2015年江西省中考数学冲刺卷(3)
说明:本卷共有六个大题,24个小题,全卷满分120分,考试时间120分钟.
一、选择题(本大题共6个小题,每小题3分,共18分)每小题只有一个正确选项. 1.-6的绝对值是( ).
11
A .-6 B .6 C D .
66
2.如图,图中的几何体是圆柱沿竖直方向切掉一半后得到的,则该几何体的左视图是( ).
3.下列运算正确的是( ). A .2x +3y =5xy C .a -(a -b ) =-b
B .a 3-a 2=a
D .(a -1)(a +2) =a 2+a -2
4.已知一元二次方程 x 2 + x ─ 1 = 0,下列判断正确的是( ). A . 该方程有两个相等的实数根 B . 该方程有一个根为1 C . 该方程没有实数根 D . 该方程有一个根为负数
5.如图,□ABCD 中,∠C =120°,AB =AE =5, AC 与BD 交于点F ,AF =2EF ,则BC 的长为( ).
A .6 B .8 C .10 D .12
6.已知y =
ax
2+bx
的图象如图所示,则y =ax -b 的图象一定过( ). A .第一、二、三象限 C .第二、三、四象限
二、填空题(本大题共8小题,每小题3分,共24分) 7.计算:2-7=. 8.函数y =
B .第一、二、四象限 D .第一、三、四象限
x +3
自变量的取值范围是. x
9.有一组数据:30,10, 50, 70,50.它们的中位数是
10.如图,点B 是AD 的延长线上一点,DE ∥AC ,AE 平分∠CAB ,∠C =50︒,∠E =30︒,则∠CDA 的度数等于 .
11.n 边形的一条对角线,将这个n 边形分成内角和分别为180°和720°的两个多边形,则
n
12.如图,直线l 与⊙O 相切于点C ,A 、B 、D 均在⊙O 上,OA ∥l ,∠BDC =85°,则∠
BAO 的度数为
13.已知P 1(x 1,2),P 2(x 2,3)是同一个反比例函数图象上的两点,若则这个反比例函数的表达式为 .
14.如图,五边形ABCDE 中,AB=BC=CD=AE=2,∠A =∠B =∠BCD =120°,点P 在五边形上,若∠CPD =30°,则CP 的长为 . 三、(本大题共4小题,每小题6分,共24分) 15.已知x -1=
16.如图,矩形ABCD 中,点E 在BC 上,AE=CE,试分别在下列两个图中按要求使用无
刻度的直尺画图.
(1)在图1中,画出∠DAE 的平分线; (2)在图2中,画出∠AEC 的平分线.
111+=,x 1x 23
2,求代数式(x +1) 2-2(x +1) +1的值.
17.如图所示,有一电路AB 是由图示的开关控制,同时闭合a ,b ,c ,d 四个开关中的任
意两个开关,求使电路形成通路的概率.
18.某地出租车公司规定:出租车起步价(含燃油费)允许行驶的最远路程为1.5千米,超
过1.5千米的部分按每千米另收费. 小张说:“我乘这种出租车行驶了10.5千米,付了25.7元”;老刘说:“我花51元钱乘这种出租车行驶了21.5千米”.
试求这种出租车的起步价(含燃油费)是多少元?以及超过1.5千米后,每千米收费是多少元?
四、(本大题共4小题,每小题8分,共32分)
19.如图,已知A(n,-2) ,B(1,4) 是一次函数y=kx+b的图象和反比例函数y=两个交点,直线AB 与y 轴交于点C . (1)求反比例函数和一次函数的关系式; (2)求△AOC 的面积; (3)求不等式kx +b -
m
的图象的x
m
a
20.如图,一书架上的方格中放置四本厚度和长度相同的书,其中左边两本书紧贴书架方格内侧竖放,右边两本书自然向左斜放,支撑点为C ,E ,右侧书角正好靠在方格内侧上,若书架方格长BF =40,∠DCE =30°.
