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当前位置：首页 > 教育专区 > 高考资料 > 2019浙教版九年级下册数学期末高效复习专题3圆的基本性质(解析版)精品教育.doc.pdf

2019浙教版九年级下册数学期末高效复习专题3圆的基本性质(解析版)精品教育.doc.pdf

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1、第 1 页期末高效复习专题 3圆的基本性质题型一点与圆的位置关系例 12019 大冶校级月考若O 的半径为 5 cm，平面上有一点 A，OA6 cm，那么点 A 与O 的位置关系是(A)A点 A 在O 外B点 A 在O 上C点 A 在O 内D不能确定【解析】O 的半径为 5 cm，OA6 cm，dr，点 A 与O 的位置关系是点 A 在O 外变式跟进12019 宜昌在公园的 O 处附近有 E，F，G，H 四棵树，位置如图 1 所示(图中小正方形的边长均相等)现计划修建一座以O 为圆心，OA 为半径的圆形水池，要求池中不留树木，则E，F，G，H 四棵树中需要被移除的为(A)图 1 AE，F，G

2、BF，G，HCG，H，EDH，E，F【解析】OA1225，OE2OA，点 E 在O 内；OF2OA，点 F 在O 内；OG1OA，点 G 在O 内；OH22222 2OA，点 H 在O 外题型二垂径定理及其推论例 2如图 2，O 的直径 CD10，弦 AB8，ABCD，垂足为 M，则 DM的长为(D)A5 B6 C7 D8 图 2 例 2 答图【解析】连结 OA，如答图所示O 的直径 CD10，OA5，弦 AB8，ABCD，AM12AB1284，在 RtAOM 中，OMOA2AM2第 2 页52423，DMODOM538.【点悟】已知直径与弦垂直的问题中，常连半径构造直角三角形，其中斜边为圆的

3、半径，两直角边是弦长的一半和圆心到弦的距离，从而运用勾股定理来计算变式跟进2如图 3，AB 为O 的直径，弦 CDAB 于点 E，若 CD8，且 AEBE14，则 AB 的长度为(A)A10 B5 C12 D.53图 3 第 2 题答图【解析】如答图，连结 OC，设 AEx，AEBE14，BE4x，OC2.5x，OE1.5x，CDAB，CEDE12CD4，RtOCE 中，OE2CE2OC2，(1.5x)242(2.5x)2，x2，AB10.3有一座弧形的拱桥如图4，桥下水面的宽度AB 为 7.2 m，拱顶与水面的距离CD 的长为 2.4 m，现有一艘宽 3 m，船舱顶部为长方形并且高出水面2

4、m 的货船要经过这里，此货船能顺利通过这座拱桥吗？图 4 第 3 题答图解：如答图，连结 ON，OB.OCAB，D 为 AB 中点，AB7.2 m，BD12AB3.6 m.又CD2.4 m，设 OBOCONr，则 OD(r2.4)m.在 RtBOD 中，由勾股定理得r2(r2.4)23.62，解得 r3.9.CD2.4 m，船舱顶部为长方形并高出水面2 m，CE2.420.4(m)，OErCE3.90.43.5(m)，在 RtOEN 中，EN2ON2OE23.923.522.96(m2)，EN1.72(m)MN2EN21.723.44 m3，此货船能顺利通过这座弧形拱桥题型三圆周角定理的综合文

5、档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9

6、F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H

7、5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10

8、T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I

9、5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:

10、CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P1

11、0G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO

12、10Y9F9P10G1第 3 页例 32019 市南区一模如图 5，在直径为 AB 的O 中，C，D 是O 上的两点，AOD58，CDAB，则 ABC 的度数为 _61_图 5【解析】AOD58，ACDAOD29，CDAB，CABACD29，AB是直径，ACB90，ABC902961.【点悟】(1)在同圆(或等圆)中，圆心角(或圆周角)、弧、弦中只要有一组量相等，则其他对应的各组量也分别相等，利用这个性质可以将问题互相转化，达到求解或证明的目的；(2)注意圆中的隐含条件(半径相等)的应用；(3)圆周角定理及其推论，是进行圆内角度数转化与计算的主要依据，遇直径，要想到直径所对的圆周角是 90，

