2010年5月21日至2010年10月31日期间，平均每天参观上海世博会的人数约为40.21万人次. 8分
19．（本小题满分9分） （1）证明：∵AC?CD，∴弧AC与弧CD相等，∴?ABC??CBD， 又∵OC?OB，∴?OCB??OBC，∴?OCB??CBD， ∴OC∥BD. ································································································ 4分 （2）解：∵OC∥BD，不妨设平行线OC与BD间的距离为h， 又S1△OBC?2OC?h，S1△DBC?2BD?h 因为BC将四边形OBDC分成面积相等的两个三角形，即S△OBC?S△DBC
∴OC?BD， ································································································ 7分
∴四边形OBDC为平行四边形. 又∵OC?OB，∴四边形OBDC为菱形. ····························································· 9分 20．（本小题满分9分） 解：设参加活动的高中学生为x人，则初中学生为?x?4?人，根据题意，得：
6x?10?x?4?≤210 ·
··················································································· 2分 ∴16x≤170 ∴x≤10.625
所以，参加活动的高中学生最多为10人. ···························································· 5分 设本次活动植树y棵，则y关于高中学生数x的函数关系式为
y?5x?3?x?4?即：y?8x?12 ···································································· 7分 ∴y的值随x的值增大而增大. ∵参加活动的高中学生最多为10人， ∴当x?10时，y最大?8?10?12?92，
答：应安排高中学生10人，初中学生14人，最多植树92棵. ································· 9分
21．（本题满分10分） 解：设灯柱BC的长为h米，过点A作AH?CD于点
B做BE?AH于点E， ∴四边形BCHE为矩形， ∵?ABC?120°，∴?ABE?30°， 又∵?BAD??BCD?90°，∴?ADC?60°， 在Rt△AEB中，
∴AE?ABsin30°?1， BE?ABcos30°?3， ··································· 4分
∴CH?3，又CD?12，∴DH?12-3， 在Rt△AHD中，
tan?ADH?AHHD?h?112?3?3， ·································································· 8分 H，过点解得，h?123?4（米）
∴灯柱BC的高为123?4米. ······································································ 10分 22．（本题满分10分） 解：（1）设矩形广场四角的小正方形的边长为x米，根据题意，得：
??4x2??100?2x??80?2x??5200
整理，得：x?45x?350?0 ········································································· 3分 解之，得：x1?35，x2?10. 经检验，x1?35，x2?10均适合题意.
所以，要使铺白色地面砖的面积为5200平方米，则矩形广场四角的小正方形的边长为35米或10米. 5分 （2）设铺矩形广场地面的总费用为y元，广场四角的小正方形的边长为x米，则，
2y?30??4x??100?2x??80?2x???2x?100?2x??2x?80?2x??? ???20??2即：y?80x?3600x?240000
配方得，y?80?x?22.5??199500 ································································ 8分 当x?22.5时，y的值最小，最小值为199500.
所以，当矩形广场四角的小正方形的边长为22.5米时，所铺广场地面的总费用最少，最少费用为199500元. ················································································································· 10分 23．（本小题满分11分）
（1）证明：∵四边形OABC为正方形，∴OC?OA， ∵三角板OEF是等腰直角三角形，∴OE1?OF1
又三角板OEF绕O点逆时针旋转至OE1F1的位置时，?AOE1??COF1
∴△OAE1≌△OCF1. ···················································································· 3分 （2）存在. ···································································································· 4分 ∵OE?OF，
∴过点F与OE平行的直线有且只有一条，并与OF垂直， 又当三角板OEF绕O点逆时针旋转一周时，则点F在以O以OF为半径的圆上， ········································································· 5分 ∴过点F与OF垂直的直线必是圆O的切线，又点C是圆O过点C与圆O相切的直线有且只有2条，不妨设为CF1和
22为圆心，
外一点，
CF2，
此时，E点分别在E1点和E2点，满足
CF1∥OE1，CF2∥OE2，················································································ 7分
当切点F1在第二象限时，点E1在第一象限， 在直角三角形CF1O中，OC?4，OF1?2，
cos?COF1?OF11 ?，OC2，∴?COF1?60°∴?AOE1?60°
∴点E1的横坐标为：xE1?2cos60°?1， 点E1的纵坐标为：yE1?2sin60°?3，
∴点E1的坐标为1，3． ················································································· 9分 当切点F2在第一象限时，点E2在第四象限，
??，?3. 同理可求：点E2的坐标为1综上所述，三角板OEF绕O点逆时针旋转一周，存在两个位置，使得OE∥CF，此时点E的坐标为
??E11，3或E21，?3. ··············································································· 11分
24．（本题满分12分）
解：（1）因为抛物线与x轴交于点A??10，?、B?3，0?两点，设抛物线的函数关系式为： y?a?x?1??x?3?，∵抛物线与y轴交于点C?0， ?3?，∴?3?a?0?1??0?3?， ∴a?1.
所以，抛物线的函数关系式为：y?x?2x?3， ·················································· 2分 又y??x?1??4，
因此，抛物线的顶点坐标为?1 ·································································· 3分 ，?4?．（2）连结EM，∵EA、ED是⊙M，的两条切线，
∴EA?ED，EA?AM，ED?MN，∴△EAM≌△EDM
22????又四边形EAMD的面积为43，∴S△EAM?23，∴又AM?2，∴AE?23.
1 AM·AE?23，2，23或E2?1，?23. ·因此，点E的坐标为E1?1··········································· 5分
当E点在第二象限时，切点D在第一象限. 在直角三角形EAM中，tan?EMA?∴?EMA?60°∴?DMB?60° ，过切点D作DF?AB，垂足为点F，
????EA23??3， AM2，DF?3 ∴MF?1因此，切点D的坐标为2，3． ········································································ 6分
??，23、D2，3的坐标代入得 设直线PD的函数关系式为y?kx?b，将E?1?3k?????3?2k?b?3解之，得? ???b?53?23??k?b?3?所以，直线PD的函数关系式为y??????353x?. ·············································· 7分 33当E点在第三象限时，切点D在第四象限.
同理可求：切点D的坐标为2，-3，直线PD的函数关系式为y?因此，直线PD的函数关系式为
??353x?. 33y??353353x?x?. ·或y?····························································· 8分 3333（3）若四边形EAMD的面积等于△DAN的面积 又S四边形EAMD?2S△EAM，S△DAN?2S△AMD ∴S△AMD?S△EAM
∴E、D两点到x轴的距离相等，
∵PD与⊙M相切，∴点D与点E在x轴同侧， ∴切线PD与x轴平行，
此时切线PD的函数关系式为y?2或y??2.
····································································· 9分
2当y?2时，由y?x?2x?3得，x?1?6；
当y??2时，由y?x?2x?3得，x?1?2. ················································ 11分
22、P21?6，2、P31?2，?2、故满足条件的点P的位置有4个，分别是P 11?6，P41?2，?2.
????????
16数学练习试卷-2010潍坊市中考数学试题(含答案)