(1)设一本书的厚度为,则EF
(2)若书的长度AB =20cm ,求一本书的厚度(精确到0.1cm ,可用科学计算器). (参考数据:2=1. 414,=1. 732)
21.某校为了了解初中各年级学生每天的平均睡眠时间(单位:h ,精确到1h ),抽样调查了部分学生,并用得到的数据绘制了下面两幅不完整的统计图.
请你根据图中提供的信息,回答下列问题:
(1)求出扇形统计图中百分数a 的值为 ,所抽查的学生人数为 ; (2)求出平均睡眠时间为8小时的人数,并补全频数直方图; (3)求出这部分学生的平均睡眠时间的众数和平均数;
(4)如果该校共有学生1200名,请你估计睡眠不足(少于8小时)的学生数.
0) ,22.如图,将△AOB 置于平面直角坐标系中,其中点O 为坐标原点,点A 的坐标为(3,∠ABO =60.
(1)若△AOB 的外接圆与y 轴交于点D ,求D 点坐标;
,0) ,试猜想过D ,C 的直线与△AOB 的外接圆的位置关系,并(2)若点C 的坐标为(-1
加以说明.
五、(本大题共10分)
23.如图,抛物线y =x 2+bx +3顶点为P ,且与坐标轴交于A ,B 两点,点A 在x 轴的负半轴上,点B 在y 轴的正半轴上,tan ∠ABO =
1
. 3
(1)求抛物线的对称轴和函数的最小值;
(2)在抛物线的对称轴上是否存在这样的点D ,使△ABD 为直角三角形? 如果存在,求点D 的坐标;如果不存在,说明理由.
六、(本大题共12分)
24.如图1所示,△ABC 中,∠ACB =90°,AC=8,BC=6,D 为AB 的中点.如图2所示,将△ACD 沿射线AD 方向平移,得到△A′C′D′ 当点D′与点B 重合时,停止平移.在平移过程中,CB 与C′D′交于点E ,A′C′与CD 、CB 分别交于点F 、P ,阴影部分为两个三角形的重叠部分.
(1)在如图2中,猜想FD 与ED ′的数量关系,并证明你的猜想;
(2)设平移距离DD ′=x ,重叠部分面积为y ,请写出y 与x 的函数关系式,以及自变量的取值范围;
(3)对于(2)中的结论,是否存在这样的x 的值,使得阴影部分的面积是原△ABC 面积的
1
;若存在,求此时x 的值;若不存在,说明理由. 4
2015年中考数学冲刺卷(3)参考答案
一、选择题(本大题共6个小题,每小题3分,共18分)每小题只有一个正确选项. 1.B 【解析】根据绝对值的定义可直接得到结果为6.
2.A 【解析】圆柱体的左视图为矩形,切去一半应仍为矩形.
3.D 【解析】根据每个选项的特点,应用相应的运算法则,可知只有D 正确. 4.D 【解析】由方程的特点,结合根的判别式可知,方程有两个不相等的实数根,再根据求根公式可得方程的两个根都是无理数,且一个正根和一个负数,故D 正确. 5.C 【解析】 根据条件可知△ABE 是等边三角形,得到BE=5,再由AF =2EF ,利用相似形的性质即可求得BC 长为BE 的2倍.
6.C 【解析】 由已知二次函数的图象,可知a <0,b >0,故直线y =ax -b 一定过第二、三、四象限.
