鄂尔多斯中考数学试题及答案

?EF?AD?CF?AD?BC1?7??4． ············································································ 8分 22（1）证法二：如图（2）

过D点作DN∥AB交BC于N，

AD∥BN，AB∥DN，?AD?BN． ··························· 1分 EF∥AB，?DN∥EF． ················································· 2分 ?△CEF∽△CDN． ··························································· 3分 CECF??． ····························································································································· 4分 DCCN图（2）

CE1CF1··························································································· 5分 ?，??，即NF?CF． ·

DC2CN2?BF?BN?NF?AD?FC． ································································································ 6分

23．（本小题满分7分） 解：（1）AB?ACtan30° ·········································································································· 1分

?12?3．（结果也可以保留一位小数，下同） ?43≈7（米）

3答：树高约7米． ·························································································································· 2分

（2）①如图（2），B1N?AN?AB1sin45°?43?2································ 3分 ≈5（米） ·

2············································································· 4分 NC1?NB1tan60°?26?3≈8（米） ·． AC1?AN?NC1?5?8?13（米）

答：树与地面成45°角时影长约13米． ···················································································· 5分 ②如图（2）当树与地面成60°角时影长最大AC2（或树与光线垂直时影长最大或光线与半径为AB的⊙A相切时影长最大） ············································································································· 6分 ． AC2?2AB2≈14（米）

答：树的最大影长约14米． ········································································································ 7分

24．（本小题满分9分） 证明：（1）

CB?BE，

······································ 2分 ??1??2，AC?AE，AC?AE， ·

?AB?CE．············································································· 3分

CE∥BD，?AB?BD． ····················································· 4分 ?BD是⊙O的切线． ······························································· 5分 （2）连接CB．

AB是⊙O的直径，??ACB?90°． ·················································································· 6分 ?ABD?90°，??ACB??ABD． ······················································································· 7分 ?1??2，△?ACB∽△ABD． ···························································································· 8分 ACAB??，?AB2?AD·AC． ···························································································· 9分 ABAD（证法二，连接BE，证明略）

25．（本小题满分10分）

解：（1）设改造一所A类学校的校舍需资金x万元，改造一所B类学校的校舍需资金y万元，

则??x?3y?480 ··············································································· 3分（正确一个方程组2分）

3x?y?400??x?90． ························································································································ 4分

?y?130解之得?答：改造一所A类学校的校舍需资金90万元，改造一所B类学校的校舍需资金130万元．5分 （2）设A类学校应该有a所，则B类学校有(8?a)所，

则??20a?30(8?a)≥210 ································· 7分（正确一个不等式给1分）

?(90?20)a?(130?30)(8?a)≤770?a≤3．································································································································ 8分

?a≥1解得??1≤a≤3，即a?1，2，3． ······································································································· 9分

答：有3种改造方案：

方案一：A类学校1所，B类学校7所； 方案二：A类学校2所，B类学校6所； 方案三：A类学校3所，B类学校5所． ··············································································· 10分 26．（本小题满分11分） 解：如图

（1）CN?CB?15，OC?9，

········································ 1分 ?ON?152?92?12，?N(12，0)． ·又

AN?OA?ON?15?12?3，设AM?x，

···································································· 2分 ?32?x2?(9?x)2， ·

?x?4，M(15，4)． ···················································································································· 3分

（2）解法一：设抛物线l为y?(x?a)?36，

则(12?a)?36. ··························································································································· 4分

22?a1?6或a2?18（舍去）

． ······································································································ 5分 ?抛物线l:y?(x?6)2?36． ··································································································· 6分 解法二：

x2?36?0，x1??6，x2?6，

?y?x2?36与x轴的交点为(?6，0)和(6，0)． ······································································ 4分 由题意知，交点(6，0)向右平移6个单位到N点， ·································································· 5分

所以y?x2?36向右平移6个单位得到抛物线l:y?(x?6)2?36． ·································· 6分 （3）①由“三角形任意两边的差小于第三边”知，P点是直线MN与对称轴x?6的交点，

分

?设直线MN的解析式为y?kx?b，则??12k?b?0，解之得?k?4?15k?b?4??3

?b??16?y?43x?16.?P(6，?8)． ······································································································ 8分 ②DE∥OA，△?ACB∽△ABD，?m9?DE12，DE?43m． ······································ 9分

?S?12?43m?(9?8?m)??2343m2?3m． ····································································· 10分

34a??23?0，开口向下，又m??334?3172???2??3?2?9，?S有最大值， ??3??4?2S??2?17?3417289最大3???2???3?2?6． ·················································································· 11分 7