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2022年中考数学卷精析版广东珠海卷 .pdf

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1、2012 年中考数学精析系列珠海卷（本试卷满分120 分，考试时间100 分钟）一、选择题（本大题5 小题，每小题3 分，共 15 分）在每小题列出的四个选项中，只有一个是正确的，请把答题卡上对应题目所选的选项涂黑.1.（2012 广东珠海3 分）2 的倒数是【】A 2 B 2 CD【答案】C。【考点】倒数。【分析】根据两个数乘积是1 的数互为倒数的定义，因此求一个数的倒数即用1 除以这个数所以2 的倒数为 1 2=12。故选 C。2.（2012 广东珠海3 分）计算 2a2+a2的结果为【】A 3aB aC 3a2 D a2【答案】D。3.（2012 广东珠海3 分）某同学对甲、乙、丙、丁四

2、个市场二月份每天的白菜价格进行调查，计算后发现这个月四个市场的价格平均值相同、方差分别为2222S8.5S2.5S10.1S7.4乙丁甲丙，二月份白菜价格最稳定的市场是【】A甲B乙C丙D丁【答案】B。【考点】方差【分析】方差就是和中心偏离的程度，用来衡量一批数据的波动大小（即这批数据偏离平均数的大小）在样本容量相同的情况下，方差越大，说明数据的波动越大，越不稳定。因此，2222S2.5S7.4S8.5S10.1的解集是【答案】1x2。【考点】解一元一次不等式组。10.（2012 广东珠海4 分）如图，AB 是 O 的直径，弦CDAB，垂足为 E，如果 AB=26，CD=24，那么sinOCE=

3、【答案】513。【考点】垂径定理，勾股定理，锐角三角函数的定义。【分析】如图，设AB 与 CD 相交于点 E，则根据直径AB=26，得出半径OC=13；由 CD=24，CDAB，根据垂径定理得出CE=12；在 RtOCE 中，利用勾股定理求出OE=5；再根据正弦函数的定义，求出sinOCE 的度数：OE5sinOCE=OC13。三、解答题（一）（本大题5 小题，每小题6 分，共 30 分）文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7

4、A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D

5、3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL

6、5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3

7、Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X

8、5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:

9、CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 H

10、R3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O911（2012 广东珠海6 分）计算：120121+20122【答案】解：原式=2112=0。12.（2012 广东珠海6 分）先化简，再求值：2x1x+1x1xx，其中x=2【答案】解：原式=2x+1x1x1111=x x1x+1x x1x+1x。当x=2时，原式=12=22。13.（2012 广东珠海6 分）如图，在 ABC 中，AB=AC，AD 是高，AM 是 ABC 外角 CAE 的平分线（1）用尺规作图方法，作ADC 的平分线 DN；（保留作图痕迹，不写作法

11、和证明）（2）设 DN 与 AM 交于点 F，判断 ADF 的形状（只写结果）【答案】解：（1）如图所示：文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文

12、档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4

13、M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H

14、7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O

15、9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9

16、V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S

17、8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9（2）ADF 的形状是等腰直角三角形。【考点】作图（基本作图），平行的判定和性质，等腰三角形的判定。【分析】（1）作法：以 D 为圆心，以任意长为半径画弧，交

18、AD 于 G，交 DC 于 H，分别以 G、H 为圆心，以大于12GH 为半径画弧，两弧交于N，作射线DN。则 DN 即为所求。（2）设 DN 交 AM 于 F，则AB=AC，AD 是高，BAD=CAD。又 AM 是 ABC 外角 CAE 的平分线，FAD=12 180=90。AFBC。CDF=AFD。又 AFD=ADF，CDF =ADF。AD=AF。ADF 是等腰直角三角形。14.（2012 广东珠海6 分）已知关于x 的一元二次方程x2+2x+m=0（1）当 m=3 时，判断方程的根的情况；（2）当 m=3时，求方程的根【考点】一元二次方程根的判别式，因式分解法解一元二次方程。【分析】（1

19、）判断一元二次方程根的情况，只要看根的判别式=b24ac 的值的符号即可判断：当 0，方程有两个不相等的实数根；当=0，方程有两个相等的实数根；当0，方程没有实数根。（2）把 m 的值代入方程，用因式分解法求解即可。15.（2012 广东珠海6 分）某商店第一次用600 元购进 2B 铅笔若干支，第二次又用600 元购进该款铅笔，但这次每支的进价是第一次进价的54倍，购进数量比第一次少了30 支（1）求第一次每支铅笔的进价是多少元？（2）若要求这两次购进的铅笔按同一价格全部销售完毕后获利不低于420 元，问每支售价至少是多少元？文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X