二、填空题(本大题共8小题,每小题3分,共24分) 7.—5 8.x ≥-3,且x ≠0 9.50 10.70° 11.7 12.50° 13.y =
15
14.2,2,4 x
三、(本大题共4小题,每小题6分,共24分)
15.解:(x +1) 2-2(x +1) +1=(x +1-1) 2=x 2,„„„„„„„„„„„„„„2分 ∴由x -1=
2,得(x +1) 2-2(x +1) +1=x 2=(2+1) 2=3+22.„„„„6分
16.画对图1得2分,画对图2得4分
17.解:所有等可能结果用树状图表示如下:
„„„„„„„„„„3分
共12种情况,其中能使电路形成通路的有ac , ad , bc , bd , ca , cb , da , db 共八种, 所以使电路形成通路的概率为
82
=. „„„„„„„„„„„„„„„„„„6分 123
18.解:设起步价是x 元,超过1.5千米后每千米收费y 元,„„„„„„„„„„1分
根据题意得⎨
⎧x +(10. 5-1. 5) y =25. 7
„„„„„„„„„„„„„„„„„„3分
⎩x +(21. 5-1. 5) y =51
解得 ⎨
⎧x =5
.„„„„„„„„„„„„„„„„„„„„„„„„„„5分
⎩y =2. 3
答:这种出租车的起步价是5元,超过1.5千米后每千米收费2.3元. „„„6分 四、(本大题共4小题,每小题8分,共32分) 19.解:(1)将B (1,4)代入y =∵A(n,-2) 也在反比例函数y =
m 4
中,得m =4,∴y =.„„„„„„„„1分 x x
4
的图象上, x
∴n =-2.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„2分 将A(-2,-2) ,B(1,4) 代入一次函数y=kx+b, 得⎨
⎧-2k +b =-2⎧k =2
,解得⎨,
⎩k +b =4⎩b =2
∴y=2x+2.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„4分 (2)∵当x =0时,y=2,∴OC=2.
1
⨯2⨯2=2.„„„„„„„„„„„„„„„„„„„„„„„„„6分 2
(3)x <-2或0<x <1.„„„„„„„„„„„„„„„„„„„„„„„„„8分
∴S ∆AOC =20. 解:(1)
4a „„„„„„„„„„„„„„„„„„„„„„„„„„„„2分 3
(2)∵AB =CE =20cm ,∠DCE =30°,
∴DE =10cm .„„„„„„„„„„„„„„„„„„„„„„„„„„„„„3分 由(1)中的结论可得,BF =BD+DE+EF=2a +10+ 化为(6+4) a =90,a ≈7.0cm .
4a =40,„„„„„„„6分 3
答:书的厚度约为7.0cm .„„„„„„„„„„„„„„„„„„„„„„„„8分
21.解:(1)45% 60人;„„„„„„„„„„„„„„„„„„„„„„„„2分 (2)平均睡眠时间为8小时的人数为60×30%=18人,„„„„„„„„„„„„3分
„„„„„„„„„„„„„„„„4分
(3)这部分学生的平均睡眠时间的众数是7小时,„„„„„„„„„„„„„„5分 平均数为
12⨯6+27⨯7+8⨯18+9⨯3
=7.2小时;„„„„„„„„„„„„6分
60
39
×1200=780人.„„„„„„„„„„„„„„8分 60
(4)∵抽取的60名学生中,睡眠时间在8小时以下的有12+27=39人, ∴1200名学生中睡眠不足的有
22.解:(1)连结AD ,则∠ADO =∠B =600, „„„„„„„„„„„„„„„„„1分
0) , 在Rt △ADO 中,∠ADO =600,点A 的坐标为(3,
∴OD =OA ÷tan ∠ODA =3÷=.„„„„„„„„„„„„„„„„„„„2分 ∴D 点的坐标是(0,).„„„„„„„„„„„„„„„„„„„„„„„„„3分 (2)猜想是CD 与圆相切,„„„„„„„„„„„„„„„„„„„„„„„„„4分 ∴∠AOD 是直角,∴AD 是圆的直径.„„„„„„„„„„„„„„„„„„„5分
,0) , 又∵若点C 的坐标为(-1
∴tan ∠CDO=
CO
=, ∠CDO =300.„„„„„„„„„„„„„„„„„„„7分 OD
∴∠CDA=∠CDO+∠ADO=900,即CD ⊥AD .
