

=

OC

2

=3×

?

OA

3

=

OC

3

=3×

?

?/p>

2

?/p>

OA

4

=

OC

4

=3×

?

?/p>

3

,于是可

得到

OA

2014

=3×

?

?/p>

2013

,由于?/p>

2014=4×

503+2

,则可判?

?/p>

A

2014

?/p>

y

轴的正半轴上,所以点

A

2014

的纵坐标?/p>

3×

?

?

2013

?/p>

 

解答?/p>

 

解:∵∠

A

2

OC

2

=30°

?/p>

OA

1

=

OC

2

=3

?/p>

 

?/p>

OA

2

=

OC

2

=3×

?/p>

 

?/p>

OA

2

=

OC

3

=3×

?/p>

 

?/p>

OA

3

=

OC

3

=3×

?

?/p>

2

?/p>

 

?/p>

OA

3

=

OC

4

=3×

?/p>

?/p>

2

?/p>

 

?/p>

OA

4

=

OC

4

=3×

?

?/p>

3

?/p>

 

?/p>

OA

2014

=3×

?

?/p>

2013

?/p>

 

Ͼλ
新建
上传
首页
助手
最?/div>
资料?/div>
工具

数学精品复习资料

 

规律探索

 

一、选择?/p>

 

1. 

?/p>

2014?/p>

山东威海?/p>

?/p>

12

?/p>

3

分)

如图?/p>

在平面直角坐标系

xOy

中,

Rt

?/p>

OA

1

C

1

?/p>

Rt

?/p>

OA

2

C

2

?/p>

Rt

?/p>

OA

3

C

3

?/p>

Rt

?/p>

OA

4

C

4

?/p>

的斜边都在坐标轴上,

?/p>

A

1

OC

1

=

?/p>

A

2

OC

2

=

?/p>

A

3

OC

3

=

?/p>

A

4

OC

4

=?30°

?/p>

若点

A

1

的坐标为

?/p>

3

?/p>

0

?/p>

?/p>

OA

1

=

OC

2

?/p>

OA

2

=

OC

3

?/p>

OA

3

=

OC

4

?/p>

,则依此规律,点

A

2014

的纵坐标为(

 

 

?/p>

 

 

 

 

A

?/p>

 

0

 

B

?/p>

 

?/p>

3×

?/p>

?/p>

2013

 

C

?/p>

 

?/p>

2

?

2014

 

D

?/p>

 

3×

?

?/p>

2013

 

考点

?/p>

 

规律型:点的坐标

 

专题

?/p>

 

规律型.

 

分析?/p>

 

根据?/p>

30

度的直角三角形三边的关系?/p>

OA

2

=

OC

2

=3×

?

OA

3

=

OC

3

=3×

?

?/p>

2

?/p>

OA

4

=

OC

4

=3×

?

?/p>

3

,于是可

得到

OA

2014

=3×

?

?/p>

2013

,由于?/p>

2014=4×

503+2

,则可判?

?/p>

A

2014

?/p>

y

轴的正半轴上,所以点

A

2014

的纵坐标?/p>

3×

?

?

2013

?/p>

 

解答?/p>

 

解:∵∠

A

2

OC

2

=30°

?/p>

OA

1

=

OC

2

=3

?/p>

 

?/p>

OA

2

=

OC

2

=3×

?/p>

 

?/p>

OA

2

=

OC

3

=3×

?/p>

 

?/p>

OA

3

=

OC

3

=3×

?

?/p>

2

?/p>

 

?/p>

OA

3

=

OC

4

=3×

?/p>

?/p>

2

?/p>

 

?/p>

OA

4

=

OC

4

=3×

?

?/p>

3

?/p>

 

?/p>

OA

2014

=3×

?

?/p>

2013

?/p>

 

">
新建
上传
首页
助手
最?/div>
资料?/div>
工具

数学精品复习资料

 

规律探索

 

一、选择?/p>

 

1. 

?/p>

2014?/p>

山东威海?/p>

?/p>

12

?/p>

3

分)

如图?/p>

在平面直角坐标系

xOy

中,

Rt

?/p>

OA

1

C

1

?/p>

Rt

?/p>

OA

2

C

2

?/p>

Rt

?/p>

OA

3

C

3

?/p>

Rt

?/p>

OA

4

C

4

?/p>

的斜边都在坐标轴上,

?/p>

A

1

OC

1

=

?/p>

A

2

OC

2

=

?/p>

A

3

OC

3

=

?/p>

A

4

OC

4

=?30°

?/p>

若点

A

1

的坐标为

?/p>

3

?/p>

0

?/p>

?/p>

OA

1

=

OC

2

?/p>

OA

2

=

OC

3

?/p>

OA

3

=

OC

4

?/p>

,则依此规律,点

A

2014

的纵坐标为(

 

 

?/p>

 

 

 

 

A

?/p>

 

0

 

B

?/p>

 

?/p>

3×

?/p>

?/p>

2013

 

C

?/p>

 

?/p>

2

?

