(满分150分,考试时间100分钟)(2017.4)
一、选择题:(本大题共6题,每题4分,满分24分)
【下列各题的四个选项中,有且只有一个选项是正确的,选择正确项的代号并填涂在答题纸的相应位置上】
1. 如果a 表示不为0的任意一个实数,那么下列四个算式中,正确的是 ··········· ( ) (A )3a -2a =a ; (B )a ⋅a =a ; 2. 在解答“一元二次方程
3
2
3
1
3
32
(C )a ÷a =a ; (D )(a ) =a .
122
121
x -x +a =0的根的判别式为”的过程中,某班同学的作业22
1
-2a ; 4
中出现了下面几种答案,其中正确的答案是 ············································ ( ) (A )
1
-2a ≥0; 4
2
(B )
(C )1-8a ≥0; (D )1-8a .
3. 如果函数y =ax +2x +1的图像不经过第四象限,那么实数a 的取值范围为 ( ) (A )a
(B )a =0;
(C )a >0;
(D )a ≥0.
4. 从概率统计的角度解读下列诗词所描述的事件,其中属于确定事件的是 ········ ( ) (A )黄梅时节家家雨,青草池塘处处蛙; (C )水面上秤锤浮,直待黄河彻底枯;
(B )人间四月芳菲尽,山寺桃花始盛开; (D )一夜北风紧,开门雪尚飘.
5. 已知⊙A 的半径长为2,⊙B 的半径长为5,如果⊙A 与⊙B 内含,那么圆心距AB 的长度可以为 ························································ ······························ ·················· ( ) (A )0;
(B )3;
(C )6;
(D )9.
6. 将两个底边相等的等腰三角形按照图1所示的方式拼接在一起(隐藏互相重合的底边)的图形俗称为“筝形”. 假如“筝形”下个定义,那么下面四种说法中,你认为最能够描述“筝形”特征的是 ·································································································· ( )
(A )有两组邻边相等的四边形称为“筝形”; (B )有两组对角分别相等的四边形称为“筝形”; (C )两条对角线互相垂直的四边形称为“筝形”;
(D )以一条对角线所在直线为对称轴的四边形称为“筝形”. 二、填空题:(本大题共12题,每题4分,满分48分) 【请直接将结果填入答题纸的相应位置】 7. 计算:()
1
12
-1
=.
8. 已知≈1. 73,那么
)
≈ (保留两个有效数字........
3
⎧2x
9. 不等式组⎨的解集是 .
x +1>0⎩
10. 方程x +2=x 的实数解是.
2
11. 已知点A (x 1, y 1) 、点B (x 2, y 2) 在反比例函数y =-的图像上. 如果x 1
x
的大小关系为:y 1y 2(从“”中选择).
12. 某校学生综合素质评价方案中有这样一段话:“学生自评、同学互评与班级评定小组评价在学生综合素质评价中所占的权重分别为10%、30%、60%”. 如果甄聪明同学的自评分数、同学互评分数、班级评定小组给出的分数分别为96分、95分、95分,那么甄聪明同学的综合素质评价分数为 分.
13. 一名射击运动员连续打靶9次,假如他打靶命中环数的情况如图2所示,那么该射击运动员本次打靶命中环数的中位数为 环.
14. 如果非零向量a 与向量b 的方向相反,且2a =3b ,那么向量a 为 (用向量b 表示).
15. 从山底A 点测得位于山顶B 点的仰角为30︒,那么从B 点测得A 点的俯角为度. 16. 已知扇形的弧长为8,如果该扇形的半径长为2,那么这个扇形的面积为 17. 命题“相等的角不一定是对顶角”是. 18. 已知在△ABC 中,∠ACB =90︒,AB =10,cos A =
3
(如图3),将△ABC 绕着点C 旋5
B 转,点A 、B 的对应点分别记为A '、B ',A 'B '与边AB 相交于点E .如果A 'B '⊥AC ,那么线段B 'E 的长为 . 三、解答题:(本大题共7题,满分78分) 19.(本题满分10分)
421
+-先化简,再求值:2,其中x =2. x -4x +2x -2
图3
C
20.(本题满分10分)
⎧x -y =2,
解方程组:⎨2 2
⎩x -2xy -3y =0.