∴CD 切外接圆于点D .„„„„„„„„„„„„„„„„„„„„„„„„„„8分 五、(本大题共10分)
3) ,OB =3. ·23.解:(1)令x =0,则y =3.∴B 点坐标为(0,································ 1分
tan ∠AOB =
OA OA 1
==, AB 33
∴AO =1.由点A 在x 轴的负半轴上,得A 点坐标为(-1,0) . ···································· 2分
∴0=(-1) 2+b (-1) +3,解得b =4.
···································································· 3分 ∴所求的抛物线解析式为y =x 2+4x +3. ·配方,得y =(x +2) -1.
2
∴其对称轴为x =-2. ······································································································ 4分 且函数的最小值为-1. ····································································································· 5分 (3)存在这样的点D ,使△ABD 为直角三角形.„„„„„„„„„„„„„„„„6分 如图,设点D 的坐标为(-2,a ),对称轴与x 轴的交点为E (-2,0).
过点B 作BF 与对称轴垂直,垂足为F , 则有BF =2,DE =a ,OE =1,DF =3-a .
2
∴DA=a 2+1,BD =(3-a ) +4=
a 2-6a +13,AB =.
在图1中,若∠DAB=90°,
22222
则有BD =AD +AB ,a -6a +13=a +1+10,解得a =
1
;„„„„„„7分 3
在图2中,若∠ADB=90°,
22222
则有BD +AD =AB ,a -6a +13+a +1=10,解得a =1或2;„„„„„8分
在图3中,若∠DBA=90°,
AD 2=BD 2+AB 2,a 2+1=a 2-6a +13+10,解得a =
综上可知,存在这样的点D ,使△ABD 为直角三角形, 且点D 的坐标为(-2,
11
.„„„„„„„„9分 3
111),(-2,1),(-2,2),(-2,).„„„„„„„„„„10分 33
六、(本大题共12分) 24.(1)FD 与ED ′.„„„„„„„„„„„„„„„„„„„„„„„„„„„„1分 ∵CD ∥C ′D ′,∴∠2=∠C ′=∠1,∠5=∠C . 又因为∠ACB =90︒,CD 是斜边上的中线,
∴在图1中,DC =DA =DB ,即在图2中,C ′D ′=AD ′.
∴∠B =∠C =∠5,∠A ′=∠C ′=∠2.„„„„„„„„„„„„„„„„„„„„„2分 ∴A ′D =DF ,B D′=ED ′.
由△ACD 沿射线AD 方向平移,得到△A′C′D′,可知A ′D =D′ B.
∴DF =ED ′.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„3分
(2)因为在Rt ∆ABC 中,AC =8, BC =6,
∴AB =10.
即AD ′=C ′D ′=CD =DB =5.
又∵DD ′=x ,
∴A ′D =DF =D ′E =D ′B =5-x .∴CF=C′E= x.„„„„„„„„„„„„„„„„„„4分 在△BCD 中,点C 到BD 的距离就是∆ABC 的AB 边上的高,为24. 5
设△EBD′ 的边BD ′边上的高为h ,由D′E,得△E D′ B∽△CDB , 24(5-x ) h 5-x , h =. =255
5
112/(5-x ) 2. „„„„„„„„„„„„„„„„„„„6分 ∴S ∆ED /B =⨯BD ⨯h =225∴
又∵∠A ′+∠B =90°, ∴∠A ′PB =90°.
又∵∠C =∠B ,∴sin B =sin C =
∴CP =43, cos B =cos C =. 5534x , PF =x . 55
162x . „„„„„„„„„„„„„„„„„„„„„„7分 ∴S ∆CPF =⨯PC ⋅PF =225
11262(5-x ) 2-x . 即y =S ∆CDB -S ∆CFP -S ∆ED /B =S ∆ABC -22525
18224x +x (0≤x ≤5) . „„„„„„„„„„„„„„„„„„„„„9分 所以y =-255
(1) 存在. „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„10分
(2) 当y =11824S ∆ABC 时,即-x 2+x =6, 4255
52整理,得3x -20x +25=0. 解得,x 1=, x 2=5. 3
51即当x =或x =5时,重叠部分的面积等于原∆ABC 面积的. „„„„„„„„12分 34
2015年江西省中考数学冲刺卷(3)
说明:本卷共有六个大题,24个小题,全卷满分120分,考试时间120分钟.
