资源简介 2025年八年级(下)数学期末考试数学参考答案及评分意见一、选择题(36 分)题号 1 2 3 4 5 6 7 8 9 10 11 12答案 C A C B A B C D A C A B二、填空题(12 分)613. -1 14. < 3 15. 10 1 16. 5三、解答题117. (8 分)解:(1)解: (2 6 6) 6 3 6 6 ·······················································································2 分 18 ·································································································4 分2(2)解: 7 + 5 7 5 + 2 2= 7 5 + 8························································································ 2 分= 10.······························································································ 4 分18.(10 分)(1)设一次函数解析式为 y kx b,·······································1 分∵图象经过 A 3,5 ,B 0,2 两点, 5 3k b∴ ··············································································· 3分 2 b解得: k 1,b 2 ········································································ 4分∴一次函数解析式为 y x 2;·························································· 5 分(2)当 y 0时,0 x 2,∴ x 2,··························································································6 分∴C 2,0 ······················································································· 7分1∴ S△AOC OC1 yA 2 5 5,2 2故△AOC的面积为 5.······································································ 10 分19.(1)△ 是等腰三角形,理由如下:······················································1 分∵∠ = 30°,∠ = 60°,∴∠ = ∠ ∠ = 60° 30° = 30°,········································· 3 分∴∠ = ∠ ,2025年八年级(下)数学答案 第 1页(共 4页)∴ = ,······················································································4 分∴△ 是等腰三角形.·······································································5 分(2)由(1)可知 = = 10,∠ = 60°,∠ = 90°,·················· 6 分∴∠ = 30°,················································································· 7 分 = 1∴ = 5,·············································································· 9 分2在 Rt△ 中, = 2 2 = 5 3米.··········································10 分20.(10 分)解:解:(1)将甲组的成绩按照从小到大的顺序排列为:3,7,7,7,8,9,10,10;位于中间的两个数为 7和 8, = 7+8故中位数为: = 7.5;·································································3 分2在乙组中,出现次数最多的是 7分;故 = 7;·························································································· 5 分(2)乙 ····························································································7 分(3)(答案不唯一)甲组的初赛成绩较好;··············································8 分理由;①从中位数的角度看,甲组成绩比乙组好;②从优秀率的角度看,甲组成绩比乙组好;············································ 10 分21.(10 分)解:所作图形,如图:·····························································3 分(2)证明:∵ AE AB,∴ ABE AEB,∵四边形 ABCD为平行四边形,∴ AD∥CB,∴ AEB EBC,1 ABE EBC ABC∴ 2 ,······························································· 5 分∵ AF 平分 BAD, BAF DAF 1 BAD∴ 2 ,······························································· 6 分∵四边形 ABCD为平行四边形,∴ AD∥CB,∴ DAB ABC 180 ,·····································································7 分1 ABC 1 BAD 90 ∴ 2 2 .即 ABE BAO 90.········································································9 分∵在 ABO中, BAO ABE AOB 180 .∴ AOB 90 .················································································10 分2025年八年级(下)数学答案 第 2页(共 4页)22.(12 ) 1 m 2 3 1 3 1,n 2 1 1 3 1分 解:( )故答案为:1;1 ·················································································4 分(2)解:如图,·················································································6 分(3)解:根据图像得:当 x=-1 时函数 y 2 | x 1| 3有最小值,最小值为 y= 3;······································· 8 分4 2 x 1 3 x 1( )解:方程 的解为: x1 2,x2 2,····························10 分理由如下:y 2 x 1 3画出函数 和 y x 1的图象,如图所示:y 2 x 1 3函数 和 y x 1D 2, 1 ,E 2,3 的图象交点坐标分别为 ,x 2 x 1 3 x 1∴关于 的方程 的解为: x1 2,x2 2.