2023年春季学期阶段性自主评估训练（二)
9.图，洫船甲以8海甲/小时的速度从港口O向东北方向航行，渔船乙同时
以6海里时的速度从港口0向西北方向航行，一个半小时后，两渔船相距为
八年级数学
A.12海里B.13海里C.14海里D.15海里
（考试时间：120分钟满分：120分)
10.在平面直角坐标系中，若点A(一a,b)在第三象限，则函数y=ax十b的
注意事项：
图象大致是
(第9题图)
1.答题前，考生务必将姓名、准考证号、座位号填写在试卷和答题卡上。
2.考生作答时，请在答题卡上作答（答题注意事项见答题卡），在本试卷上作答无效。
第I卷
并。开开
·、选择题（共12小题，每小题3分，共36分.在每小题给出的四个选项中只有一项是符合要求的，
11.如图，四边形ABCD为.正方形，点A的坐标为(0,2)，点B的坐标为(4,0)，
用2B铅笔把答题卡上对应题目的答案标号涂黑.)
点E为对角线的交点，点F与点E关于y轴对称，则点F的坐标为
1.下列式了中，属于最简次根式的是
B
A.(-2,3)B.(3,-3)C.(-3,2)D.(-3,3)
(第11题图)
A.0.5B.√4C.V6D.
12.小玮、小华两人相约沿同一路线从学校出发，以不同的速度匀速骑行前往体
2.下列各组线段长能组成直角：角形的是
育馆，小华比小玮早出发3分钟，小华骑行23分钟后，小玮以原速的1.4
A.1,2,3B.1,1,V3C.6,7,8D.5,12,13
倍继续骑行.小玮先到达体育馆，小华一直保持原速前往.在此过程中，
小玮、小华两人相距的路程y(单位：米)与小华骑行的时间x(单位：
3.若式子√x一3在实数范国内有意义，则x的取值范围是
分钟)之间的关系如图所示，下列结论：①小华的速度为320米/分钟；
1900
A.x≤3B.x≥3C.x>3D.x+3
900
②23分钟后，小玮的速度为350米1分钟：③总路程为29000米：④小华
0Λ
4.下列选项中，y是x的正比例函数的是
032391x
比小玮晚5分钟到达体育馆.其中正确的是
(第12题图)
A.y=x2
B.y=2x
C.y D.y
A.②③B.②④C.①③④
D.②③④
5.下列运算正确的是
D
第卷
A.2+V3=2√3
B.5-2=V3
c5x方-
二、填空题（本大题共6小题，每小题2分，共12分.）
D.V2÷V3…V5
(第6题图)
13.计算：V(-2)2=★·
（第14题图)
6.如图，下列四组条件中，不能判定四边形ABCD是平行四边形的是
A.AB∥CD,AD=BCB.AB∥CD,AD∥BC
14.如图，在□ABCD中，∠B十∠D=100°，则∠A的大小为太·
15.一次函数y=5x+1的图象与x轴的交点的坐标为★·
C.AD//BC,AD=BC D.AB=CD,AD-BC
、、1
16.如图，在边长为1的正方形网格中，A,B,C均在格点上，点D为AB的
7.下列说法正确的是
(第8题图)
(第16题图)
A.对角线相等的四边形是平行四边形B.对角线垂直的四边形是菱形
中点，则线段CD的长为★一·
17.已知等腰二角形的周长为16，那么底边长y关于腰长x的函数解析
C.三：个角都是直角的四边形是矩形D.一组邻边相等的平行四边形是正方形
式为★·
8.如图，在矩形ABCD中，AB=5,AD=25,将矩形沿EF折叠，使点D与点B重合，则BE的长为
18.如图，正方形ABCD是由四个全等的直角三角形围成的，若AE=5,BE=13,
A.8B.10C.12D.13
则EF的长为★一·
(第18题图)
八年级数学试卷第1页（共4页）
八年级数学试卷第2页（共4页）2023 年春季学期阶段性自主评估训练（二）
八年级数学参考答案
一、选择题（本大题共 12小题，每小题 3分，共 36分．）
题号 1 2 3 4 5 6 7 8 9 10 11 12
答案 C D B B C A C D D A D B
二、填空题（本大题共 6小题，每小题 2分，共 12分．）
1
13．2 14．130° 15．( ，0) 16 26． 17．y＝－2x＋16 18．
5 8 22
三、解答题（本大题共 8小题，共 72分．）
19．（本题满分 6分）
解：（1） 6 ÷ 3 × 2
＝ 2 2 ·······················································································2分
＝2；····························································································· 3分
1
（2） 18－ 8＋ 2
＝ 3 2 2 2 2 ·················································································· 4分
2
1
＝(3－2＋ ) 2 ·····················································································5分2
3 2
＝ ．······························································································ 6分
2
20．（本题满分 6分）
解：∵x＝1－ 5 ，y＝1＋ 5 ，
∴x2－y2＝(x＋y)(x－y)····················································································2分
＝(1－ 5 ＋1＋ 5 )[(1－ 5 )－(1＋ 5 )]···············································4分
＝2×(－2 5 )·····················································································5分
＝－4 5 ．·······················································································6分
21．（本题满分 10分）
解：（1）△ABD是直角三角形．理由如下：····················································· 1分
∵BD＝15，CD＝8，BC＝17，
∴152＋82＝172．
八年级数学参考答案（第 1页 共 5页）
∴BD2＋CD2＝BC2．······················································································2分
∴△BDC是直角三角形，且∠BDC＝90°．···································3分
∴∠ADB＝180°－∠BDC＝90°．················································ 4分
∴△ABD是直角三角形．··························································5分
（2）设 AD＝x．·····································································6分
∵AB＝AC，
∴AB＝AC＝x＋8．······················································································· 7分
在 Rt△ABD中，BD2＋AD2＝AB2，
∴152＋x2＝(x＋8)2．····················································································· 8分
161
∴x＝ ．································································································· 9分
16
289
∴AB＝x＋8＝ ．··················································································· 10分
16
22．（本题满分 10分）
解：（1）证明：∵∠A＝∠F，
∴DF∥AC．···································································· 1分
∴∠C＝∠FEC．······························································ 2分
又∵∠C＝∠D，
∴∠FEC＝∠D．··························································································3分
∴DB∥EC．································································································4分
∴四边形 BCED是平行四边形．······································································ 5分