21.(本题满分10分,每小题5分)
将大小相同,形状也相同的三个菱形按照图4的方式拼接在一起(其中,点B 、C 、F 、G 在同一条直线上),AB =3. 联结AG ,AG 与EF 相交于点P . (1)求线段EP 的长;
(2)如果∠B =60︒,求△APE 的面积.
22.(本题满分10分,第(1)小题6分;第(2)小题4分)
某种型号的家用车在高速公路上匀速行驶时,测得部分数据如下表:
图4
F
H
G
(1)如果该车的油箱内剩余油量y (升)与行驶路程x (千米)之间是一次函数关系,求y 关
于x 的函数解析式(不需要写出它的定义域);
(2)张老师租赁该型号的家用车也在该高速公路的相同路段以相同的速度匀速行驶300千米(不
考虑小轿车载客的人数以及堵车等因素). 假如不在高速公路上的服务区加油,那么在上高速公路之前,张老师这辆车的油箱内至少需要有多少升汽油?请根据题目中提供的相关信息..简要说明理由.
23.(本题满分12分,每小题6分)
已知:正方形ABCD ,点E 在边CD 上,点F 在线段BE 的延长线上,且∠FCE =∠CBE . (1)如图5,当点E 为CD 边的中点时,求证:CF =2EF ; (2)如图6,当点F 位于线段AD 的延长线上,求证:
B
图5
图6
EF DE
=. BE DF
F
D F
24.(本题满分12分,每小题4分)
在平面直角坐标系xOy (如图7)中,已知点A 的坐标为(3, 1),点B 的坐标为(6,5),点C 的坐标为(0, 5);某二次函数的图像经过点A 、点B 与点C . (1)求这个二次函数的解析式;
(2)假如点Q 在该函数图像的对称轴上,且△ACQ 是等腰三角形,直接写出点Q 的坐标; ..(3)如果第一象限内的点P 在(1)中求出的二次函数的图像上,且tan ∠PCA =1
2
,求∠PCB 的正弦值.
25.(满分14分,第(1)小题5分,第(2)小题5分、第(3)小题4分)
已知:AB =8,⊙O 经过点A 、B . 以AB 为一边画平行四边形ABCD ,另一边CD 经过点
O (如图8). 以点B 为圆心,BC 为半径画弧,交线段OC 于点E (点E 不与点O 、点C 重合).
(1)求证:OD =OE ;
(2)如果⊙O 的半径长为5(如图9),设OD =x ,BC =y ,求y 关于x 的函数解析式,并写
出它的定义域;
(3)如果⊙O 的半径长为5,联结AC ,当BE ⊥AC 时,求OD 的长.
图8
图9
备用图
2016学年嘉定区九年级第二次质量调研
数学试卷参考答案
一、选择题:(本大题共6题,每题4分,满分24分) 1、C ;2、B ;3、D ;4、C ;5、A ;6、D.
二、填空题:(本大题共12题,每题4分,满分48分)
33 7、2;8、0. 58;9、-1;12、95. 1;13、9环;14、a =-b ;
22
24
15、30︒;16、8;17、真命题;18、.