一、选择题(本大题共6个小题,每小题3分,共18分)每小题只有一个正确选项. 1.-6的绝对值是( ).
11
A .-6 B .6 C D .
66
2.如图,图中的几何体是圆柱沿竖直方向切掉一半后得到的,则该几何体的左视图是( ).
3.下列运算正确的是( ). A .2x +3y =5xy C .a -(a -b ) =-b
B .a 3-a 2=a
D .(a -1)(a +2) =a 2+a -2
4.已知一元二次方程 x 2 + x ─ 1 = 0,下列判断正确的是( ). A . 该方程有两个相等的实数根 B . 该方程有一个根为1 C . 该方程没有实数根 D . 该方程有一个根为负数
5.如图,□ABCD 中,∠C =120°,AB =AE =5, AC 与BD 交于点F ,AF =2EF ,则BC 的长为( ).
A .6 B .8 C .10 D .12
6.已知y =
ax
2+bx
的图象如图所示,则y =ax -b 的图象一定过( ). A .第一、二、三象限 C .第二、三、四象限
二、填空题(本大题共8小题,每小题3分,共24分) 7.计算:2-7=. 8.函数y =
B .第一、二、四象限 D .第一、三、四象限
x +3
自变量的取值范围是. x
9.有一组数据:30,10, 50, 70,50.它们的中位数是
10.如图,点B 是AD 的延长线上一点,DE ∥AC ,AE 平分∠CAB ,∠C =50︒,∠E =30︒,则∠CDA 的度数等于 .
11.n 边形的一条对角线,将这个n 边形分成内角和分别为180°和720°的两个多边形,则
n
12.如图,直线l 与⊙O 相切于点C ,A 、B 、D 均在⊙O 上,OA ∥l ,∠BDC =85°,则∠
BAO 的度数为
13.已知P 1(x 1,2),P 2(x 2,3)是同一个反比例函数图象上的两点,若则这个反比例函数的表达式为 .
14.如图,五边形ABCDE 中,AB=BC=CD=AE=2,∠A =∠B =∠BCD =120°,点P 在五边形上,若∠CPD =30°,则CP 的长为 . 三、(本大题共4小题,每小题6分,共24分) 15.已知x -1=
16.如图,矩形ABCD 中,点E 在BC 上,AE=CE,试分别在下列两个图中按要求使用无
刻度的直尺画图.
(1)在图1中,画出∠DAE 的平分线; (2)在图2中,画出∠AEC 的平分线.
111+=,x 1x 23
2,求代数式(x +1) 2-2(x +1) +1的值.
17.如图所示,有一电路AB 是由图示的开关控制,同时闭合a ,b ,c ,d 四个开关中的任
意两个开关,求使电路形成通路的概率.
18.某地出租车公司规定:出租车起步价(含燃油费)允许行驶的最远路程为1.5千米,超
过1.5千米的部分按每千米另收费. 小张说:“我乘这种出租车行驶了10.5千米,付了25.7元”;老刘说:“我花51元钱乘这种出租车行驶了21.5千米”.
试求这种出租车的起步价(含燃油费)是多少元?以及超过1.5千米后,每千米收费是多少元?
四、(本大题共4小题,每小题8分,共32分)
19.如图,已知A(n,-2) ,B(1,4) 是一次函数y=kx+b的图象和反比例函数y=两个交点,直线AB 与y 轴交于点C . (1)求反比例函数和一次函数的关系式; (2)求△AOC 的面积; (3)求不等式kx +b -
m
的图象的x
m
a
20.如图,一书架上的方格中放置四本厚度和长度相同的书,其中左边两本书紧贴书架方格内侧竖放,右边两本书自然向左斜放,支撑点为C ,E ,右侧书角正好靠在方格内侧上,若书架方格长BF =40,∠DCE =30°.
