数学试题答案及评分参考
评分说明：
1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分参考制定相应的评分细则．
2．对于计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分．
3．解答右端所注分数，表示考生正确做到这一步应得的累加分数． 4．只给整数分数．选择题和填空题不给中间分．
一、选择题：共10小题，每小题4分，满分40分；在每小题给出的四个选项中，只有一项是符合题目要求的，请
在答题卡的相应位置填涂． 1．A 2．C 3．B 4．D 5．B 6．D 7．C 8．B 9．B 10．A
二、填空题：共6小题，每小题4分，满分24分，请在答题卡的相应位置作答． 11．50 12．34 13．＞ 14．70 15．6 16．6
三、解答题：共9小题，满分86分，请在答题卡的相应位置作答． 17．（本小题满分8分）
解：原式?22?1?2?22?1 ························································································ 6分
?2． ················································································································· 8分
18．（本小题满分8分）
解：（1）将（1，4）代入y?kx?2（k≠0），
得k?2?4， ······································································································· 2分 解得k?2， ········································································································· 3分 则该一次函数的解析式为y?2x?2． ········································································ 4分
x 0 ?1 y 0 2 该一次函数的图象如图所示：
y43y?2x?221－2－1
－1－2O 123x
································································ 6分
（2）由图象可得，当y≤0时，x≤?1． ·········································································· 8分
八年级数学试题答案及评分参考第1页（共4页）
19．（本小题满分8分）
解：（1）
M AC DO B ································································ 3分
如图，射线BC，线段BD即为所求作； ···································································· 4分 C M （2）解：由（1）得BD∥OA，BD?OA， DA∴四边形OBDA是平行四边形． ······································································· 5分
∵OA?OB，
∴平行四边形OBDA是菱形， ·········································································· 6分 E ∴DE?1OD，AB⊥OD． ················································································ 7分 N BO 2∵OD?8，AB?6， ∴DE?4，
∴△ABD的面积?1AB?DE?1?6?4?12． ······················································ 8分
2220．（本小题满分8分）
解：如图，依题意得AD?10，FG?1，∠EGD?90°． ···························································· 1分
∵G为AD的中点， F A D ∴GD?1AD?5． ····································································································· 2分 G 2设这根芦苇的长度为x尺，则水池的深度为(x?1)尺． ······························································ 3分 在Rt△DGE中，
根据勾股定理可得EG2?DG2?DE2，
B E C
(x?1)2?52?x2， ················································································································ 5分 解得x?13， ······················································································································· 6分 ∴x?1?12， ······················································································································ 7分 答：水的深度为12尺，这根芦苇的长度为13尺． ···································································· 8分
21．（本小题满分8分）
证法一：∵将△ABO平移得到△DCE，
∴△ABO≌△DCE， ····························································································· 1分 ∴AO?DE，BO?CE． ························································································· 2分 ∵四边形ABCD是平行四边形， ·············································································· 3分 ∴AO?CO，BO?DO， ························································································· 4分 ∴DE?CO，CE?DO， ························································································· 5分 ∴四边形OCED是平行四边形． ············································································· 6分
A D ∵12?22?5?(5)2，
即在△ABO中，OB2?OA2?AB2，
E ∴△ABO是直角三角形，∠AOB?90°， ··································································· 7分
O ∴∠COD?90°，
∴平行四边形OCED是矩形． ················································································ 8分 C B 证法二：∵将△ABO平移得到△DCE，
∴AD∥OE∥BC，AD?OE?BC，····································································································· 1分 ∴四边形AOED，四边形OBCE都是平行四边形， ······················································ 3分 ∴DE∥AO，CE∥BO， ························································································· 5分 ∴四边形OCED是平行四边形． ············································································· 6分 ∵12?22?5?(5)2，
即在△ABO中，OB2?OA2?AB2，
∴△ABO是直角三角形，∠AOB?90°， ··································································· 7分 ∴∠COD?90°，
∴平行四边形OCED是矩形． ················································································ 8分
八年级数学试题答案及评分参考第2页（共4页）
N
22．（本小题满分10分）
解：（1）依题意，得y?(60?5?8?2)x?(68?5?10?2)(30?x) ················································ 2分
?2x?480． ························································································ 4分 ?
?x?0，?
