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★保密材料 2022～2023学年度第二学期七年级期末考试
阅卷使用 数学试题参考答案和评分标准
说明：本评分标准每题给出的解法仅供参考，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 A D D B A D A C C B
二、填空题（本大题共 8小题，第 11～12题每小题 3分，第 13～18题每小题 4分，共 30分）
11． 6 12．12 13．AB＝DE（答案不唯一） 14．(0，5)
15．(7，2) 16．87° 17．7≤a＜8 18．1≤z＜6
三、解答题（本大题共 8小题，共 90分）
19．（本小题满分 12分）
（1）解：原式＝2－(2－ 3 )··············································································· 2分
＝2－2＋ 3 ··················································································4分
＝ 3． ················································································ 6分
（2）解：①×4得 8x－4y＝20③···································································· 8分
②＋③得 11x＝22．········································································8分
解得 x＝2．················································································ 10分
代入①得 y＝－1．······································································· 10分
x＝2
∴方程组的解为 ．·····························································12分
y＝－1
20．（本小题满分 10分）
解：解不等式①，得 x≤1．··········································································· 2分
解不等式②，得 x＜4．··········································································· 4分
∴不等式组的解集为 x≤1．····································································· 6分
∵x为非负整数，
∴x＝1或 0．······················································································· 10分
21．（本小题满分 10分）
解：∵∠BAD＝80°，∠CAD＝50°，
∴∠CAB＝∠BAD－∠CAD＝80°－50°＝30°．·············································2分
∵AD∥BE，
∴∠BAD＋∠ABE＝180°，······································································4分
∴∠ABE＝180°－∠BAD＝100°，····························································· 5分
∵∠EBC＝40°，··················································································· 6分
∴∠ABC＝∠ABE－∠EBC＝60°，····························································8分
∴在△ABC中，∠ACB＝180°－∠ABC－∠CAB＝90°．································ 9分
答：从 B岛看 A，C两岛的视角∠ABC是 60°，
从 C岛看 A，B两岛的视角∠ACB是 90°，········································· 10分
22.（本小题满分 10分）
解：（1） 频数/人
30 27
24
18 18
12 9
6 6
············································ 4分
0 20 40 60 80 100 成绩/分
七年级数学试题参考答案和评分标准 第 1 页（共 4页）
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（2） 720 18 =216 ；·················································································· 7分60
答：估计有 216名学生的测试成绩不低于 80分．········································8分
（3）通过调查发现 80分以上的学生占比不高，建议学校加大航天科技知识的宣传普及
力度．
（注：答案不唯一，只要合理就给分）．·················································· 10分
23.（本小题满分 10分）
解：（1）∠ACD，······················································································· 2分
SAS．···························································································4分
（2）∵∠CBE＝2∠ABE，∠ACB＝2∠ACD，
∴设∠ABE＝x°，∠ACD＝y°，
则∠CBE＝2x°，∠ACB＝2y°，
∴∠ABC＝∠ABE＋∠CBE＝3x°．
∵△CAD≌△ABE，
∴∠BAE＝∠ACD＝y°，∠ADC＝∠BEA．···········································6分
∵在△ABC中，∠BAC＋∠ABC＋∠ACB＝180°，
∴y＋3x＋2y＝180，
∴x＋y＝60．················································································· 8分
∵∠AEB＝∠CBE＋∠ACB，
∴∠AEB＝2x°＋2y°＝2(x＋y)°＝120°，