(1)设一本书的厚度为,则EF
(2)若书的长度AB =20cm ,求一本书的厚度(精确到0.1cm ,可用科学计算器). (参考数据:2=1. 414,=1. 732)
21.某校为了了解初中各年级学生每天的平均睡眠时间(单位:h ,精确到1h ),抽样调查了部分学生,并用得到的数据绘制了下面两幅不完整的统计图.
请你根据图中提供的信息,回答下列问题:
(1)求出扇形统计图中百分数a 的值为 ,所抽查的学生人数为 ; (2)求出平均睡眠时间为8小时的人数,并补全频数直方图; (3)求出这部分学生的平均睡眠时间的众数和平均数;
(4)如果该校共有学生1200名,请你估计睡眠不足(少于8小时)的学生数.
0) ,22.如图,将△AOB 置于平面直角坐标系中,其中点O 为坐标原点,点A 的坐标为(3,∠ABO =60.
(1)若△AOB 的外接圆与y 轴交于点D ,求D 点坐标;
,0) ,试猜想过D ,C 的直线与△AOB 的外接圆的位置关系,并(2)若点C 的坐标为(-1
加以说明.
五、(本大题共10分)
23.如图,抛物线y =x 2+bx +3顶点为P ,且与坐标轴交于A ,B 两点,点A 在x 轴的负半轴上,点B 在y 轴的正半轴上,tan ∠ABO =
1
. 3
(1)求抛物线的对称轴和函数的最小值;
(2)在抛物线的对称轴上是否存在这样的点D ,使△ABD 为直角三角形? 如果存在,求点D 的坐标;如果不存在,说明理由.
六、(本大题共12分)
24.如图1所示,△ABC 中,∠ACB =90°,AC=8,BC=6,D 为AB 的中点.如图2所示,将△ACD 沿射线AD 方向平移,得到△A′C′D′ 当点D′与点B 重合时,停止平移.在平移过程中,CB 与C′D′交于点E ,A′C′与CD 、CB 分别交于点F 、P ,阴影部分为两个三角形的重叠部分.
(1)在如图2中,猜想FD 与ED ′的数量关系,并证明你的猜想;
(2)设平移距离DD ′=x ,重叠部分面积为y ,请写出y 与x 的函数关系式,以及自变量的取值范围;
(3)对于(2)中的结论,是否存在这样的x 的值,使得阴影部分的面积是原△ABC 面积的
1
;若存在,求此时x 的值;若不存在,说明理由. 4
2015年中考数学冲刺卷(3)参考答案
一、选择题(本大题共6个小题,每小题3分,共18分)每小题只有一个正确选项. 1.B 【解析】根据绝对值的定义可直接得到结果为6.
2.A 【解析】圆柱体的左视图为矩形,切去一半应仍为矩形.
3.D 【解析】根据每个选项的特点,应用相应的运算法则,可知只有D 正确. 4.D 【解析】由方程的特点,结合根的判别式可知,方程有两个不相等的实数根,再根据求根公式可得方程的两个根都是无理数,且一个正根和一个负数,故D 正确. 5.C 【解析】 根据条件可知△ABE 是等边三角形,得到BE=5,再由AF =2EF ,利用相似形的性质即可求得BC 长为BE 的2倍.
6.C 【解析】 由已知二次函数的图象,可知a <0,b >0,故直线y =ax -b 一定过第二、三、四象限.