（2）依题意，得?30?x?0， ··················································································································· 6分
?1x?(30?x)，??2解得0＜x≤10． ····································································································································· 7分 ∵2＞0，
∴当0＜x≤10时y随x的增大而增大，·································································································· 8分 ∴当x?10时，y取得最大值， ············································································································· 9分 此时y?2?10?480?500． ················································································································ 10分 ∴8斤装的礼盒数x为10时这30盒水蜜桃售出的利润最大，且利润的最大值为500．
23．（本小题满分10分）
解：（1）该家庭未使用节水龙头的日用水量的平均数约为
0.1?1?0.3?2?0.5?7?0.7?13?0.9?6?1.1?1?0.66． ································································ 4分
30
（2）该家庭使用该节水龙头的日用水量的平均数约为
0.1?1?0.3?6?0.5?14?0.7?7?0.9?2?0.52． ············································································ 8分
30∴估计该家庭使用节水龙头后，一年能节省水365?(0.66?0.52)?51.1 t． ···················· 10分 答：估计该家庭使用节水龙头后，一年能节省水51.1吨水．
24．（本小题满分12分）
解：（1）设直线l1的解析式为y?ax?b． ······································································································· 1分
?b?2，
将A（0，2），B（1，0）代入y?ax?b得? ····································································· 2分
a?b?0，?
?a??2，
解得? ········································································································································· 3分
b?2，?
∴直线l1的解析式为y??2x?2．···································································································· 4分 （2）依题意得y?k(x?1)， ························································································································ 5分
当x?1?0时，k无论取何值都有y?0， ························································································· 6分 此时x??1，
∴直线l2必经过一定点，且该定点坐标为（?1，0）． ·································································· 7分 （3）∵线段AB平移得到线段EF，
∴点A向右平移m个单位，向上平移(n?2)个单位得到点E， ··················································· 8分 ∴F（m?1，n?2）． ··························································································································· 9分 将F（m?1，n?2）代入y?kx?k， 得k(m?1)?k?n?2，
整理得n?km?2k?2． ····················································································································· 10分 当m??2时，n??2k?2k?2?2， ······························································································ 11分 ∴点（?2，2）在n关于m的函数图象上． ················································································· 12分
25．（本小题满分14分）
（1）解：∵四边形ABCD是正方形，
∴∠ADC?90°，DA?DC， ∴∠ADE?90°?α． ··························································································· 1分 ∵△DCE是等腰三角形， ∴DE?DC，
∴DE?DA， ···································································································· 2分
180??(90???)
∴∠DEA??45???； ································································· 3分
22八年级数学试题答案及评分参考第3页（共4页）
（2）证明：∵DC?DE，
∴∠DEC?180????90°??， ······································································ 4分
22∴∠AEF?∠DEF?∠DEA?45°．
∵AD?ED，DG⊥AE， ··················································································· 5分 ∴DF垂直平分AE，
∴FA?FE， ·································································································· 6分 ∴∠EAF?∠AEF?45°， ················································································· 7分 ∴∠AFE?90°，FA?FE， ∴△AEF是等腰直角三角形． ··········································································· 8分
（3）解：过点E作EH⊥CD于点H，连接AC，
∴∠EHD?∠EHC?90°． ∵四边形ABCD是正方形，
∴AB?BC?CD?AD?5，∠ABC?90°， ∴DE?5．
在Rt△ABC中，AC?AB2?BC2?10． ··························································· 9分 在Rt△AFC中，∠AFC?90°，CF?2，
∴AF?AC2?CF2?22， ············································································· 10分 ∴EF?AF?22，
AD ∴EC?2． ·································································································· 11分
解法一：在Rt△EHD和Rt△EHC中，
EH2?ED2?DH2?EC2?CH2． ································································ 12分 G 设CH?x，则DH?5?x， H E 2222∴(5)?(5?x)?(2)?x，
C B 5解得x?， ····················································································· 13分 F 5∴EH?
35， ···················································································· 14分 535． 5∴点E到CD边的距离为解法二：在Rt△AEF中，AE?
AF2?EF2?4．
由（2）得FG?EG?AG?1AE?2．
2在Rt△ADG中，DG?AD2?AG2?1， ∴DF?3， ·························································································· 12分 ∴S△DEF?1DF?EG?3．
2又S△DEF?S△DCF?S△DEC?1CF?h?1CE?h?2S△DEC， ································ 13分
22∴2?1CD?EH?5EH?3，
235∴EH?． ···················································································· 14分
5∴点E到CD边的距离为
35． 5八年级数学试题答案及评分参考第4页（共4页）
2019—2020学年度福州市八年级下学期期末质量检测数学试题答案及评分参考