13、从而获得到直角三角形；遇到弧所对的圆周角与圆心角，要想到同弧所对的圆心角等于圆周角的2 倍以及同弧所对的圆周角相等变式跟进4如图 6，O 是正方形 ABCD 的外接圆，点 P 在O 上，则APB_45_图 6 第 4 题答图【解析】如答图，连结OA，OB.根据正方形的性质，得AOB90.再根据圆周角定理，得 APB45.52019 永嘉二模如图 7，已知 AB 是半圆 O 的直径，OCAB 交半圆于点 C，D 是射线 OC 上一点，连结 AD 交半圆 O 于点 E，连结 BE，CE.(1)求证：EC 平分 BED；(2)当 EBED 时，求证：AECE.图 7 第 5 题答图证明：(1)AB

14、是半圆 O 的直径，AEB90，DEB90.OCAB，AOCBOC90，BEC45，DEC45.BECDEC，即 EC 平分BED；(2)如答图，连结 BC，OE，在BEC与DEC 中，BEDE，BECDEC，ECEC，文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I

15、5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:

16、CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P1

17、0G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO

18、10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L1

19、0R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1

20、HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L

21、3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1第 4 页BECDEC，CBECDE.CDE90AABE，ABECBE.AOECOE，AECE.题型四弧长的计算例 4如图 8，ABC 是正三角形，曲线 CDEF 叫做“正三角形的渐开线”，其中，CD，DE，EF，圆心依次按 A，B，C循环，它们依次相连结若AB1，则曲线 CDEF 的长是 _4_(结果保留)图 8【解析】CD的长是120118023，

22、DE的长是120218043，EF的长是12031802，则曲线 CDEF 的长是234324.变式跟进6一个扇形的半径为8 cm，弧长为163 cm，则扇形的圆心角为 _120_【解析】设扇形的圆心角为n，根据题意得163n8180，解得 n120，扇形的圆心角为 120.题型五扇形的面积计算例 52019 河南如图 9，在扇形 AOB 中，AOB90，以点 A 为圆心，OA 的长为半径作 OC交AB于点 C，若 OA2，则阴影部分的面积是313图 9 例 5 答图【解析】如答图，连结 OC，AC，OAC 是等边三角形，扇形OBC 的圆心角是30，阴影部分的面积等于扇形OBC 的面积减去弓形

23、OC 的面积 S扇形OBC302236013，S弓形OC60223603422233，S阴影13233 313.【点悟】求不规则图形的面积，常转化为易解决的基本图形，然后求出各图形的面积，通过面积的和差求出结果变式跟进7 若扇形的半径为 3 cm，扇形的面积为 2 cm2，则该扇形的圆心角为 _80_，文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1

24、 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2

25、L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1

26、文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y

27、9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2

28、H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ1

29、0T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6

30、I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1第 5 页弧长为 _43_cm.【解析】由n323602，解得 n80，由 212l3，解得 l43.8如图 10，以 AB 为直径的 O 经过 AC 的中点 D，DEBC 于点 E，若 DE1，C30，则图中阴影部分的面积是4933图 10【解析】C30，DE1，DEC

31、90，DC2，ODBC，ODA30，ODOA，OADODA30，AOD120，OA2 33，S阴影1202 332360122334933.题型六圆锥例 62019 西湖区校级三模一个圆锥的侧面展开图是圆心角为120且半径为 6 的扇形，则这个圆锥的底面半径为(B)A2 B2 C2.5 D3【解析】设这个圆锥的底面半径为r，根据题意，得2r1206180，解得 r2.【点悟】(1)圆锥侧面展开图是一个扇形；(2)圆锥的底面周长是其侧面展开图的弧长；(3)圆锥的母线就是其侧面展开扇形的半径变式跟进9一个圆锥的底面半径是5 cm，其侧面展开图是圆心角为150的扇形，则圆锥的母线长为(B)A9 c