二、填空题(本大题共8小题,每小题3分,共24分) 7.—5 8.x ≥-3,且x ≠0 9.50 10.70° 11.7 12.50° 13.y =
15
14.2,2,4 x
三、(本大题共4小题,每小题6分,共24分)
15.解:(x +1) 2-2(x +1) +1=(x +1-1) 2=x 2,„„„„„„„„„„„„„„2分 ∴由x -1=
2,得(x +1) 2-2(x +1) +1=x 2=(2+1) 2=3+22.„„„„6分
16.画对图1得2分,画对图2得4分
17.解:所有等可能结果用树状图表示如下:
„„„„„„„„„„3分
共12种情况,其中能使电路形成通路的有ac , ad , bc , bd , ca , cb , da , db 共八种, 所以使电路形成通路的概率为
82
=. „„„„„„„„„„„„„„„„„„6分 123
18.解:设起步价是x 元,超过1.5千米后每千米收费y 元,„„„„„„„„„„1分
根据题意得⎨
⎧x +(10. 5-1. 5) y =25. 7
„„„„„„„„„„„„„„„„„„3分
⎩x +(21. 5-1. 5) y =51
解得 ⎨
⎧x =5
.„„„„„„„„„„„„„„„„„„„„„„„„„„5分
⎩y =2. 3
答:这种出租车的起步价是5元,超过1.5千米后每千米收费2.3元. „„„6分 四、(本大题共4小题,每小题8分,共32分) 19.解:(1)将B (1,4)代入y =∵A(n,-2) 也在反比例函数y =
m 4
中,得m =4,∴y =.„„„„„„„„1分 x x
4
的图象上, x
∴n =-2.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„2分 将A(-2,-2) ,B(1,4) 代入一次函数y=kx+b, 得⎨
⎧-2k +b =-2⎧k =2
,解得⎨,
⎩k +b =4⎩b =2
∴y=2x+2.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„4分 (2)∵当x =0时,y=2,∴OC=2.
1
⨯2⨯2=2.„„„„„„„„„„„„„„„„„„„„„„„„„6分 2
(3)x <-2或0<x <1.„„„„„„„„„„„„„„„„„„„„„„„„„8分
∴S ∆AOC =20. 解:(1)
4a „„„„„„„„„„„„„„„„„„„„„„„„„„„„2分 3
(2)∵AB =CE =20cm ,∠DCE =30°,
∴DE =10cm .„„„„„„„„„„„„„„„„„„„„„„„„„„„„„3分 由(1)中的结论可得,BF =BD+DE+EF=2a +10+ 化为(6+4) a =90,a ≈7.0cm .
4a =40,„„„„„„„6分 3
答:书的厚度约为7.0cm .„„„„„„„„„„„„„„„„„„„„„„„„8分
21.解:(1)45% 60人;„„„„„„„„„„„„„„„„„„„„„„„„2分 (2)平均睡眠时间为8小时的人数为60×30%=18人,„„„„„„„„„„„„3分
„„„„„„„„„„„„„„„„4分
(3)这部分学生的平均睡眠时间的众数是7小时,„„„„„„„„„„„„„„5分 平均数为
12⨯6+27⨯7+8⨯18+9⨯3
=7.2小时;„„„„„„„„„„„„6分
60
39
×1200=780人.„„„„„„„„„„„„„„8分 60
(4)∵抽取的60名学生中,睡眠时间在8小时以下的有12+27=39人, ∴1200名学生中睡眠不足的有
22.解:(1)连结AD ,则∠ADO =∠B =600, „„„„„„„„„„„„„„„„„1分
0) , 在Rt △ADO 中,∠ADO =600,点A 的坐标为(3,
∴OD =OA ÷tan ∠ODA =3÷=.„„„„„„„„„„„„„„„„„„„2分 ∴D 点的坐标是(0,).„„„„„„„„„„„„„„„„„„„„„„„„„3分 (2)猜想是CD 与圆相切,„„„„„„„„„„„„„„„„„„„„„„„„„4分 ∴∠AOD 是直角,∴AD 是圆的直径.„„„„„„„„„„„„„„„„„„„5分
,0) , 又∵若点C 的坐标为(-1
∴tan ∠CDO=
CO
=, ∠CDO =300.„„„„„„„„„„„„„„„„„„„7分 OD
∴∠CDA=∠CDO+∠ADO=900,即CD ⊥AD .
