2007年广东省中考数学试卷答案

2007年广东省初中毕业生学业考试 数学试卷参考答案及评分说明

说明：

1．提供的答案除选择题外，不一定是唯一答案，对于与此不同的答案，只要是正确的，同样给分．

2．评分说明只是按照一种思路与方法给出作为参考．在阅卷过程中会出现各种不同情况，可参照评分说明，定出具体处理办法，并相应给分． 一、选择题（本大题共5小题，每小题3分，共15分） 1．B 2．B 3．C 4．A 5．D

二、填空题（本大题共5小题，每小题4分，共20分） 6．90 7．?ABC 8．10000 9．2 10．52 三、解答题（本大题共5小题，每小题6分，共30分） 11．解：原式?1?4??2············································································ 4分 ?1?2?2， ·

2········································································································· 6分 ?1?22?22?1． ·

12．解：不等式变形整理得 3x?8?m， ·········································································· 2分

8?m ································································································· 4分 38?m因为不等式的解集是x?3，所以， ·········································································· 5分

3解得m??1． ······················································································································· 6分

两边同除以3，得x?13．解：设直线l对应的函数解析式为y?kx?b， ····· 1分

y0)，B(3，2)，得C(0，2)． ·依题意A(3，························· 2分 0)，C(0，2)在直线l上， 由A(3，CB OAx?3k?b?0，得? ························································································································ 4分

b?2?2?k??，?解得?······················································································································· 5分 3 ·

??b?2.直线l对应的函数解析式为y??2x?2． ·········································································· 6分 314．解：（1）作图正确得2分（不保留痕痕迹的得1分）， ················································ 2分 （2）因为直线l垂直平分线段AC，所以CE?AE， 又因为BC?AC，所以DE∥BC，

第 1 页 共 5 页

1BC． ···················································· 3分 23因为在Rt△ABC中，AB?5，cosA?，

53所以AC?ABcosA?5??3，···························· 4分

5所以DE?由BC?·················· 5分 AB?AC?5?3?4，

2222B

D

C E

A

A

得DE?2． ··························································································································· 6分 15．解：在△AOF和△COE中，?AFO??CEO?90，

??AOF??COE，所以?A??C， ····································· 1分

连接OD，则?A??ODA，?C??ODC， ····················· 2分 所以?A??ODA??ODC， ················································ 3分

因为?A??ODA??ODC?90，

所以?ODC?30， ································································· 4分

??F O C E B D

所以DE?OD?cos30??3， ·························································································· 5分 2·············································································································· 6分 CD?2DE?3． ·

四、解答题（本大题共4小题，每小题7分，共28分）

16．解：设该文具厂原来每天加工x套画图工具， ····························································· 1分

2500?10002500?1000??5， ···································································· 4分

x1.5x解方程，得x?100， ············································································································ 5分 经检验x?100是原方程的根． ····························································································· 6分

依题意，有

答：该文具厂原来每天加工100套画图工具． ····································································· 7分 17．解：（1）全等三角形：△B1EO≌△BFO，△AC1E≌△ACF， ························ 2分 1相似三角形：△AEC1∽△ABC， △AEC1∽△A1BC11， ················ 4分 △A1FC∽△ABC，△A1FC∽△A1B1C1． ·（2）（以△AC1E≌△ACF为例）因为AC?AC111，

A

所以AC1?AC······························································ 5分 1， 又因为?A??A···· 6分 ?90， ·1?30，?AC1F??ACF1所以Rt△AC1E≌Rt△ACF． ······································ 7分 1??B1

E

O B F

C1

C

A1

第 2 页 共 5 页

(4)，在反比例函数y?18．解：（1）点A1因为B(3，m)也在y?所以m?k24的图象上，所以k2?xy?1?4?4，故有y?． xx4的图象上， x4?4?，即点B的坐标为B?3，?， ············································································ 1分 33???k1?b?4，4???，4)，B?3，?两点，所以?一次函数y?k1x?b过A(1································ 2分 4 ·

3k1?b?，?3??3?4?k??，?416?13解得?，所以所求一次函数的解析式为y??x?． ···································· 3分