20、5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:

21、CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 H

22、R3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN

23、8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编

24、码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8

25、 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7

26、ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9（2）设售价为y 元，根据题意列不等式为：6006005y4+y442054444解得，y6。答：每支售价至少是6 元。【考点】分式方程和一元一次不等式组的应用。【分析】（1）方程的应用解题关键是找出等量关系，列出方程求解。设第一次每支铅笔进价为x 元，由第二次每支铅笔进价为54x 元。本题等量关系为：第一次购进数量第二次购进数

27、量=30 600 x6005x4=30。（2）设售价为y 元，求出利润表达式，然后列不等式解答。利润表达式为：第一次购进数量第一次每支铅笔的利润第二次购进数量第二次每支铅笔的利润6004y46005445y44。四、解答题（二）（本大题4 小题，每小题7 分，共 28 分）16（2012 广东珠海7 分）如图，水渠边有一棵大木瓜树，树干DO（不计粗细）上有两个木瓜A、B（不计大小），树干垂直于地面，量得AB=2 米，在水渠的对面与O 处于同一水平面的C 处测得木瓜A 的仰角为 45、木瓜 B 的仰角为30 求 C 处到树干 DO 的距离 CO（结果精确到1 米）（参考数据：31.73，21

28、.41）文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W

29、9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7

30、A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D

31、3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL

32、5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3

33、Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X

34、5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9【答案】解：设 OC=x，在 RtAOC 中，ACO=45，OA=OC=x。在 RtBOC 中，BCO=30，3OBOC?tan30 x3。AB=OAOB=3xx=23，解得x=3+31+1.73=4.735。OC=5 米。答：C 处到树干DO 的距离 C

35、O 为 5 米。17.（2012 广东珠海7 分）某学校课程安排中，各班每天下午只安排三节课（1）初一（1）班星期二下午安排了数学、英语、生物课各一节，通过画树状图求出把数学课安排在最后一节的概率；（2）星期三下午，初二（1）班安排了数学、物理、政治课各一节，初二（2）班安排了数学、语文、地理课各一节，此时两班这六节课的每一种课表排法出现的概率是136已知这两个班的数学课都有同一个老师担任，其他课由另外四位老师担任求这两个班数学课不相冲突的概率（直接写结果）【答案】解：（1）画树状图如下：三节课安排共有6 种等可能情况，数学科安排在最后一节有2 种情况，数学科安排在最后一节的概率是21=63。

36、文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V

37、4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8

38、H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10

39、O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E

40、9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8

41、S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H

42、10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9（2）两个班数学课不相冲突的概率为242=363。【考点】树状图法，概率。【分析】（1）画出树状图，然后根据概率公式列式计算即可得解。（2）画树状图，然后根据概率公式列式计算即可得解：画树状图如下：所有等可能情况共有6 6=36 种。初二（1）班的 6 种情况，在

43、对应初二（2）班的 6 种情况时，有2 种情况数学课冲突，其余4种情况不冲突。例如，初二（1）班（数学，物理，政治）对应初二（2）班的 6 种情况时，与初二（2）班的（数学，语文，地理）和（数学，地理，语文）冲突。初二（1）班（物理，数学，政治）对应初二（2）班的 6 种情况时，与初二（2）班的（语文，数学，地理）和（地理，数学，语文）冲突。不冲突的情况有4 6=24。两个班数学课不相冲突的概率为242=363。18.（2012 广东珠海7 分）如图，把正方形ABCD 绕点 C 按顺时针方向旋转45 得到正方形ABCD（此时，点 B落在对角线AC 上，点 A落在 CD 的延长线上），A B 交

44、 AD 于点 E，连接 AA、CE求证：（1）ADA CDE；（2）直线 CE 是线段 AA的垂直平分线文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档

45、编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M

46、8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7

47、 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9

48、文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V

49、4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8

50、H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 ZN8X5D3H10O9【答案】证明：（1）四边形ABCD 是正方形，AD=CD，ADC=90。ADE=90。根据旋转的方法可得：EAD=45，AED=45。A D=DE。在 AD A

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