∴CD 切外接圆于点D .„„„„„„„„„„„„„„„„„„„„„„„„„„8分 五、(本大题共10分)
3) ,OB =3. ·23.解:(1)令x =0,则y =3.∴B 点坐标为(0,································ 1分
tan ∠AOB =
OA OA 1
==, AB 33
∴AO =1.由点A 在x 轴的负半轴上,得A 点坐标为(-1,0) . ···································· 2分
∴0=(-1) 2+b (-1) +3,解得b =4.
···································································· 3分 ∴所求的抛物线解析式为y =x 2+4x +3. ·配方,得y =(x +2) -1.
2
∴其对称轴为x =-2. ······································································································ 4分 且函数的最小值为-1. ····································································································· 5分 (3)存在这样的点D ,使△ABD 为直角三角形.„„„„„„„„„„„„„„„„6分 如图,设点D 的坐标为(-2,a ),对称轴与x 轴的交点为E (-2,0).
过点B 作BF 与对称轴垂直,垂足为F , 则有BF =2,DE =a ,OE =1,DF =3-a .
2
∴DA=a 2+1,BD =(3-a ) +4=
a 2-6a +13,AB =.
在图1中,若∠DAB=90°,
22222
则有BD =AD +AB ,a -6a +13=a +1+10,解得a =
1
;„„„„„„7分 3
在图2中,若∠ADB=90°,
22222
则有BD +AD =AB ,a -6a +13+a +1=10,解得a =1或2;„„„„„8分
在图3中,若∠DBA=90°,
AD 2=BD 2+AB 2,a 2+1=a 2-6a +13+10,解得a =
综上可知,存在这样的点D ,使△ABD 为直角三角形, 且点D 的坐标为(-2,
11
.„„„„„„„„9分 3
111),(-2,1),(-2,2),(-2,).„„„„„„„„„„10分 33
六、(本大题共12分) 24.(1)FD 与ED ′.„„„„„„„„„„„„„„„„„„„„„„„„„„„„1分 ∵CD ∥C ′D ′,∴∠2=∠C ′=∠1,∠5=∠C . 又因为∠ACB =90︒,CD 是斜边上的中线,
∴在图1中,DC =DA =DB ,即在图2中,C ′D ′=AD ′.
∴∠B =∠C =∠5,∠A ′=∠C ′=∠2.„„„„„„„„„„„„„„„„„„„„„2分 ∴A ′D =DF ,B D′=ED ′.
由△ACD 沿射线AD 方向平移,得到△A′C′D′,可知A ′D =D′ B.
∴DF =ED ′.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„3分
(2)因为在Rt ∆ABC 中,AC =8, BC =6,
∴AB =10.
即AD ′=C ′D ′=CD =DB =5.
又∵DD ′=x ,
∴A ′D =DF =D ′E =D ′B =5-x .∴CF=C′E= x.„„„„„„„„„„„„„„„„„„4分 在△BCD 中,点C 到BD 的距离就是∆ABC 的AB 边上的高,为24. 5
设△EBD′ 的边BD ′边上的高为h ,由D′E,得△E D′ B∽△CDB , 24(5-x ) h 5-x , h =. =255
5
112/(5-x ) 2. „„„„„„„„„„„„„„„„„„„6分 ∴S ∆ED /B =⨯BD ⨯h =225∴
又∵∠A ′+∠B =90°, ∴∠A ′PB =90°.
又∵∠C =∠B ,∴sin B =sin C =
∴CP =43, cos B =cos C =. 5534x , PF =x . 55
162x . „„„„„„„„„„„„„„„„„„„„„„7分 ∴S ∆CPF =⨯PC ⋅PF =225
11262(5-x ) 2-x . 即y =S ∆CDB -S ∆CFP -S ∆ED /B =S ∆ABC -22525
18224x +x (0≤x ≤5) . „„„„„„„„„„„„„„„„„„„„„9分 所以y =-255
(1) 存在. „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„10分
(2) 当y =11824S ∆ABC 时,即-x 2+x =6, 4255
52整理,得3x -20x +25=0. 解得,x 1=, x 2=5. 3
51即当x =或x =5时,重叠部分的面积等于原∆ABC 面积的. „„„„„„„„12分 34