5
三、解答题:(本大题共7题,满分78分) 19.(本题满分10分) 解:
42142(x -2) x +2
+-=+- ······ 3分 x 2-4x +2x -2(x +2)(x -2) (x +2)(x -2) (x +2)(x -2)
=
4+2x -4-x -2(x -2) 1
==. ···································· 2+2+1分
(x +2)(x -2) (x +2)(x -2) x +2
当x =2时,原式=
12
. ···················································· 2分 =1-22+2
20.(本题满分10分)
22
解:x -2xy -3y =0可以化为:(x -3y )(x +y ) =0,
所以:x -3y =0或x +y =0. ································································· 2分
原方程组可以化为:⎨
⎧x -y =2,
(Ⅰ)与
x -3y =0⎩⎧x -y =2,
(Ⅱ) ·························· 2分 ⎨
x +y =0⎩
⎧x =3,
解(Ⅰ)得⎨; 解(Ⅱ)得
y =1⎩
所以,原方程组的解为:⎨
⎧x =1,
················································ 2+2分 ⎨y =-1⎩
⎧x 1=3, ⎧x 2=1, 与⎨ ················································ 2分 y =1; y =-1. ⎩1⎩2
21.(本题满分10分,每小题5分)
解:(1)由题意得四边形ABGH 、ABFE 是平行四边形. ·································· 1分 ∴ AE ∥FG . ···················································································· 1分
EP AE
=. ······················································································ 1分 FP FG
EP EP 2
=2,即= 将AE =6,FG =3代入,得 ································· 1分 FP EF 3
又∵四边形ABFE 是平行四边形,AB =3,∴EF =AB =3. ∴EP =2. ··········· 1分 (2)过点P 作PH ⊥AE ,垂足为H (如图4). ········································ 1分 ∵四边形ABFE 是平行四边形,∠B =60︒,∴∠PEH =∠B =60︒. ············ 1分 在Rt △PEH 中,∠PHE =90︒,∠PEH =60︒,EP =2,
∴
∴PH =EP ⋅sin 60︒=2⨯∴△APE 的面积为
22.(本题满分10分)
解:(1)设油箱内剩余油量y (升) 与行驶路程x (千米)之间的函数
C
图4
3
····················································· 2分 =. ·
2
11
AE ⋅PH =⨯6⨯3=33. ··································· 1分 22
E
F
关系式为y =kx +b . ······················································································· 1分
分别将x =100,y =52;x =150,y =48代入上式,得⎨
⎧100k +b =52,
······· 2分
⎩150k +b =48.
2⎧
⎪k =-, 解得:⎨·················································································· 2分 25 ·
⎪⎩b =60.
2
x +60 ························································ 1分 25
(2)方法1:由题意可得,该型号的汽车在该路段行驶时,每行驶100耗油8升. ·· 2分 设行驶300公里时需要耗油x 升,可得300:100=x :8,解得x =24升. ············· 1分
2
x +60,得y =36. ·方法2:将x =300代入y =-····································· 2分 25
60-36=24. ··············································································· 1分
答:张老师的这辆车的油箱内至少需要有24升汽油. ······································· 1分 ..
∴所求的函数关系式为y =-
备注:学生若是在得到24升油的基础上又考虑了其它因素(如离开高速公路之后还需要再行驶一段路程才可以抵达目的地(或寻找到加油站),因此给出了大于24升油的其它数据,只要能够自圆其说,且符合生活实际情况,那么可以酌情评分. 23.(本题满分12分,每小题6分)
(1)证明:∵四边形ABCD 是正方形,∴CD =BC . ·········································· 1分
11
CD =BC . ·································· 1分 22
∵∠FCD =∠CBE ,∠F =∠F ,∴△FCE ∽△FBC . ···························· 2分 EF CE
=∴. ·················································································· 1分 CF BC
1EF 1
=. 即CF =2EF . ·又∵CE =BC ,∴··········································· 1分
2CF 2
(2)∵四边形ABCD 是正方形,∴DE ∥AB ,AD ∥BC ,AD =CD . ················ 1分
∵点E 为CD 边的中点,∴CE =
∵点F 位于线段AD 的延长线上,DE ∥AB ,∴ 又∵AD =CD ,∴
EF DF
=. ························ 1分 BE AD
EF DF
=. (1) ··························································· 1分 BE CD
∵AF ∥BC ,∴∠DFE =∠CBE .