32、m B12 cm C15 cm D18 cm【解析】设圆锥的母线长为l，根据题意得 25150l180，解得 l12.即圆锥的母线长为 12 cm.过关训练1一个圆锥形的圣诞帽底面半径为12 cm，母线长为 13 cm，则圣诞帽的侧面积文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:C

33、P2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10

34、G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO1

35、0Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10

36、R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 H

37、J10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3

38、F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档

39、编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1第 6 页为(B)A312 cm2B156 cm2C78 cm2D60 cm2【解析】圆锥的底面周长是12224，则圆锥的侧面积是122413156(cm2)22019 连云港三模一个滑轮起重装置如图1 所示，滑轮的半径是15 cm，当重物上升 15 cm 时，滑轮的一条半径 OA 绕轴心 O 按顺时针方向旋转的角度约为(取 3.14，

40、结果精确到 1)(C)图 1 A115B60C57D29【解析】根据题意得 15n15180，解得 n18057，OA 绕轴心 O 按顺时针方向旋转的角度约为57.3一个隧道的横截面如图2 所示，它的形状是以点 O 为圆心，5 为半径的圆的一部分，M 是O 中弦 CD 的中点，EM 经过圆心 O 交O 于点 E.若 CD6，则隧道的高(ME 的长)为(D)图 2 A4 B6 C8 D9【解析】M 是O 弦 CD 的中点，根据垂径定理：EMCD，又 CD6，则有CM12CD3，设 OM 是 x，在 RtCOM 中，有 OC2CM2OM2，即 5232x2，解得 x4，EM549.42019 大庆

41、模拟如图 3 是圆内接正方形ABCD，分别将 AB，BC，CD，DA沿边长 AB，BC，CD，DA 向内翻折，已知 BD2，则阴影部分的面积为 _4_图 3【解析】由圆内接正方形的性质知，正方形的边长等于半径的2倍，阴影部分的面积(2)2(2)24.文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10

42、G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO1

43、0Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10

44、R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 H

45、J10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3

46、F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档

47、编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F

48、9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1第 7 页52019 贵港如图 4，在 RtABC 中，C90，BAC60，将 ABC 绕点 A 逆时针旋转 60后得到 ADE，若 AC1，则线段 BC 在上述旋转过程中扫过部分(阴影部分)的面积是 _2_(结果保留)图 4【解析】C90，BAC60，AC1，AB2，S扇形BAD602236023，S扇形CAE60123

49、606，则 S阴影S扇形DABSABCSADES扇形ACE2362.6将一盛有不足半杯水的圆柱形玻璃水杯拧紧杯盖后放倒，水平放置在桌面上，水杯的底面如图 5 所示，已知水杯的半径是4 cm，水面宽度 AB 是 4 3 cm.(1)求水的最大深度(即 CD)是多少？(2)求杯底有水部分的面积(阴影部分)图 5 解：(1)ODAB，AB4 3 cm，BC12AB124 32 3(cm)，在 RtOBC 中，OB4 cm，BC2 3(cm)，OCOB2BC242（2 3）22(cm)，DCODOC422(cm)水的最大深度(即 CD)是 2 cm；(2)OC2，OB4，OC12OB，ABO30，OA

50、OB，BAOABO30，AOB120，SAOB12ABOC124 324 3，S扇形OAB12042360163，S阴影S扇形SAOB1634 3cm2.72019 苏州一模如图 6，已知 RtABD 中，A90，将斜边 BD 绕点 B 顺文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编码:CP2L3F6I5G1 HJ10T9L10R2H5 ZO10Y9F9P10G1文档编

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