发布时间 : 星期日 文章内蒙古鄂尔多斯市中考真题更新完毕开始阅读
7cm ·························································································································· 8分 5解法二：（1）四边形ACED是等腰梯形． ··········································································· 1分 证明：过点D，点E作AC的垂线，垂足分别G，H， E 则DG∥EH ··································································· 2分 ?△ACE≌△ACB≌△CAD D C O ??ECA??DAC，AD?CE H ················· 3分 ?Rt△ADG≌Rt△CEH，?DG?EH ·G ······················ 4分 ?四边形DGHE是矩形，?DE∥GH ·B A 第22题图 ?四边形ACED是等腰梯形 （2）?DC?AB?4cm，AD?3cm， ?AC?5cm 在Rt△ADC中，?DG?AC?AD?DC 即：5DG?3?4 ·············································· 5分 12···························································································································· 6分 ?DG? ·5?DE??AG?9?12?AD2?DG2?32???? ········································································· 7分 55??297····································································· 8分 ?DE?GH?AC?2AG?5?2??cm ·55（解法三：延长AD，CE相交于点P，利用全等、勾股定理、相似等解答，类似于解法一．略） 23．（本小题满分7分） 解：（1）设?BOC的度数为x 则2πr?C O xπl ············································································· 2分 180····························· 3分 ?l?2r，?x?180?，即?BOC?180? ·（2）?BOC?120，?BOC?90 ··············· 5分（每问1分） ??A H B 第23题图 360?（3）?BOC? ············································································································· 7分 n24．（本小题满分9分） 解：（1）y乙????10t(0≤t≤1)????1分 ?30t?20(1?t≤5)??3分?（2）290，甲，20． ························································································ 6分（每空1分） （3）在5月17日，甲厂生产帐篷50顶，乙厂生产帐篷30顶． ······································ 7分 设乙厂每天生产帐篷的数量提高了x%，则30(1?x%)?50?5·········································· 8分 ?x?50． 答：乙厂每天生产帐篷的数量提高了50%． ········································································ 9分 25．（本小题满分9分） 解：（1）连接CE ························································· 1分 C 证明：连接DE． ······················ 2分 ??ABC?90?，?CE是?O的直径 ···························································· 3分 ??CDE?90? ·A D F O 又?AD?CD，?AE?CE． ································· 4分 （还可以连接OD，利用中位线定理证AE等于?O的直径，或连接BD，利用“直角三角形斜边上的中线等于斜边的一半”证AD?BD，?A??DBA，?DBA??ACE．） （2）?EF是?O的切线，?EF?EC ············································································ 5分 ············································································································· 6分 ?△CEF∽△EDF ·EFFC2，即EF?FD?FC ························································································ 7分 ??FDEF1122（3）?AF?DF，AD?CD，?FD?FC，?EF?FC 33B E 第25题图 ?EF33?，?sin?ACE? ······················································································· 8分 3FC33 ······························································ 9分 3又?EA?EC，??ACE??A，?sinA?26．（本小题满分12分） 解：（1）A(6，，·················· 2分 0)D(6，，2)E(3，，4)C(0，4) ·答：不等于． ······································································ 3分 理由：连接OE，OD，ED． y C E B D O A 第26题图 x ?OE2?25，ED2?13，OD2?40． ?OE2?ED2?OD2． ······································ 4分 ?OE与DE不垂直，点O到直线ED的距离不是线段OE的长 ·（证明方法很多，①△ODE的面积为9，求出DE边上的高h?1813与OE?5的长比13181313较；②在直线DE与x，y轴围成的三角形中，利用等积法，求点O到直线DE的距离?与OE比较；③证明△OCE和△EBD不相似，则?OED?90；④延长ED交x轴于P，在Rt△DAP中，tan?EPO?2:3，而在△QEP中，OE:EP?2:3，则?OED?90．） （2）解法一：延长ED较交x轴于点H． ? 由已知得△EBD≌△HAD ?AH?EB?3，?HO?9设OG?m，则HG?9?m． y C H O E F B D H x HAAD由△HAD∽△HGF可得， ?HGGF32即 ?9?mGF22································· 5分 ?GF?(9?m)??m?6 ·33G A 图1 2?2?···································· 6分 S矩形OGFH?OG?GF?m??m?6???m2?6m(3≤m≤6) ·3?3?（不写定义域不扣分） 当m??b??2a69························································· 7分 ?时，S矩形OGFH最大． ·?2?22?????3?29GF????6?3． 32?9?····················································································································· 8分 3?． ·?点F?，2??（解法三：过点D作DM?GF于M，利用△DMF∽△EBD解答．方法同解法一类似，略） （解法四：过点E作EN?HF于N，利用△ENF∽△BDE解答．方法同解法一类似，略） 解法二：设直线ED的解析式为y?kx?b，由图象经过E，D两点可得： 2?k???2?6k?b2?解得：····································································· 5分 ?y??x?6 ·3??3?4?3k?b??b?6设点F的坐标为F(m，n)，由点F在线段ED上可得：n??2m?6 3?FG?x轴于点G，FH?y轴于点H．?FG?n，FH?m． 2?2??S矩形OGFH?mn?m??m?6???m2?6m(3≤m≤6)（不写定义域不扣分） ···· 6分 3?3?当m??b??2a69?时，S矩形OGFH最大． ·························································· 7分 ?2?22?????3?29?9?3?． ·············································································· 8分 ?n????6?3，?点F?，32?2? （3）设这个抛物线的解析式为y?ax?bx?c(a?0)由内接矩形的定义可知： 此抛物线经过B，C两点，对称轴为x??2b··········································· 9分 ?3，且c?4 ·2a····································································· 10分 ?这个抛物线的解析式为y?ax2?6ax?4 ·如图2，设矩形SPQR是这个抛物线的任一内接矩形，且点R(x，y) 由对称性可知点S(6?x，y)，?RS?2x?6，RQ?y 又?点R在这个抛物线上，?y?ax?6ax?4． 2?C矩形SPQR?2(2x?6?y) y C ?2(2x?6?ax2?6ax?4)?2ax2?(4?12a)x?4 ···· 11分 由已知可知当x?6时，C矩形SPQR取得最大值． S E R B D 4?12a1???6，?a??． 4a3因此，所求抛物线的解析式为y?? O P Q A 图2 x 12··························································· 12分 x?2x?4·3