又∵∠DCF =∠CBE ,∴∠DFE =∠DCF . ················································ 1分 又∵∠FDE =∠CDF ,∴△FDE ∽△CDF . ················································· 1分
DE DF EF DE
== ∴(2). 由(1)、(2)得 . ······································· 1分 DF CD BE DF
B
图5
图6
F
F
2
24.(本题满分12分,每小题4分)
解:(1)设所求二次函数的解析式为y =ax +bx +c ,将A (3, 1)、B (6,5)、C (0, 5)
⎧9a +3b +c =1,
48⎪
代入,得 ⎨36a +6b +c =5, 解得 a =,b =-,c =5. ································ 3分
93⎪c =5.
⎩428
x -x +5. ·········································· 1分 93
25
(2)Q 1(3, 6) ,Q 2(3, -4) ,Q 3(3, 9) ,Q 4(3, ) . ··········································· 4分
8
所以,这个二次函数的解析式为y =
(3)由题意得,该二次函数图像的对称轴为直线x =3. ···································· 1分
联结PC 交直线x =3于点M ,过点M 作MN ⊥AC ,垂足为N (图7-1) . 将直线x =3与BC 的交点记为H ,易得CH =3,AH =4,AC =5.
CH 3
= ········································································ 1分 CA 5
1
故可设MN =3k ,则AM =5k ,AM =4k . 又∵tan ∠PCA =,则CN =6k .
2
1553
由题意得方程:4k +6k =5. 解得k =,AM =,MH =4-= ·········· 1分
2222
∴sin ∠CAH =∴CM =3+() =
2
3
2
2
3MH . ···························· 1分 =5. ∴sin ∠PCB =
CM 52
25.(满分14分,第(1)小题5分,第(2)小题5分,第(3)小题4分)
解:(1)联结OA 、OB (如图8-1) ,易得OA =OB ,∠OAB =∠OBA . ··················· 1分
∵四边形ABCD 是平行四边形,∴AB ∥CD ,AD =BC .
∵BE =BC ,AD =BC ,∴AD =BE . ····················································· 1分 又 ∵AB ∥CD ,∴四边形ABED 是等腰梯形. ∴∠DAB =∠EBA . ····················· 1分 又 ∵∠OAB =∠OBA ,∴∠DAB -∠OAB =∠EBA -∠OBA .
即 ∠O A D =∠O B E . ·················································································· 1分
在△AOD 和△BOE 中,∵OA =OB ,∠OAD =∠OBE ,AD =BE ,
∴△AOD ≌△BOE. ∴OD =OE . ·····················
··
··· 1分
图8-1
图8-2
图8-3
方法2:∵∠ADE =∠BED ,∠DAO =∠EBO ,AD =BE ,∴△AOD ≌△BOE. „„ 方法3:∵∠ADE =∠BED ,∠DAO =∠EBO ,OA =OB ,∴△AOD ≌△BOE. „„ 方法4:如图8-2,过点O 作OH ⊥AB ,过点D 作DG ⊥AB ,过点E 作EI ⊥AB . „„ 方法5:如图8-3,过点O 作OH ⊥AB ,垂足为H ,联结DH 、EH . „„
(2)方法1:如图9-1,过点O 作OH ⊥AB ,垂足为H ,过点D 作DG ⊥AB ,垂足为G . 联结OB ,OH =3,AH =BH =4,得1分;得到DG =OH =3,得2分;在Rt △ADG 中,
201704嘉定区初三二模试卷-----------------------------p11
222写出AG =4-x ,AD =BC =y ,得1分;利用AD =DG +AG 得到y =x 2-8x +25,
得1分,函数定义域0
(3)如图10-1,过点O 作OM ⊥AC ,交AC 于点M ,交AB 于点N . 证明四边形ONBE 是平行四边形,得1分;利用BN =OE =OD ,AB =CD 得到OC =AN ,得1分;利用△AMN ≌
AM AN =得到AM =CN ,进而得到OM 是AC 的垂直平分线,OC =OA =5,CM CO
得1分;利用CD =AB =8,OC =5得到OD =3,得1分. △CMO 或
方法2. 如图10-,2;方法3:如图10-3;方法4(利用圆周角,略).
