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七年级数学
试题参考答案与评分标准
说明：本评分标准每题只给出了一种解法供参考，如果考生的解法与本解答不同，参照本
评分标准给分．
一、选择题（本大题共 10 小题，每小题 3 分，共 30 分）
题号 1 2 3 4 5 6 7 8 9 10
选项 B D B D B B A A C C
二、填空题（本大题共 6 小题，11~12 每小题 3 分，13~16 每小题 4 分，共 22 分）
11．2 12．－5 13．15
2 4
14．36（写 37 也对） 15．－2≤a <8 16．（1） ；（2）
5 3
三、解答题（本大题共 9 小题，共 98 分）
17．（本小题满分 10 分）
（1）解：原式＝3+1·················································································2 分
＝4 ···················································································5 分
（2）解：原式＝2√6－2√7－（2√6－√7） ····················································7 分
＝2√6－2√7－2√6 + √7·························································8 分
＝－√7 ···········································································10 分
18．（本小题满分 10 分）
3x－5y＝3 ①
解：（1）{ ，
2x－y＝16 ②
5×②得，10x－5y=80 ③，
③-①得，7x=77，
解得 x＝11，···················································································2 分
把 x＝11 代入②，得 y＝6， ······································································4 分
x＝11
所以方程组的解是{ ． ·······································································5 分
y＝6
2x+5
－1<2－x ①
解：（2）{ 3 ，
2x+3≥ x+11 ②
4
解不等式①得：x< ， ··············································································7 分
5
解不等式②得：x≥8，···············································································9 分
则不等式组无解．···················································································10 分
七年级年级数学答案卷第1页（共7页）
19．（本小题满分 10 分）
解：（1）C1（3，－1），··············································································3 分
（2）△A1B1C1 如图所示，
y
4
A
3
B
2
A1 1
C
－4 －3－ 2 －1O B1 1 2 3 4 x
－1 C1
－2
－3
························································6 分
－4
（3）△AOA1 如图所示，
1 1 1
△AOA1 的面积＝ ×（1+3）×5－ ×1×3－ ×4×1
2 2 2
＝10－1.5－2
＝6.5 ．···············································································10 分
20．（本小题满分 10 分）
解：选择（1）假命题················································································4 分
反例为：若 a＝2，b＝1，c＝－3
则 ac2＝18，bc2＝9
显然 ac2>bc2
∴ac2（举例不唯一，只要言之有理都对）···························································10 分
或：
选择（2）真命题······················································································4 分
证明：∵△ABC 的三个外角的度数之比是 3：4：5
设三个外角的度数依次为 3x，4x，5x
由三角形的外角和等于 360°可得 3x+4x+5x＝360°
∴x＝30°·············································6 分
∴3x＝90°，4x＝120°，5x＝150°
∴对应的三个内角的度数为 90°，60°，30°·························································8 分
∴△ABC 是直角三角形． ··········································································10 分
21．（本小题满分 10 分）
解：（1）100； ························································································2 分
补全条形统计图如下：
七年级年级数学答案卷第2页（共7页）
·················································4分
（2）B； ·······························································································7 分
4
（3）1050× =84（人），·······························································8 分
6+32+8+4
2
（2000－1050）× =38（人），·······················································9 分
4+39+5+2
84+38=122（人）
答：估计全校体重指数为“肥胖 D”的学生约为 122 人． ································10 分
22．（本小题满分 10 分）
解：（1）由题意得：设∠ABC＝x，则∠C＝x+10°
∴∠BAC＝180°－∠ABC－∠C＝180°－x－（x+10°）＝170°－2x，
∵AD 平分∠BAC，
1 1
∴∠BAD= ∠BAC= (170°－2x)=85°－x ，
2 2
∵∠PDE是△ABD 的外角
∴∠PDE＝∠ABD+∠BAD＝x+（85°－x）＝85° ，··········································2 分
∵PE⊥AD，
∴∠E＝90°，
∴∠P＝180°－∠E－∠PDE＝180°－90°－85°＝5°；·······································4 分
1
（2）∠P= (∠C－∠ABC) ···········································································5 分
2
证明：∵AD 平分∠BAC，
1 1 1
∴∠BAD= ∠BAC＝ (180°－∠ABC－∠C) ＝90°－ (∠ABC+∠C) ，
2 2 2
∵∠PDE是△ABD 的外角
∴∠PDE＝∠ABC+∠BAD
1
＝∠ABC + 90°－ (∠ABC+∠C)，
2
七年级年级数学答案卷第3页（共7页）
1
＝90° － (∠C－∠ABC) ·······························································7 分
2
∵PE⊥AD，
∴∠E＝90°，
∴∠P＝180°－∠E－∠PDE
1
＝180°－90°－［90° － (∠C－∠ABC)］
2
1
