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2017-2018年江苏省扬州市邗江区梅岭中学初一上学期期末数学试卷与答案

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(2)如果∠1=∠2,且∠3=115°,求∠ACB的度数.

【解答】解:(1)CD与EF平行.理由如下: ∵CD⊥AB,EF⊥AB,

∵垂直于同一直线的两直线互相平行, ∴CD∥EF;

(2)∵CD∥EF, ∴∠2=∠BCD, ∵∠1=∠2, ∴∠1=∠BCD, ∴DG∥BC,

∴∠ACB=∠3=115°.

25.(10分)魔术师为大家表演魔术.他请观众想一个数,然后将这个数按以下步骤操作:

魔术师立刻说出观众想的那个数.

(1)如果小明想的数是﹣1,那么他告诉魔术师的结果应该是 4 ;

(2)如果小聪想了一个数并告诉魔术师结果为93,那么魔术师立刻说出小聪想的那个数是 88 ;

(3)观众又进行了几次尝试,魔术师都能立刻说出他们想的那个数,请你说出其中的奥妙.

第17页(共21页)

【解答】解:(1)(﹣1×3﹣6)÷3+7=4; 故填:4;

(2)设这个数为x, (3x﹣6)÷3+7=93; 解得:x=88;

(3)设观众想的数为a..

因此,魔术师只要将最终结果减去5,就能得到观众想的数了.

26.(12分)A、B两地相距560千米,甲、乙两车分别从A、B两地相向而行,甲车先以每小时120千米的速度出发1小时后,乙车出发,若乙车出发2小时后,两车相遇,并以各自的速度继续匀速行驶. (1)求乙车的速度;

(2)乙车出发多长时间后两车相距400千米?

【解答】解:(1)设乙车的速度为x千米/小时,根据题意得: 120(2+1)+2x=560, 解得:x=100,

答:乙车的速度是100千米/小时;

(2)若两车在相遇前相距400千米,

设乙车出发x小时两车相距400千米,根据题意得: 120+(120+100)x+400=560, 解得:x=

若两车相遇后相距400千米, 则120+(120+100)x﹣400=560, 解得:x=

第18页(共21页)

但驾车到达B地的时间为因为

﹣1,

小时,

所以不成立,

即甲车到达B地后,甲乙两车才可能相距400千米, 即100x=400, 解得:x=4, 答:乙车出发

27.(14分)在同一平面内有n条直线,任何两条不平行,任何三条不共点.当n=1时,如图(1),一条直线将一个平面分成两个部分;当n=2时,如图(2),两条直线将一个平面分成四个部分;则:当n=3时,三条直线将一个平面分成 7 部分;当n=4时,四条直线将一个平面分成 11 部分;若n条直线将一个平面分成an个部分,n+1条直线将一个平面分成an+1个部分.试探索an、an+1、n之间的关系.

或4小时,两车相距400千米.

【解答】解:当n=1时,分成2部分, 当n=2时,分成4=2+2部分,

当n=3时,分成7=4+3部分,…(2分) 当n=4时,分成11=7+4部分,…(4分)

规律发现,有几条线段,则分成的部分比前一种情况多几部分, an、an+1、n之间的关系是:an+1=an+(n+1).…(8分) 故答案为:7,11,an+1=an+(n+1).

第19页(共21页)

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(2)如果∠1=∠2,且∠3=115°,求∠ACB的度数. 【解答】解:(1)CD与EF平行.理由如下: ∵CD⊥AB,EF⊥AB, ∵垂直于同一直线的两直线互相平行, ∴CD∥EF; (2)∵CD∥EF, ∴∠2=∠BCD, ∵∠1=∠2, ∴∠1=∠BCD, ∴DG∥BC, ∴∠ACB=∠3=115°. 25.(10分)魔术师为大家表演魔术.他请观众想一个数,然后将这个数按以下步骤操作: 魔术师立刻说出观众想的那个数. (1)如果小明想的数是﹣1,那么他告诉魔术师的结果应该是 4 ; (2)如果小聪想了一个数并告诉魔术师结果为93,那么魔术师立刻说出小聪想的那个数是 88 ; <

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