图
9-1 图
10-1 图
10-2 图10-3
(满分150分,考试时间100分钟)(2017.4)
一、选择题:(本大题共6题,每题4分,满分24分)
【下列各题的四个选项中,有且只有一个选项是正确的,选择正确项的代号并填涂在答题纸的相应位置上】
1. 如果a 表示不为0的任意一个实数,那么下列四个算式中,正确的是 ··········· ( ) (A )3a -2a =a ; (B )a ⋅a =a ; 2. 在解答“一元二次方程
3
2
3
1
3
32
(C )a ÷a =a ; (D )(a ) =a .
122
121
x -x +a =0的根的判别式为”的过程中,某班同学的作业22
1
-2a ; 4
中出现了下面几种答案,其中正确的答案是 ············································ ( ) (A )
1
-2a ≥0; 4
2
(B )
(C )1-8a ≥0; (D )1-8a .
3. 如果函数y =ax +2x +1的图像不经过第四象限,那么实数a 的取值范围为 ( ) (A )a
(B )a =0;
(C )a >0;
(D )a ≥0.
4. 从概率统计的角度解读下列诗词所描述的事件,其中属于确定事件的是 ········ ( ) (A )黄梅时节家家雨,青草池塘处处蛙; (C )水面上秤锤浮,直待黄河彻底枯;
(B )人间四月芳菲尽,山寺桃花始盛开; (D )一夜北风紧,开门雪尚飘.
5. 已知⊙A 的半径长为2,⊙B 的半径长为5,如果⊙A 与⊙B 内含,那么圆心距AB 的长度可以为 ························································ ······························ ·················· ( ) (A )0;
(B )3;
(C )6;
(D )9.
6. 将两个底边相等的等腰三角形按照图1所示的方式拼接在一起(隐藏互相重合的底边)的图形俗称为“筝形”. 假如“筝形”下个定义,那么下面四种说法中,你认为最能够描述“筝形”特征的是 ·································································································· ( )
(A )有两组邻边相等的四边形称为“筝形”; (B )有两组对角分别相等的四边形称为“筝形”; (C )两条对角线互相垂直的四边形称为“筝形”;
(D )以一条对角线所在直线为对称轴的四边形称为“筝形”. 二、填空题:(本大题共12题,每题4分,满分48分) 【请直接将结果填入答题纸的相应位置】 7. 计算:()
1
12
-1
=.
8. 已知≈1. 73,那么
)
≈ (保留两个有效数字........
3
⎧2x
9. 不等式组⎨的解集是 .
x +1>0⎩
10. 方程x +2=x 的实数解是.
2
11. 已知点A (x 1, y 1) 、点B (x 2, y 2) 在反比例函数y =-的图像上. 如果x 1
x
的大小关系为:y 1y 2(从“”中选择).
12. 某校学生综合素质评价方案中有这样一段话:“学生自评、同学互评与班级评定小组评价在学生综合素质评价中所占的权重分别为10%、30%、60%”. 如果甄聪明同学的自评分数、同学互评分数、班级评定小组给出的分数分别为96分、95分、95分,那么甄聪明同学的综合素质评价分数为 分.
13. 一名射击运动员连续打靶9次,假如他打靶命中环数的情况如图2所示,那么该射击运动员本次打靶命中环数的中位数为 环.
14. 如果非零向量a 与向量b 的方向相反,且2a =3b ,那么向量a 为 (用向量b 表示).
15. 从山底A 点测得位于山顶B 点的仰角为30︒,那么从B 点测得A 点的俯角为度. 16. 已知扇形的弧长为8,如果该扇形的半径长为2,那么这个扇形的面积为 17. 命题“相等的角不一定是对顶角”是. 18. 已知在△ABC 中,∠ACB =90︒,AB =10,cos A =
3
(如图3),将△ABC 绕着点C 旋5
B 转,点A 、B 的对应点分别记为A '、B ',A 'B '与边AB 相交于点E .如果A 'B '⊥AC ,那么线段B 'E 的长为 . 三、解答题:(本大题共7题,满分78分) 19.(本题满分10分)
421
+-先化简,再求值:2,其中x =2. x -4x +2x -2
图3
C
20.(本题满分10分)
⎧x -y =2,
解方程组:⎨2 2
⎩x -2xy -3y =0.
