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07-13年数学中考题东莞

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AB为⊙O的直径,且AB?10,

?当OF?5时,CD与⊙O相切于F点,

即3103m?5,m?, ········································································································· 6分 23?当m?103时,CD与⊙O相切. ····················································································· 7分 319.解:(1)在Rt△ABC中,

BC?AC2?AB2?202?122?16,

S矩形ABCD?AB?BC?12?16?192. ························································································ 2分

(2)?矩形ABCD,对角线相交于点O,

?SABCD?4S△OBC. ···················································································································· 3分

?四边形OBB1C是平行四边形,

?OB∥CB1,OC∥BB1,

??OBC??B1CB,?OCB??B1BC.

又?BC?CB,

?△OBC≌△B1CB,

?SOBB1C?2S△OBC?1························································································ 5分 SABCD?96, ·

2111同理,SA1B1C1C?SOBB1C???SABCD?48, ······································································ 6分

2221第6个平行四边形的面积为6SABCD?3. ··············································································· 7分

2五、解答题(三)(本大题3小题,每小题9分,共27分) 20.证明:(1)如图1,连结OA,OC, 因为点O是等边三角形ABC的外心,

所以Rt△OFC≌Rt△OGC≌Rt△OGA. ····································· 2分

A G E O B F

C SOFCG?2S△OFC?S△OAC,

因为S△OAC?所以SOFCGD 1S△ABC,

答案20题图(1) 31·············································································································· 4分 ?S△ABC. ·

3(2)解法一: 连结OA································· 5分 ,OB和OC,则△AOC≌△COB≌△BOA,?1??2, ·

A 不妨设OD交BC于点F,OE交AC于点G,

E

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2 3 G O 4 5 1 C

B

F D

?AOC??3??4?120°,?DOE??5??4?120°,

························································································· 7分 ??3??5.·

在△OAG和△OCF中, ??1??2,? ?OA?OC,??3??5,?··············································································································· 8分 ?△OAG≌△OCF, ·1··································································································· 9分 ?SOFCG?S△AOC?S△ABC. ·

3A 解法二: E 不妨设OD交BC于点F,OE交AC于点G, G 3 K 作OH?BC,OK?AC,垂足分别为H、K, ······················· 5分 O 2

在四边形HOKC中,?OHC??OKC?90°,?C?60°, 1 C B F H ······························ 6分 ??HOK?360°-90 ?90??60??120?, ·

D 即?1??2?120°.

答案第20题图(3) 又??GOF??2??3?120°,

······························································································································· 7分 ??1??3. ·

?AC?BC, ?OH?OK,

·············································································································· 8分 ?△OGK≌△OFH, ·

1···································································································· 9分 ?SOFCG?SOHCK?S△ABC. ·

321.解: 方程 换元法得新方程 令x?t,则解新方程 检验 求原方程的解 t1?1?0, t1?1,t2??3 ??2分 x?1,所以x?2x?3?0 t2?2t?3?0 ??1分 令x?2?t,t2??3?0(舍去) ??3分 x?1. ??4分 t1?1?0, t1?1,t2??2 ??7分 x?x?2?4?0 2则x?2?1,所以t?t?2?0 ??6分 t2??2?0(舍去) ??8分 x?2?1,x?3. ??9分 22.解:(1)在正方形ABCD中,AB?BC?CD?4,?B??C?90°,

?AM?MN,

A ??AMN?90°,

??CMN??AMB?90°.

在Rt△ABM中,?MAB??AMB?90°, ??CMN??MAB,

······················································ 2分 ?Rt△ABM∽Rt△MCN. ·

- 34 -

D

N C

B

M

答案22题图

(2)?Rt△ABM∽Rt△MCN,

?ABBM4x, ?,??MCCN4?xCN?x2?4x, ···················································································································· 4分 ?CN?4?y?S梯形ABCN?1??x2?4x11???4??4??x2?2x?8??(x?2)2?10, 2?422?当x?2时,y取最大值,最大值为10. ··················································································· 6分 (3)??B??AMN?90°,

?要使△ABM∽△AMN,必须有

由(1)知

AMAB, ································································ 7分 ?MNBMAMAB, ?MNMC?BM?MC,

··································· 9分 ?当点M运动到BC的中点时,△ABM∽△AMN,此时x?2. ·

(其它正确的解法,参照评分建议按步给分)

★ 机密·启用前

2008年广东省初中毕业生学业考试

数 学

说明:1.全卷共4页,考试用时100分钟,满分为120分.

2.答卷前,考生务必用黑色字迹的签字笔或钢笔在答题卡填写自己的准考证号,姓名、试室号、座位号.用2B铅笔把对应该号码的标号涂黑.

3.选择题每小题选出答案后,用用2B铅笔把答题卡上对应题目选项的答案信息点涂黑,如需改动,用橡皮擦干净后,再选涂其他答案,答案不能答在试题上. 4.非选择题必须用黑色字迹钢笔或签字笔作答、答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液.不按以上要求作答的答案无效.

5.考生务必保持答题卡的整洁.考试结束时,将试卷和答题卡一并交回.

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.?1的值是 2A.?11 B. C.?2

22D.2

2.2008年5月10日北京奥运会火炬接力传递活动在美丽的海滨城市汕头举行,整个火炬传递

- 35 -

路线全长约40820米,用科学计数法表示火炬传递路程是 A.408.2?10米 B.40.82?10米 C.4.082?10米 D.0.4082?10米 3.下列式子中是完全平方式的是

A.a?ab?b B.a?2a?2

D.a?2a?1

22224523

C.a?2b?b

224.下列图形中是轴对称图形的是

5.下表是我国部分城市气象台对五月某一天最高温度的预报,当天预报最高温度数据的中位

数是 城市 最高温度 (℃) 北京 26 上海 25 杭州 29 苏州 29 武汉 31 重庆 32 广州 28 汕头 27 珠海 28 深圳 29 A.28 B.28.5 C.29 D.29.5

二、填空题(本大题5小题,每小题4分,共20分)请将下列各题的正确答案填写在答题

卡相应的位置上.

6.?2 的相反数是__________;

7.经过点A(1,2)的反比例函数解析式是_____ _____; 8.已知等边三角形ABC的边长为3?3,则ΔABC的周长是____________; 9.如图1,在ΔABC中,M、N分别是AB、AC的中点,且∠A +∠B=120°, 则∠AN M= °;

10.如图2,已知AB是⊙O的直径,BC为弦,∠A BC=30°过圆心O作OD⊥BC交弧

BC于点D,连接DC,则∠DCB= °.

A M B

图1

D B C

- 36 -

C A

N O 图2

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