＝ (∠C－∠ABC) ·················································································10 分
2
23．（本小题满分 12 分）
解：（1）设购进 A 型 x 辆、B 型 y 辆，
x+y=20
由题意可得：{ ， ··························································3 分
0.8x+1.4y=22
x=10
解得：{ ，··························································································5 分
y=10
答：购进 A 型 10 辆，B 型 10 辆；·······························································6 分
（2）设购进 A 型新能源汽车 a 辆，······························································7 分
27 + 24.4 × （20－ ）＞508.8
由题意可得：{ ·················································9 分
0.8 + 1.4 × （20－ ）＞20.5，
解得：8＜a＜12.5，·················································································10 分
∵a 是正整数
∴a 的值为 9,10,11,12·············································································11 分
答：共有 4种购进方案，
第 1种购进 A 型 9辆，B 型 11 辆；
第 2种购进 A 型 10辆，B 型 10 辆；
第 3种购进 A 型 11辆，B 型 9 辆；
第 4种购进 A 型 12辆，B 型 8 辆．····························································12 分
24．（本小题满分 13 分）
解：（1）连接 AP
∵PO∥AB
∴S△AOP＝S△BOP，且点 P 在第三象限
1 1
∴ ×3×（－m）＝ ×2×3
2 2
七年级年级数学答案卷第4页（共7页）
∴m＝－2
∴P（－2，－3）·····················································································3 分
（2）将△ABO 沿 AB 方向平移，使点 A 与 B 重合，点 B 与 C 重合得到△BCD，
则 C（－4，－3），D（－2，－3），∴点 D 在 PC 上，BD=3
∵A（0，3），P（m，－3）且 m＞－４
∴PC＝ m－(－4) ＝ m+4
设射线 CP 交 y 轴于点 E，
AE＝ 3－(－3) ＝ 6··············································································4分
1 1
S△PAC＝ PC·AE＝ （m+4）×6
2 2 y
＝3m+12·························5 分
1 1 A
S△PBC＝ PC·BD＝ （m+4）×3
2 2
3
＝ m+6·······················6 分 B
2
O x
3
S△PAB＝S△PAC－S△PBC＝（3m+12）－（ m+6）
2
C －3 P
3
＝ m+6 －4 D E
2
3
∴n＝ m+6 ．······················································································8 分
2
3
（3）由（2）得 S△PAB＝ m+6
2
当－４＜m＜0 时，
1 3
S△POA＝ ×3×（－m）＝－ m ，
2 2
3
∵S△PAB≥2S△POA ， ∴ m+6≥－3m ，
2
4
∴m≥－ ，·····················································································9 分
3
4
即－ ≤m＜0 ；················································································10 分
3
当 m＞0 时，
1 3
S△POA＝ ×3×m＝ m ，
2 2
3
∵S△PAB≥2S△POA ， ∴ m+6≥3m ，
2
∴m≤4，···························································································11 分
即 0＜m≤4．·····················································································12 分
七年级年级数学答案卷第5页（共7页）
4
综上所述，m 的取值范围是－ ≤m＜0 或 0＜m≤4．··································13 分
3
25．（本小题满分 13 分）
解：（1）设 S△BOC＝m
∵D 是 BC 的中点
1 1
∴S△OCD=S△OBD= S2 △BOC
= m，
2
∵F 是 AC 的中点，
∴S△AOF＝S△FOC，S△ABF＝S△FBC，
∴S△ABF S△AOF＝S△FBC S△FOC，
∴S△AOB＝S△BOC＝m，
△
∴ = 1 = 2， △
2
∵△AOB 与△OBD 同高，
△
∴ = = 2； ·············································································3 分
△
（2）如图②，连接 BO 并延长交 AC 于点 D，延长 AO 交 BC 于点 E， A
BO AO
∵点 O 是△ABC 的重心，由（1）可知 = =2，
OD OE M
N D
S△ABO BO S= = ， △ABO
O
∴ 2 = = 2，
S△AOD DO S△EOB
设 S△BOE＝a，则 S△AOB＝2a，S△AOD＝a， B E C
图②
∵D，E 分别是 AC，BC 的中点，
∴S△COD＝S△AOD＝a，S△EOC＝S△OBE＝a，
∴S△ABC＝S△AOB+S△AOD+S△EOC+S△COD+S△OBE＝6a，
∴6a＝12，a＝2，
∴S△AOB＝2a＝4，······················································································5 分
1
∵点 M 是 AO 的中点，∴S△BOM＝ S△AOB＝2，
2
又∵CN 的延长线过 AB 的中点，∴N 是△ABO 的重心，
BN 1 2
∴由（1）可知 =2，∴S△MON＝ S△BOM＝ ．·············································8 分
MN 3 3
（3）如图③，连接 AN，MP，OP，延长 CN 交 AB 于点 D
七年级年级数学答案卷第6页（共7页）
∵Q 是 NP 的中点，
A
∴设 S△MQN＝S△MQP＝x，S△OQN＝S△OQP＝y
M
P
∵点 M 是 AO 的中点， D Q
N
O
∴S△AON＝2S△MON＝2x+2y ，S△AOP＝2S△MOP＝2x+2y
B C
∵由（2）可知点 O 是△ABC 的重心，N 是△ABO 的重心，
图③
NO CO OC
∴由（1）可知 =2， =2，∴ =3，·······················································10 分
DN OD ON
∴S△AOC＝3S△AON＝6x+6y，∴S△POC＝S△AOC－S△AOP＝4x+4y
S△PON ON + 1 1
∴ = = = ，即 =
S△POC OC 4 +4 3 2 +2 3
∴y＝2x·······················································12 分
OQ OQ
∴ = ＝2，即 的值为 2. ·································································13 分
MQ MQ
七年级年级数学答案卷第7页（共7页）七年级数学
注意事项
考生在答题前请认真阅读本注意事项：
1,本试卷共6页，满分为150分，考试时间为120分钟，考试结束后，请将本试卷和答
题卡一并交回.
