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2013年全国名校中考模拟试卷分类汇编40阅读理解型问题

2013年全国名校中考模拟试卷分类汇编40阅读理解型问题
2013年全国名校中考模拟试卷分类汇编40阅读理解型问题

阅读理解型问题

一、解答题

1、(2013·湖州市中考模拟试卷1)阅读理解:对于任意正实数a,b ,

20

≥,∴0a b +≥,∴a =b 时,等号成立.

结论:在a,b 均为正实数)中,若ab 为定值p ,则a b

+,当且仅当a

=b ,a+b 有最小值2.根据上述内容,回答下列问题:

(1)若x ﹥0,只有当x = 时,4x x

+

有最小值 . (2)探索应用:如图,已知A(-2,0),B(0,-3),点P 为双曲线6(0)y x x =>上的任意一点,过点P 作PC ⊥x 轴于点C ,PD ⊥y 轴于点D .求四边形ABCD 面积的最小值,并说明

此时四边形ABCD 的形状.

答案:(1)2 …………………………‥2分,

4 …………………‥‥4分 设6(,)P x x

, 则6(,0),(0,)Cx D x ,62,3C Ax D B x ∴=+=+, ∴116(2)(3)22A B C D S C A D Bx x

=?=+?+四边形,化简得:112(3)6,2S x x =++ ‥‥‥8分, 当且仅当123,2x x x

==即时,等号成立. ‥‥‥9分 ∴S≥×12+6=12

∴S 四边形ABCD 有最小值12. ‥‥‥10分

∵OA =OC ,OD =OB

∴四边形ABCD 是平行四边形. ………………‥‥11分

又AC ⊥BD

∴四边形ABCD 是菱形. (12)

2、(2013·湖州市中考模拟试卷8)阅读材料:如图,△ABC 中,AB =AC ,P 为底边BC 上

任意一点,点P 到两腰的距离分别为21,r r ,腰上的高为h ,连结AP ,则

ABC ACP ABP S S S ???=+ ,即:h AB r AC r AB ?=?+?21212121 ,h r r =+∴21

(1)理解与应用 如果把“等腰三角形”改成“等边三角形”,那么P 的位置可以由“在底

边上任一点”放宽为“在三角形内任一点”,即:已知边长为2的等边△ABC 内任意一点P 到

各边的距离分别为1r ,2r ,3r ,试证明:123r r r ++.

(2)类比与推理 边长为2的正方形内任意一点到各边的距离的和等于 ;

(3)拓展与延伸 若边长为2的正n 边形A 1A 2…An 内部任意一点P 到各边的距离为

n r r r ,,21,请问n r r r ++21是否为定值(用含n 的式子表示),如果是,请合理猜测

出这个定值。

答案:(第1小题4分,2、3小题各3分,共10分)

(1)分别连接AP ,BP ,CP ,由A B P B C P A C P A B C

S S S S ????++=可证得123r r r h ++=,再

.

(2) 4. (3)0

180tan(90)n n -.

3、(2013·湖州市中考模拟试卷8的最小值,小

明巧妙的运用了“数形结合”思想.具体方法是这样的:如图,C 为线段BD 上一动点,分

别过点B 、D 作,AB BD ED BD ⊥⊥,连结AC 、EC .已知AB =1,DE =5,BD =8,设BC =x .则

A C ,CE =则问题即转化成求AC +CE 的最小值.

(1)我们知道当A 、C 、E 在同一直线上时, AC +CE 的值最小,于是可求得

的最小值等于 ,此时x = ;

(2)请你根据上述的方法和结论,试构图...

答案:(第1小题每空3分,第二小题图形2分,结论2分,共10分)

(1)10,4

3

(2) 13.

4、(2013年河南西华县王营中学一摸)(9分)小强和爸爸上山游玩,两人距地面的高度y (米)与小强登山时间x(分)之间的函数图象分别如图中折线OAC和线段DE所示,根据函数图象进行以下探究:

信息读取[ (1)爸爸登山的速度是每分钟__米;

(2)请解释图中点B的实际意义;

图象理解

(3)求线段DE所表示的y与x之间的函数关系式,并写出自变量x的取值范围

(4)计算、填空:m=____;

问题解决

(5)若小强提速后,他登山的速度是爸爸速度的3倍,间:小强登山多长时间时开始提速?此时小强距地面的高度是多少米?

解:(1)10;1分

(2)图中点B的实际意义是:距地面高度为165米时人相遇(或小强迫上爸爸);2分

(3)∵D(0,100),E(20,300)

∴线段DE的解析式为

110100(020)

y x x

=+≤≤4分(4)m=6.5 6分

(5)由图知300165

6.5

t

-

-

=3×10 ∴t=11. 7分

∴B(6.5,165),C(11,300),∴直线AC的解析式为y2=30x-30.又∵线段OA过点(1,15),直线OA的解析式为

y3=15x8分

由{153030

y x

y x

=

=-解之得:{230x y==∴A(2,30)

即登山2分钟时小强开始提速,此时小强距地面的高度是30米,9分

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