···························12 分23.(12 分)解:(1)证明:如图1,作 EP CD于 P, EQ BC于Q,·············1 分∵ DCA BCA 45 ,∴ EQ EP,2025年八年级(下)数学答案 第 3页(共 4页)∵ QEF PEF 90 , PED PEF 90 ,∴ QEF PED ,·············································································2 分在Rt EQF和Rt△EPD中, QEF PED EQ EP EQF EPD,Rt EQF≌Rt EPD ASA ∴ ,······························································· 4 分∴ EF ED。····················································································· 5 分(2)解:连接 CG∵四边形 ABCD是正方形, AB 3,∴ AD CD 3, ADC 90 , AC 2AD 3 2,∵CE 2 2,∴ AE 2,······················································································7 分∵ EF ED∴四边形DEFG是正方形,∴DE DG, EDG 90 ADC,∴ ADE CDG, ADE≌ CDG SAS ∴ ,······································································9 分∴CG AE 2;·············································································10 分(3)30 或120 ···················································································· 12 分2025年八年级(下)数学答案 第 4页(共 4页)2025年春季学期期末检测试卷八年级(下)数学注意:1.本试题卷分第Ⅰ卷和第Ⅱ卷,满分为120分,考试用时120分钟。2.考生必须在答题卷上作答,在本试题卷上作答无效。考试结束,将答题卡交回。第Ⅰ卷(选择题,共36分)一、选择题(本大题共12小题,每小题3分,共36分.)请考生用2B铅笔在答题卷上将选定的答案标号涂黑.1.使代数式有意义的的取值范围是( )A. B. C. D.2.某校举行健美操比赛,甲、乙、丙三个班各选10名学生参加比赛.若参赛学生的平均身高都是1.65米,方差分别是则参赛学生身高比较整齐的班级是( )A.甲班 B.乙班 C.丙班 D.同样整齐3.下列曲线中,能表示y是x的函数的是( )A. B. C. D.4.如图,公路,互相垂直,公路的中点与点被湖隔开,若测得的长为,则,两点间的距离为( )A. B.C. D.距离不确定5.以下列各组数为边长,能构成直角三角形的是( )A.13,12,5 B.3,3,4 C.3,6,4 D.4,8,56.一次函数的图象不经过( )A.第一象限 B.第二象限 C.第三象限 D.第四象限7.菱形具有而平行四边形不一定具有的性质是( )A.对角线互相平分 B.对角线相等C.对角线互相垂直 D.对边平行且相等8.某校规定学生的体育成绩由三部分组成:早锻炼及体育课外活动表现占成绩的10%,体育理论测试占20%,体育技能测试占70%.小亮的上述三项成绩依次是:90分,85分,80分,则小亮这学期的体育成绩是( )分.A. 80 B. 84 C. 85 D. 829.已知点,,都在直线上,则,,的大小关系是( )A. B. C. D.10.如图,有六根长度相同的木条,小明先用四根木条制作了能够活动的菱形学具,他先将该活动学具调成图1所示菱形,测得,对角线,接着将该活动学具调成图2所示正方形,最后用剩下的两根木条搭成了如图3所示的图形,连接,则图3中的面积为( ) A. B. C. D.11.如图,均匀地向一个如图所示的容器中注水,最后把容器注满,在注水过程中水面高度h随时间变化的函数图象大致是( )A.B.C.D.12.如图,四边形是边长为9的正方形纸片,将其沿折叠,使点落在边上的处,点的对应点为,且,则的长是( )A. B.2 C. D.第11题图 第12题图(第10题图)第Ⅱ卷(非选择题,共84分)二、填空题(本大题共4小题,每小题3分,共12分。请把答案写在答题卡上对应的答题区域内。)13. 计算: .14.如图,直线 与直线 交于点A,当时,x的取值范围是 .15.如图,长方形中,,,边在数轴上,表示的数为,若以点为圆心,对角线的长为半径作弧交数轴的正半轴于,则点表示的数为 .16.如图,在中,,点E是中点,作于点F,已知,,则的长为 ..第14题图 第15题图 第16题图三、解答题:(本大题共7小题,共72分。解答应写出文字说明、证明过程或运算步骤。请将解答写在答题卡上对应的答题区域内。)17.(每小题4分,共8分)计算:(1). (2).18.(10分)一次函数图象经过点和两点.(1)求出该一次函数的表达式;(2)若直线AB与x轴交于点C,求的面积.19.(10分)在人教版八下数学教材第36页数学活动一《测量学校旗杆高度》中,聪聪想到了一种新颖的求解方式,聪聪从点C观察旗杆顶端的仰角为(即),接着往前走10米到达点D,观察旗杆顶端的仰角为(即).(1)请你帮助聪聪判断的形状,并说明理由;(2)根据聪聪的方法请你求出旗杆的高度.(人的身高忽略不计,结果保留根号)20.(10分)为激发青少年崇尚科学、探索未知的热情,学校开展“科学小达人”知识竞赛。各班以小组为单位组织初赛,规定满分为分,分及以上为优秀.数据整理:小明将本班甲、乙两组同学(每组人)初赛成绩整理为如下统计图:数据分析:小明对这两个小组的成绩进行了如下分析:平均数(分) 中位数(分) 众数(分) 方差 优秀率甲组乙组请认真阅读上述信息,回答下列问题:(1)求出表格中a、b的值;(2)小亮同学说:“这次竞赛我得了分,在我们小组中略偏上!”观察上面表格判断,小亮可能是___________组的学生(填“甲”或“乙”);(3)结合以上信息,你认为哪个小组的初赛成绩较好?并说出两条理由.21.(10分)如图,在平行四边形中,,在取一点E,使得,连接.(1)用尺规完成以下基本作图:作的角平分线交于点F,交于点O;(保留作图痕迹,不写作法和结论)(2)根据 (1)中作图,经过学习小组讨论发现,请你证明学习小 组发现的结论.22.(12分)综合与实践同学,还记得学习研究一次函数的路径吗?请结合一次函数的学习经验探究函数的图象.x … 0 1 2 …y … 3 m n 3 …(1)列表:表格中_____________,_____________;(2)在下面的平面直角坐标系中画出该函数的图象; (3)观察(2)中所画函数的图象,求出当x取何值时该函数的有最小值.最小值是多少?(4)写出关于的方程的解,并利用(2)中的图像简单说明此方程的解是如何得到的.23.(12分)四边形为正方形,点为线段上一点,连接,过点作,交射线于点,以为邻边作矩形.(1)如图1,求证:;(2)在图1中连接CG,若,,求的长度;(3)当线段与正方形的某条边的夹角是时,直接写出的度数.图1 展开更多...... 收起↑ 资源列表 广西河池市宜州区2024-2025学年八年级下学期期末检测道数学试卷.docx 广西河池市宜州区2024-2025学年八年级下学期期末检测道数学试卷答案.pdf