2.答题前，请务必将自己的姓名、考试证号用0.5毫米黑色字迹的签字笔填写在试卷
及答题卡上指定的位置，
3.
答案必须按要求填涂、书写在答题卡上，在试卷、草稿纸上答题一律无效
一、选择题（本大题共10小题，每小题3分，共30分.在每小题给出的四个选项中，恰有
一项是符合题目要求的，请将正确选项的字母代号填涂在答题卡相应位置上)
1.实数2，V5,0,一√5中，最大的数是
A.2
B.V5
C.0
D.-5
2.若关于x的不等式的解集在数轴上表示为如图所示，则该不等式可能是
A.+2<0
B.x+2>0
C.x+2≤0
D.x+2≥0
-2
0
3.点P(一1,2)所在的象限是
(第2题)
A.第一象限
B.第二象限
C.第三象限
D.第四象限
4,要了解全校学生每周用于体育锻炼的时间，下列选取调查对象的方式中最合适的是
A.随机选取一个班的学生
B.在全校女生中随机选取100人
C.随机选取一个体育队的学生
D,在全校学生中随机选取100人
5。为估计池塘两岸A,B间的距离，如图，小明在池塘一侧选取了一点O,测得
OA=10m,OB=6.5m,那么A,B间的距离可能是
A.3m
B.12m
C.17m
D,20m
x=4
6.若方程mx+2y=6有一组解
则m的值为
(第5题)
y=1
A.-2
B.
C.2
D.4
7.如图，面积为6的正方形ABCD的顶点A在数轴上表示数一1，以点A为圆心，AB长
为半径作圆弧交数轴于点E,若点E在点A的右侧，则点E表示的数为
A.-1+6
B.-1-V6
C.v6
D,-1+V3
-4-3-2-101234→
(第7题)
七年级数学试卷第1页（共6页）
0000000
8.(九章算术》是人类科学史上应用数学的“算经之首“，书中有这样一个问题：若2人坐
一辆车，则9人需要步行，若““.问：人与车各多少？小明同学设有x辆车，人
数为为根据愿意可列方程组02根据已有信息，题中用…”发示的锁失条
件应补为
A,三人坐一辆车，则有两辆空车
B.三人坐一辆车，则2人需要步行
C.三人坐一辆车，则有一车少坐2人D,三人坐一辆车，则还缺两辆车
9.若关于a,b,c的方程组g2b3c满足a+bc<4,则6的取值范围为
3a-2btc=-31
A.b<1.5
B.b>2
C.b<2.5
D.b>3
10.在平面直角坐标系中，点A(2m,0),B(4m+1,0),P(3m+2,0),P2Lx轴，点2
的纵坐标为2.则以下说法正确的是
A.当m=一3时，点P是线段AB的中点
B.无论m取何值，线段BP的长度恒为1
C.存在唯一一个m的值，使得AB=PQ
D.存在唯一一个m的值，使得AB=3PQ
二、填空题（本大题共6小题，第11~12题每小题3分，第13~16题每小题4分，
共22分不需写出解答过程，请把答案直接填写在答题卡相应位置上)
11.4的算术平方根等于▲。
卫。小明正确求得方程组红的解为4则表示的致为上，
13.如图，从点A处观测C处的仰角∠CAD=30°，从B处观测C处的仰角LCBD=45°.从
C处观测A,B两处的视角LACB是▲度，
踢毽子个数/个
40…
35
25
20
15
5F-
D
0
】23456天数/天
D
(第13题)
(第14题)
(第16题)
14.某校计划举办一场一次不间断踢毽子比赛（即毽子不落地），体育老师将丽丽连续5
天一次不间断踢毽子的训练情况绘制成如图所示的趋势图，根据所绘制的趋势图估
计丽丽第6天一次不间断可踢毽子▲个.
15.已知实数，b满足2a-3b=4,且a≥-2，b<4,则a的取值范围是▲
16.如图，将三角形纸片ABC折叠，使点B与C重合，折痕交边AB于点D,交边BC于
点E,展开并复位.连接CD,AE交于点F,己知BD=4AD,设△BDB的面积为S0.
(1)设△ABC的面积为S,若S0=,则k-▲：
(2)设△CEF的面积为S1,△ADF的面积为S2,若So=n(S1一S2),则=▲，
七年级数学试卷第2页（共6页）
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