2022年中考数学卷精析版广东珠海卷 .pdf

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=【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 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 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: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 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 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文档编码:CL5W9E9V4M8 HR3Q7A8S8H7 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: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 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 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: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（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）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 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: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 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 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: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）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：31.73，21.41）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: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 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 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: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 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】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 CO 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。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: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 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 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: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 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）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 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: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 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 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: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【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