21.(本题满分10分,每小题5分)
将大小相同,形状也相同的三个菱形按照图4的方式拼接在一起(其中,点B 、C 、F 、G 在同一条直线上),AB =3. 联结AG ,AG 与EF 相交于点P . (1)求线段EP 的长;
(2)如果∠B =60︒,求△APE 的面积.
22.(本题满分10分,第(1)小题6分;第(2)小题4分)
某种型号的家用车在高速公路上匀速行驶时,测得部分数据如下表:
图4
F
H
G
(1)如果该车的油箱内剩余油量y (升)与行驶路程x (千米)之间是一次函数关系,求y 关
于x 的函数解析式(不需要写出它的定义域);
(2)张老师租赁该型号的家用车也在该高速公路的相同路段以相同的速度匀速行驶300千米(不
考虑小轿车载客的人数以及堵车等因素). 假如不在高速公路上的服务区加油,那么在上高速公路之前,张老师这辆车的油箱内至少需要有多少升汽油?请根据题目中提供的相关信息..简要说明理由.
23.(本题满分12分,每小题6分)
已知:正方形ABCD ,点E 在边CD 上,点F 在线段BE 的延长线上,且∠FCE =∠CBE . (1)如图5,当点E 为CD 边的中点时,求证:CF =2EF ; (2)如图6,当点F 位于线段AD 的延长线上,求证:
B
图5
图6
EF DE
=. BE DF
F
D F
24.(本题满分12分,每小题4分)
在平面直角坐标系xOy (如图7)中,已知点A 的坐标为(3, 1),点B 的坐标为(6,5),点C 的坐标为(0, 5);某二次函数的图像经过点A 、点B 与点C . (1)求这个二次函数的解析式;
(2)假如点Q 在该函数图像的对称轴上,且△ACQ 是等腰三角形,直接写出点Q 的坐标; ..(3)如果第一象限内的点P 在(1)中求出的二次函数的图像上,且tan ∠PCA =1
2
,求∠PCB 的正弦值.
25.(满分14分,第(1)小题5分,第(2)小题5分、第(3)小题4分)
已知:AB =8,⊙O 经过点A 、B . 以AB 为一边画平行四边形ABCD ,另一边CD 经过点
O (如图8). 以点B 为圆心,BC 为半径画弧,交线段OC 于点E (点E 不与点O 、点C 重合).
(1)求证:OD =OE ;
(2)如果⊙O 的半径长为5(如图9),设OD =x ,BC =y ,求y 关于x 的函数解析式,并写
出它的定义域;
(3)如果⊙O 的半径长为5,联结AC ,当BE ⊥AC 时,求OD 的长.
图8
图9
备用图
2016学年嘉定区九年级第二次质量调研
数学试卷参考答案
一、选择题:(本大题共6题,每题4分,满分24分) 1、C ;2、B ;3、D ;4、C ;5、A ;6、D.
二、填空题:(本大题共12题,每题4分,满分48分)
33 7、2;8、0. 58;9、-1;12、95. 1;13、9环;14、a =-b ;
22
24
15、30︒;16、8;17、真命题;18、.
