2013年1海淀、西城区丰台、昌平初二期末数学试题及答案 下载本文

五、解答题(本题6分)

26.在△ABC中,AD是△ABC的角平分线. (1)如图1,过C作CE∥AD交BA延长线于点E,若F为CE的中点,连结AF,求证:AF⊥AD;

(2)如图2,M为BC的中点,过M作MN∥AD交AC于点N,若AB=4, AC=7, 求NC的长. 图1

(1) 证明:

(2)解:

图2

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北京市西城区(北区)2012–2013学年度第一学期期末试卷 八年级数学附加题 2013.1

一、填空题(本题共6分)

1.在平面直角坐标系xOy中,横、纵坐标都为整数的点称为整点.已知一组正方形的四个顶点恰好落在两坐标轴上,请你观察每个正方形四条边上的整点的个数的变化规律.

回答下列问题:

(1)经过x轴上点(5,0)的正方形的四条边上的整点个数是 ;

(2)经过x轴上点(n,0)(n为正整数)的正方形的四条边上的整点个数记为m,则m与

n之间的函数关系是 .二、解答题(本题共14分,

第2题8分,第3题6分)

2.在平面直角坐标系xOy中,直线y?x?6与x轴交于点A,与y轴交于点B.

(1)求∠BAO的度数; (2)如图1,P为线段AB上一点,在AP上方以AP为斜边作等腰直角三角形APD.点Q

在AD上,连结PQ,过作射线PF⊥PQ交x轴于点F,作PG⊥x轴于点G. 求证:PF=PQ ; (3)如图2,E为线段AB上一点,在AE上方以AE为斜边作等腰直角三角形AED.若P

为线段EB的中点,连接PD、PO,猜想线段PD、PO有怎样的关系?并说明理由.

图1

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图2

3.在Rt△ABC中,∠ACB=90°,∠A=30°,BD是△ABC的角平分线, DE⊥AB于点E.

(1)如图1,连接EC,求证:△EBC是等边三角形;

(2)点M是线段CD上的一点(不与点C,D重合),以BM为一边,在BM的下方作

∠BMG=60°,MG交DE延长线于点G.请你在图2中画出完整图形,并直接写出MD,DG与AD之间的数量关系; (3)如图3,点N是线段AD上的一点,以BN为一边,在BN的下方作∠BNG=60°,

NG交DE延长线于点G.试探究ND,DG与AD数量之间的关系,并说明理由.

(1)证明: 图1

(2)结论: ;

(3)证明 :

图2

图3

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北京市西城区(北区)2012 — 2013学年度第一学期期末试卷

八年级数学参考答案及评分标准 2013.1

一、选择题(本题共30分,每小题3分)

题号 1 2 3 4 答案 C D A B 11 x?2 5 D 6 C 13 7 C 8 A 9 B 14 6 18 k?2 10 D 二、填空题(本题共24分,第13题4分,第18题2分,其余各题每小题3分)

12 1?2,? (,0),(0,?1) 2每空2分 17 15 -5或1 15x?16 151.2x?12 72 或0

20.解:

(1m?3?1m?3)?2mm?6m?9(m?3)2m22

==

2m(m?3)(m?3)?···············································································3分

m?3m?3. ···································································································4分

9?39?3?12当m?9时,原式=. ···································································5分

21.解:方程两边同乘(x?1)(x?1),得

x(x?1)?3(x?1)?(x?1)(x?1).

···································································2分

分 分

化简,得x?3x?3?解得

?1.··············································································4

x?2. ·······························································································5

?2时,(x?1)(x?1)?0,

检验:当x∴x?2是原分式方程的解. ·········································································6分

22.解:(1)∵AE∥BF,

∴∠A=∠FBD. ···················································································· 1分 又∵AB= CD, ∴AB+BC = CD+BC. 即AC=BD. ··································· 3分 在△AEC和△BFD中, 16