1633?b??3?（2）解法一：过点A作x轴的垂线，交BO于点F． 因为B?3，?，所以直线BO对应的正比例函数解析式为y???4?3?4x， ································· 4分 9当x?1时，y?4?4?，即点F的坐标为F?1，?， 9?9?y 所以AF?4?432?， ············································ 5分 99O A(1，4) F 所以S△AOB?S△OAF?S△ORF

B(3，m) 13213216??1???(3?1)??， 2929316即△AOB的面积为． ······································································································ 7分

3解法二：过点A分别作x轴，y轴的垂线，垂足分别为A?，A??，过点B作x轴的垂线，垂足为B?．

则S△AOB?S矩形OA?AA???S梯形A?ABB??S△OAA???S△OBB? ····························································· 4分

x

1?4?114···························································· 6分 ?1?4???4???(3?1)??1?4??3? ·

2?3?223?1616，即△AOB的面积为． ························································································· 7分 33解法三：过A，B分别作x，y轴的垂线，垂足分别为E，F．

第 3 页 共 5 页

，4)，B?3，?，得E(0，4)，F(3，0)． 由A(1······································································ 4分

设过AB的直线l分别交两坐标轴于C，D两点． 由过AB的直线l表达式为y????4?3?416?16?x?，得C(4，0)，D?0，?． 33?3?由S△AOB?S△COD?S△AOD?S△BOC， 得S△AOB?111?OC?OD??AE?OD??OC?BF ···················································· 6分 2221161161416??4???1???4??． ········································································ 7分 232323319．解：（1）填18，0.55 ······································································································· 2分 （2）画出正确图形 ················································································································· 5分 （3）给出猜想的概率的大小为0.55?0.1均为正确． ························································· 7分 五、解答题（本大题共3小题，每小题9分，共27分）

?33?3OA1???OA?20．解：（1）OA2?······························································ 2分

?， ·22?2??33?OA?a； ····················································································································· 4分 44?3??3?333（2）依题意，得OA1?， OA1??OAOA?OA?OA，OA，OA2????32????222?2??2? ················································································································································· 6分

23?3?2727以此类推，OA6??····························································· 8分 OA?OA?a， ·

?2??6464??8181l△OA6B6?3OA6?a，即OA6B6的周长为a． ···························································· 9分

646421．解：过M作与AC平行的直线，与OA，FC分别相交于H，N．

?（1）在Rt△OHM中，?OHM?90，OM?5，

6HM?OM?sin??3． ··············································· 1分 所以OH?4． ································································ 2分 MB?HA?5?4?1（单位） ······································· 3分

O H A P N C ? M B 1?5?5(cm)，所以铁环钩离地面的高度为5cm； ····· 4分

?（2）因为?MOH??OMH??OMH??FMN?90，?FMN??MOH??，

所以

FN33?sin??，即得FN?FM， ······································································· 5分 FM55?在Rt△FMN中，?FNM?90，

第 4 页 共 5 页

MN?BC?AC?AB?11?3?8（单位）， ····································································· 6分

?3?222由勾股定理FM?FN?MN，即FM2??FM??82， ········································· 7分

5??解得FM?10（单位）， ········································································································ 8分

210?5?50(cm)，所以铁环钩的长度FM为50cm． ·························································· 9分

22．解：（1）连接FH，则FH∥BE且FH?BE， ······················································· 1分

在Rt△DFH中，DF?3a?a?2a，

FH?a，?DFH?90?， ············································ 2分

所以，DH?··························· 3分 DF2?FH2?5a．

A D F H G

（2）设BE?x，△DHE的面积为y， ····················· 4分 依题意，y?S△CDE?S梯形CDHG?S△EGH， ··················· 5分

B

E C

111??3a?(3a?x)??(3a?x)?x??3a?x ································································ 6分 222139?x2?ax?a2 ·············································································································· 7分 2221391?3?27y?x2?ax?a2??x?a??a2．

2222?2?8当x?231a，即BE?BC， 22E是BC的中点时，y取最小值． ······················································································· 8分

△DHE的面积y的最小值为

272a． ················································································· 9分 8第 5 页 共 5 页

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