5
三、解答题:(本大题共7题,满分78分) 19.(本题满分10分) 解:
42142(x -2) x +2
+-=+- ······ 3分 x 2-4x +2x -2(x +2)(x -2) (x +2)(x -2) (x +2)(x -2)
=
4+2x -4-x -2(x -2) 1
==. ···································· 2+2+1分
(x +2)(x -2) (x +2)(x -2) x +2
当x =2时,原式=
12
. ···················································· 2分 =1-22+2
20.(本题满分10分)
22
解:x -2xy -3y =0可以化为:(x -3y )(x +y ) =0,
所以:x -3y =0或x +y =0. ································································· 2分
原方程组可以化为:⎨
⎧x -y =2,
(Ⅰ)与
x -3y =0⎩⎧x -y =2,
(Ⅱ) ·························· 2分 ⎨
x +y =0⎩
⎧x =3,
解(Ⅰ)得⎨; 解(Ⅱ)得
y =1⎩
所以,原方程组的解为:⎨
⎧x =1,
················································ 2+2分 ⎨y =-1⎩
⎧x 1=3, ⎧x 2=1, 与⎨ ················································ 2分 y =1; y =-1. ⎩1⎩2
21.(本题满分10分,每小题5分)
解:(1)由题意得四边形ABGH 、ABFE 是平行四边形. ·································· 1分 ∴ AE ∥FG . ···················································································· 1分
EP AE
=. ······················································································ 1分 FP FG
EP EP 2
=2,即= 将AE =6,FG =3代入,得 ································· 1分 FP EF 3
又∵四边形ABFE 是平行四边形,AB =3,∴EF =AB =3. ∴EP =2. ··········· 1分 (2)过点P 作PH ⊥AE ,垂足为H (如图4). ········································ 1分 ∵四边形ABFE 是平行四边形,∠B =60︒,∴∠PEH =∠B =60︒. ············ 1分 在Rt △PEH 中,∠PHE =90︒,∠PEH =60︒,EP =2,
∴
∴PH =EP ⋅sin 60︒=2⨯∴△APE 的面积为
22.(本题满分10分)
解:(1)设油箱内剩余油量y (升) 与行驶路程x (千米)之间的函数
C
图4
3
····················································· 2分 =. ·
2
11
AE ⋅PH =⨯6⨯3=33. ··································· 1分 22
E
F
关系式为y =kx +b . ······················································································· 1分
分别将x =100,y =52;x =150,y =48代入上式,得⎨
⎧100k +b =52,
······· 2分
⎩150k +b =48.
2⎧
⎪k =-, 解得:⎨·················································································· 2分 25 ·
⎪⎩b =60.
2
x +60 ························································ 1分 25
(2)方法1:由题意可得,该型号的汽车在该路段行驶时,每行驶100耗油8升. ·· 2分 设行驶300公里时需要耗油x 升,可得300:100=x :8,解得x =24升. ············· 1分
2
x +60,得y =36. ·方法2:将x =300代入y =-····································· 2分 25
60-36=24. ··············································································· 1分
答:张老师的这辆车的油箱内至少需要有24升汽油. ······································· 1分 ..
∴所求的函数关系式为y =-
备注:学生若是在得到24升油的基础上又考虑了其它因素(如离开高速公路之后还需要再行驶一段路程才可以抵达目的地(或寻找到加油站),因此给出了大于24升油的其它数据,只要能够自圆其说,且符合生活实际情况,那么可以酌情评分. 23.(本题满分12分,每小题6分)
(1)证明:∵四边形ABCD 是正方形,∴CD =BC . ·········································· 1分
11
CD =BC . ·································· 1分 22
∵∠FCD =∠CBE ,∠F =∠F ,∴△FCE ∽△FBC . ···························· 2分 EF CE
=∴. ·················································································· 1分 CF BC
1EF 1
=. 即CF =2EF . ·又∵CE =BC ,∴··········································· 1分
2CF 2
(2)∵四边形ABCD 是正方形,∴DE ∥AB ,AD ∥BC ,AD =CD . ················ 1分
∵点E 为CD 边的中点,∴CE =
∵点F 位于线段AD 的延长线上,DE ∥AB ,∴ 又∵AD =CD ,∴
EF DF
=. ························ 1分 BE AD
EF DF
=. (1) ··························································· 1分 BE CD
∵AF ∥BC ,∴∠DFE =∠CBE .