（2）∵BN平分∠DBC，
∴∠DBN＝∠CBN．······················································································6分
∵BD∥EC，
∴∠DBN＝∠BNC．······················································································7分
∴∠CBN＝∠BNC．
∴CN＝BC．································································································8分
又∵四边形 BCED是平行四边形，
∴ BC＝DE＝3．··························································································9分
∴CN＝3．·································································································10分
23．（本题满分 10分）
解：（1）作图如右，点 E为所求；······································ 3分
（2）连接 EC，ED，························································· 4分
八年级数学参考答案（第 2页 共 5页）
设 AE＝xkm，则 EB＝(2.5－x)km．······································· 5分
∵AC2＋AE2＝EC2，EB2＋BD2＝ED2，··································· 6分
又∵EC＝ED，
∴AC2＋AE2＝EB2＋BD2．··················································· 7分
∴1.52＋x2＝(2.5－x)2＋12．·············································································8分
解得 x＝1．································································································· 9分
即 AE的距离为 1km．··················································································10分
24．（本题满分 10分）
1
解：（1）∵直线 y x经过点 B，点 B的横坐标为 2，
2
1
∴当 x＝2时，y＝ ×2＝1．················································1分
2
∴点 B的坐标是（2，1）．·················································2分
将点 A，B的坐标分别代入 y＝kx＋b得到，
b 5，
2k b 1. ·································································································· 3分
k 2，
解得 b 5. ································································································5分
∴点 B的坐标为（2，1），k＝－2，b＝5．······················································· 7分
（2）x＜2．·······························································································10分
25．（本题满分 10分）
证明：（1）∵D，E分别是 AC，AB的中点，
∴DE是△ABC的中位线．
1
∴DE∥BC，DE＝ BC．···············································································1分
2
∴EF∥BC．································································································ 2分
∵BF∥CE，
∴四边形 BCEF是平行四边形．······································································ 3分
1
又∵DE＝ CE
2
∴BC＝CE．·····································································4分
∴平行四边形 BCEF是菱形．·············································· 5分
（2）如图，过点 E作 EG⊥BC于点 G．································6分
由（1）知 BC＝CE．
八年级数学参考答案（第 3页 共 5页）
∵∠BCE＝60°，
∴△BCE是等边三角形．··············································································· 7分
∴BE＝CE＝BC＝2．
∵EG⊥BC，
1
∴BG＝ BC＝1．························································································ 8分
2
在 Rt△BGE中，由勾股定理得：
EG＝ BE 2 BG 2 ＝ 22 12 ＝ 3．································································9分
∴S 菱形BCEF＝BC EG＝2× 3＝2 3．·······························································10分
26．（本题满分 10分）
解：（1）设 A种纪念品单价为 a元，B种纪念品单价为 b元．
10a 5b 1000，
根据题意，得 5a 3b 550 ····························································· 1分 ．
a 50，
解得
b 100
·······························································································2分
．
答：购进 A，B两种纪念品的单价分别为 50元、100元．···································· 3分
（2）设该商店购进 A种纪念品 x个，购进 B种纪念品 y个．
根据题意，得 50x＋100y＝10000．···································································4分
1
即 y＝100－ x．
2
x 6 100 1 x
≥ 2
，①
由题意，得 ·····································································5分
100
1
x≥20．②
2
由①，得 x≥150．
由②，得 x≤160．
∴150≤x≤160．·························································································· 6分
∵x，y均为正整数，
∴x可取的正整数值是 150，152，154，156，158，160．
与 x相对应的 y可取的正整数值是 25，24，23，22，21，20．
∴共有 6种进货方案．·················································································· 7分
（3）设总利润为 W元．
1
则 W＝20x＋30y＝20x＋30(100－ x)＝5x＋3000．·············································· 8分
2
∵5＞0，
八年级数学参考答案（第 4页 共 5页）
∴W随 x的增大而增大．
∴当 x＝160时，W有最大值为 5×160＋3000＝3800（元）．·································9分
∴当购进 A种纪念品 160件，B种纪念品 20件时，可获得最大利润，最大利润是 3800
元．········································································································· 10分
八年级数学参考答案（第 5页 共 5页）