2014

 

D

?/p>

 

3×

?

?/p>

2013

 

考点

?/p>

 

规律型:点的坐标

 

专题

?/p>

 

规律型.

 

分析?/p>

 

根据?/p>

30

度的直角三角形三边的关系?/p>

OA

2

=

OC

2

=3×

?

OA

3

=

OC

3

=3×

?

?/p>

2

?/p>

OA

4

=

OC

4

=3×

?

?/p>

3

,于是可

得到

OA

2014

=3×

?

?/p>

2013

,由于?/p>

2014=4×

503+2

,则可判?

?/p>

A

2014

?/p>

y

轴的正半轴上,所以点

A

2014

的纵坐标?/p>

3×

?

?

2013

?/p>

 

解答?/p>

 

解:∵∠

A

2

OC

2

=30°

?/p>

OA

1

=

OC

2

=3

?/p>

 

?/p>

OA

2

=

OC

2

=3×

?/p>

 

?/p>

OA

2

=

OC

3

=3×

?/p>

 

?/p>

OA

3

=

OC

3

=3×

?

?/p>

2

?/p>

 

?/p>

OA

3

=

OC

4

=3×

?/p>

?/p>

2

?/p>

 

?/p>

OA

4

=

OC

4

=3×

?

?/p>

3

?/p>

 

?/p>

OA

2014

=3×

?

?/p>

2013

?/p>

 

Ͼλ">
Ͼλ
Ŀ

2019年中考数学试卷分类汇编:规律探索(含答? - 百度文库
新建
上传
首页
助手
最?/div>
资料?/div>
工具

数学精品复习资料

 

规律探索

 

一、选择?/p>

 

1. 

?/p>

2014?/p>

山东威海?/p>

?/p>

12

?/p>

3

分)

如图?/p>

在平面直角坐标系

xOy

中,

Rt

?/p>

OA

1

C

1

?/p>

Rt

?/p>

OA

2

C

2

?/p>

Rt

?/p>

OA

3

C

3

?/p>

Rt

?/p>

OA

4

C

4

?/p>

的斜边都在坐标轴上,

?/p>

A

1

OC

1

=

?/p>

A

2

OC

2

=

?/p>

A

3

OC

3

=

?/p>

A

4

OC

4

=?30°

?/p>

若点

A

1

的坐标为

?/p>

3

?/p>

0

?/p>

?/p>

OA

1

=

OC

2

?/p>

OA

2

=

OC

3

?/p>

OA

3

=

OC

4

?/p>

,则依此规律,点

A

2014

的纵坐标为(

 

 

?/p>

 

 

 

 

A

?/p>

 

0

 

B

?/p>

 

?/p>

3×

?/p>

?/p>

2013

 

C

?/p>

 

?/p>

2

?

2014

 

D

?/p>

 

3×

?

?/p>

2013

 

考点

?/p>

 

规律型:点的坐标

 

专题

?/p>

 

规律型.

 

分析?/p>

 

根据?/p>

30

度的直角三角形三边的关系?/p>

OA

2

=

OC

2

=3×

?

OA

3

=

OC

3

=3×

?

?/p>

2

?/p>

OA

4

=

OC

4

=3×

?

?/p>

3

,于是可

得到

OA

2014

=3×

?

?/p>

2013

,由于?/p>

2014=4×

503+2

,则可判?

?/p>

A

2014

?/p>

y

轴的正半轴上,所以点

A

2014

的纵坐标?/p>

3×

?

?

2013

?/p>

 

解答?/p>

 

解:∵∠

A

2

OC

2

=30°

?/p>

OA

1

=

OC

2

=3

?/p>

 

?/p>

OA

2

=

OC

2

=3×

?/p>

 

?/p>

OA

2

=

OC

3

=3×

?/p>

 

?/p>

OA

3

=

OC

3

=3×

?

?/p>

2

?/p>

 

?/p>

OA

3

=

OC

4

=3×

?/p>

?/p>

2

?/p>

 

?/p>

OA

4

=

OC

4

=3×

?

?/p>

3

?/p>

 

?/p>

OA

2014

=3×

?

?/p>

2013

?/p>

 



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