又∵∠DCF =∠CBE ,∴∠DFE =∠DCF . ················································ 1分 又∵∠FDE =∠CDF ,∴△FDE ∽△CDF . ················································· 1分
DE DF EF DE
== ∴(2). 由(1)、(2)得 . ······································· 1分 DF CD BE DF
B
图5
图6
F
F
2
24.(本题满分12分,每小题4分)
解:(1)设所求二次函数的解析式为y =ax +bx +c ,将A (3, 1)、B (6,5)、C (0, 5)
⎧9a +3b +c =1,
48⎪
代入,得 ⎨36a +6b +c =5, 解得 a =,b =-,c =5. ································ 3分
93⎪c =5.
⎩428
x -x +5. ·········································· 1分 93
25
(2)Q 1(3, 6) ,Q 2(3, -4) ,Q 3(3, 9) ,Q 4(3, ) . ··········································· 4分
8
所以,这个二次函数的解析式为y =
(3)由题意得,该二次函数图像的对称轴为直线x =3. ···································· 1分
联结PC 交直线x =3于点M ,过点M 作MN ⊥AC ,垂足为N (图7-1) . 将直线x =3与BC 的交点记为H ,易得CH =3,AH =4,AC =5.
CH 3
= ········································································ 1分 CA 5
1
故可设MN =3k ,则AM =5k ,AM =4k . 又∵tan ∠PCA =,则CN =6k .
2
1553
由题意得方程:4k +6k =5. 解得k =,AM =,MH =4-= ·········· 1分
2222
∴sin ∠CAH =∴CM =3+() =
2
3
2
2
3MH . ···························· 1分 =5. ∴sin ∠PCB =
CM 52
25.(满分14分,第(1)小题5分,第(2)小题5分,第(3)小题4分)
解:(1)联结OA 、OB (如图8-1) ,易得OA =OB ,∠OAB =∠OBA . ··················· 1分
∵四边形ABCD 是平行四边形,∴AB ∥CD ,AD =BC .
∵BE =BC ,AD =BC ,∴AD =BE . ····················································· 1分 又 ∵AB ∥CD ,∴四边形ABED 是等腰梯形. ∴∠DAB =∠EBA . ····················· 1分 又 ∵∠OAB =∠OBA ,∴∠DAB -∠OAB =∠EBA -∠OBA .
即 ∠O A D =∠O B E . ·················································································· 1分
在△AOD 和△BOE 中,∵OA =OB ,∠OAD =∠OBE ,AD =BE ,
∴△AOD ≌△BOE. ∴OD =OE . ·····················
··
··· 1分
图8-1
图8-2
图8-3
方法2:∵∠ADE =∠BED ,∠DAO =∠EBO ,AD =BE ,∴△AOD ≌△BOE. „„ 方法3:∵∠ADE =∠BED ,∠DAO =∠EBO ,OA =OB ,∴△AOD ≌△BOE. „„ 方法4:如图8-2,过点O 作OH ⊥AB ,过点D 作DG ⊥AB ,过点E 作EI ⊥AB . „„ 方法5:如图8-3,过点O 作OH ⊥AB ,垂足为H ,联结DH 、EH . „„
(2)方法1:如图9-1,过点O 作OH ⊥AB ,垂足为H ,过点D 作DG ⊥AB ,垂足为G . 联结OB ,OH =3,AH =BH =4,得1分;得到DG =OH =3,得2分;在Rt △ADG 中,
201704嘉定区初三二模试卷-----------------------------p11
222写出AG =4-x ,AD =BC =y ,得1分;利用AD =DG +AG 得到y =x 2-8x +25,
得1分,函数定义域0
(3)如图10-1,过点O 作OM ⊥AC ,交AC 于点M ,交AB 于点N . 证明四边形ONBE 是平行四边形,得1分;利用BN =OE =OD ,AB =CD 得到OC =AN ,得1分;利用△AMN ≌
AM AN =得到AM =CN ,进而得到OM 是AC 的垂直平分线,OC =OA =5,CM CO
得1分;利用CD =AB =8,OC =5得到OD =3,得1分. △CMO 或
方法2. 如图10-,2;方法3:如图10-3;方法4(利用圆周角,略).
图
9-1 图
10-1 图
10-2 图10-3