∴∠ADC＝120°．··········································································10分
24.（本小题满分 12分）
解：（1）设 A型电脑购进 x台，B型电脑购进(45－x)台．···································1分
6000x＋4000(45－x)＝210000，·························································· 3分
解得 x＝15．
∴45－x＝30．················································································5分
答：A型电脑购进 15台，B型电脑购进 30台．····································6分
（2）设 A型电脑购进 a台，B型电脑购进 b台，C型电脑购进 c台．
a＋b＋c＝60，
······················································ 8分
6000a＋4000b＋3000c＝190000．
解得 3a＋b＝10．············································································9分
∵a，b是正整数，
∴ a＝1，b＝7；a＝2，b＝4；a＝3，b＝1．
当 a＝1，b＝7时，c＝52；
当 a＝2，b＝4时，c＝54；
当 a＝3，b＝1时，c＝56．
共有 3种方案：A型电脑购进 1台，B型电脑购进 7台，C型电脑购进 52台；
A型电脑购进 2台，B型电脑购进 4台，C型电脑购进 54台；
A型电脑购进 3台，B型电脑购进 1台，C型电脑购进 56台．
··································································································12分
七年级数学试题参考答案和评分标准 第 2 页（共 4页）
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25.（本小题满分 13分）
（1）证明：设∠ACP＝x°
∵∠BAP＝2∠ACP，
B A ∴∠BAP＝2x°．
∵AP平分∠BAC，
P ∴∠CAP＝∠BAP＝2x°．
∵AP⊥PC，
∴∠ACP＋∠BAP＝90°，即 x＋2x＝90．
D C
∴x＝30．···················································· 2分
（第 25题答图 1） ∵AB//CD，
∴∠BAC＋∠ACD＝180°．
∴∠PCD＝30°．··········································· 4分
（2）解：过点 P作 PH⊥AC
∵AB//CD，MN⊥CD，AP平分∠BAC，
∴PH＝PM．················································ 6分
B M A ∵AB//CD，
H ∴∠BAC＋∠ACD＝180°．
P ∵AP平分∠BAC，
CAP 1∴∠ ＝ ∠BAC．
D N C
2
∵AP⊥PC，
（第 25题答图 2）
∴∠CAP＋∠ACP＝90°，
1
∴∠ACP＝ ∠ACD．
2
∴PH＝PN．·················································8分
∴PM＝PN．················································ 9分
（3）解：在 AC上截取 AQ＝AM，
∵AP平分∠BAC，
∴∠CAP＝∠BAP．
∴△AMP≌△AQP．····································· 11分
B M A ∴∠AMP＝∠AQP．
∵AB//CD，
Q ∴∠AMP＋∠MNC＝180°．
P
∴∠AQP＝∠MNC．
∵∠AQP＋∠PQC＝180°．
D N C ∴∠PQC＝∠MNC．
1
（第 25题答图 3） ∵∠ACP＝ ∠ACD．
2
∴△CNP≌△CQP．
∴CQ＝CN．
∴AQ＋CQ＝AM＋CN．
即 AC＝AM＋CN．······································· 13分
七年级数学试题参考答案和评分标准 第 3 页（共 4页）
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26．（本小题满分 13分）
解：（1）Q2．····························································································· 4分
（2）根据定义得： m 1＝4．
∴m＝5或 m＝－3．········································································ 8分
（3）根据定义得，① n 1＝4．
∴n＝3或 n＝－5（舍去）．···········································10分
② 2n 3 ＝4．
1 7
∴n＝－ 或 n＝ （舍去）．············································12分
2 2
1
综上所述：n＝3或 n＝－ ．····························································13分
2
七年级数学试题参考答案和评分标准 第 4 页（共 4页）
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25.(本小题满分13分)
如图，已知AB/CD,AP平分∠BAC,APLPC,.点M是射线AB上一动点，MP交射
线CD于点N.
(1)当∠BAP=2∠ACP时，求∠PCD的度数：
(2)当MN⊥CD时，求证：MP=PN;
布京米圆翼好塑溶培史音
味(3)试探究线段AM,NC,AC之间的数量关系，并说明理由.，
D
D
的，中数个四的
(第25题)
(备用图)
大本〉藏超
置身
条母分自华的能数头指，色朱数日各件学高
26.(本小题满分13分)
士.日
在平面直角坐标系xOy中，对于不同的两点M,N,若点M到x轴，y轴的距离的较
大值等于点N到x轴，y轴的距离的较大值，则称点M,互为“方格点”
例如：点(3，-4)，(4，一2)互为“方格点”；点(2，一2)，（一2,0）互为“方格点”.
已知点P(1,-4).
、百
(1)在点2(4，-6)，22(-4,4)，03(-3,5)中，是点P的“方格点”的是△：
(2)若点9(m-1,3)与点P互为“方格点”，求m的值：
(3)若点2(n十1,2n-3)与点P互为“方格点”，求n的值.
头的改鬼宁一失增不识映0，<。容。
地十相台
七年级数学试卷第6页（共6页）

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