四川省资阳市安岳县2021-2022学年九年级上学期期末数学试卷(图片版含答案)

2021—2022 学年度第一学期期末义务教育九年级学情诊断
数学学科参考答案及评分意见
一、选择题（共 10小题，每小题 4 分，共 40 分）
题号 1 2 3 4 5 6 7 8 9 10
答案 A A B C D C B D B D
二、填空题（共 6小题，每小题 4分，共 24 分）
11． 4 212． 13．-2
3
(48 24) 2414. 等边 15． ， 16．
5 5 5
三、解答题（共 86分）
17．解：（1）8- 2 3 ·························································································4分
3
（2）原式= ······························································································ 7分
x y
x 3 0 y x 3 3 x 2，∴ ，∴x=3，y=2·············································8分
3 x 0
3
当 x=3 ，y=2时，原式= =3··············································································9分
3 2
18．解：（1） x1 3，x2 0.5 ··········································································5分
x 2 6 x 2 6（2） 1 = ， 2 . ···································································10分2 2
19．解：（1）7 卡片中共有两张卡片写有数字 1，∴从中任意抽取一张卡片，卡片上写有数字 1
2
的概率是 ········································································································ 2分
7
（2）列表或画树状图略························································································8分
7
这个两位数大于 22的概率为 ············································································· 10分
12
20．解: 1 x 2（ ）∵关于 的方程 x 4x k 1 0有两实数根,
∴△=（ 4）2 4 1 （k 1) 0 ············································································2分
解得. k 3 ········································································································3分
（2）由根与系数的关系可得 x1 x2 4, x1.x2 k 1················································· 4分
九年级上数学参考答案及评分意见 第 1页（共 4页） 1
3 3 x x 4 3(x1 x2 )∵ 1 2 ∴ x1x2 4 ········································· 5分x1 x2 x1x2
12
∴ k 1 4 ······························································································ 6分
k 1
解得： k1 3，　k2 5 ····················································································· 8分
又∵ k 3，经检验 k=-3是原方程的根，故 k2 5舍去···············································9分
∴ k 3 ···········································································································10分
21．解：如图 1所示，延长 DA交水平虚线于 F，过 E作 EH⊥BF于 H，····················· 1分
∵∠BAF=90°，∠ABF=37°，
∴Rt△ABF中，AF=tan37°×AB≈0.75×8=6（米），…………3分
∴EF=AF+AD+DE=8.5，………………………………………5分
∵∠EHF=90°=∠BAF，∠BFA=∠EFH，
∴∠E=37°，……………………………………………………7分
∴Rt△EFH中，EH=cos37°×EF≈0.80×8.5=6.8（米），……9分
图 1
又∵底边 AB离地面的距离为 1.3米，
∴点 E离地面的高度为 6.8+1.3=8.1（米）．……………………………………………………….10分
答：此时档板最高点离地面的高为 8.1米．……………………………………………………….11分
22．解：（1）设每件衬衫应降价 x元，则每件盈利（40-x）元，每天可售出（20+2x）件，
由题意得:（40-x）（20+2x）=1200，··········································································2分
即：（x-10）（x-20）=0，解得：x1=10，x2=20····························································· 3分
为了扩大销售量，增加盈利，尽快减少库存，所以 x的值应为 20.
所以，若商场平均每天要盈利 1200元，每件衬衫应降价 20元·········································
（2）假设能达到，由题意，得（40-x）（20+2x）=1500，··············································
整理得：2x2-60x+700=0，
△=602-2×4×700=3600-5600＜0，··············································································
即：该方程无解， P
所以，商场平均每天盈利不能达到 1500元；······························································
Q H
（3）设每件衬衫应降价 x元，
图 2
由题意得:商场平均每天盈利为（40-x）（20+2x），·····················································
=800+80x-20x-2x2，
=-2（x2-30x+225）+450+800，
=-2（x-15）2+1250，······················································
所以当 x=15元时，商场平均每天盈利取得最大值为 1250元，
所以，商场平均每天盈利最多 1250 图 3元，达到最大值时应降价 15元．······························
23．解：（1）∵AB=10，AC=8，∴CB=6····································································
6 8
∴CD= =4.8·····································································································
10
P
九年级上数学参考答案及评分意见 第 2页（共 4页） 2
Q
图 4
（2）如图 2所示，作 PH AB，垂足为 H，BQ=t，CP=t，BP=6-t
6 t PH PH 4(6 t)∴ ，∴
6 4.8 5
∴ S 1 .t. 4(6 t) 2(6 t)t （02(6 t)t
∴ 2，解得：t=1或 5 ................. ................ ................ ..6分
5
（3）存在点 P，使以点 B、P、Q 为顶点的三角形与△ABC 相似，理由如下：分两种情况：
①当∠BQP=90°时，如图 3，此时△PQB∽△ACB，
∴ BP BQ 6 t t ，∴ …………………………………………..8分
AB BC 10 6
解得 t=2.25.................. .......................................................................9分
②当∠BPQ=90°时，如图 4，此时△QPB∽△ACB，
BP BQ 6 t t
∴ ，∴ ……………………………………..11分
BC AB 6 10
解得 t=3.75
综上可得，t=2.25 或 t=3.75．……………………………..12分
24 1 1．解：（1）如图 5，∵P是 CD边的中点，∴DP= DC．∵DC=AB，AB=AP，∴DP= AP．
2 2
∵∠D=90° DP 1，∴sin∠DAP= = ．∴∠DAP=30°．∵∠DAB=90°，∠PAO=∠BAO，∠DAP=30°，
AP 2
∴∠OAB=30°．∴∠OAB的度数为 30°．......................... ........................ .......................... ......... .. ...2分
（2）如图 5
①∵四边形 ABCD是矩形，∴AD=BC，DC=AB，∠DAB=∠B=∠C=∠D=90°．
由折叠得：AP=AB，PO=BO，∠PAO=∠BAO，∠APO=∠B．
∴∠APO=90°．∴∠APD=90°﹣∠CPO=∠POC．
∵∠D=∠C，∠APD=∠POC．∴△OCP∽△PDA．.................. ...................... .................... .. .... .. ..４分
OCP PDA 1 4 OC=OP=CP 1 1②∵△ 与△ 的面积比为 ： ，∴ = =
PD PA DA 4 2
∴PD=2OC，PA=2OP，DA=2CP．∵AD=8，∴CP=4，BC=8．.............. .. .... .. . . .................... .. ...６分
设 OP=x，则 OB=x，CO=8﹣x．
在 Rt△PCO中，∵∠C=90°，CP=4，OP=x，CO=8﹣x，∴x2=（8﹣x）2+42．解得：x=5．
∴AB=AP=2OP=10．∴边 AB的长为 10．.......................... ......................... ........................... ....... .. ..８分
（3）作 MQ∥AN，交 PB于点 Q，如图 6．
∵AP=AB，MQ∥AN，∴∠APB=∠ABP，∠ABP=∠MQP．∴∠APB=∠MQP．∴MP=MQ．
∵MP=MQ，ME⊥PQ 1，∴PE=EQ= PQ．∵BN=PM，MP=MQ，∴BN=QM．
2
∵MQ∥AN，∴∠QMF=∠BNF．.................................... ................................... .................................10分
九年级上数学参考答案及评分意见 第 3页（共 4页） 3
在△MFQ和△NFB中，
QMF BNF
1
QFM BFN．∴△MFQ≌△NFB．∴QF=BF．∴QF= QB．.......................... ............12分
2
QM BN
EF=EQ+QF=1PQ+1QB=1∴ PB．由（2）中的结论可得：PC=4，BC=8，∠C=90°．
2 2 2
∴PB= 82+42=4 5．∴EF=1PB=2 5．............................... ............. .............................. ....................13分
2
图 5 图 6
九年级上数学参考答案及评分意见 第 4页（共 4页） 4