青浦区2020学年第一学期九年级期终学业质量调研测试数学试卷答案

青浦区2021学年第一学期期终学业质量调研 九年级数学试卷
参考答案及评分说明Q 2021.1
一、选择题：
1．B ； 2．C ； 3．B ； 4．A ； 5．D ； 6．C ．
二、填空题：
7．17
； 8．6+a b ； 9．2:3； 10．22=-+y x ； 11．下降； 12．1-； 13．6； 14．32； 15．12； 16
； 17．-a b a ； 18
．3． 三、解答题：
19．解：原式
0112+223
⎛⎫-⨯ ⎪⎝⎭． ················································ （4分）
1- ······························································· （4分）
． ··················································································· （2分）
20．解：（1）∵AD =2，DB =4，AE =3，EC =6，
∴12=AD DB ，12=AE EC ．∴=AD AE DB EC
． ······································ （2分） ∴DE//BC ． ·············································································· （1分） ∴=AD DE AB BC
． ········································································ （1分） 又∵AB =6，DE =3.2， ∴2 3.26=BC
．∴BC =9.6． ···························································· （1分） （2）∵DE//BC ，∴
=AD DE AB BC ．∴13=DE BC ． ∴3=BC DE ． ······································································· （1分）
∵=DE a ，∴3=BC a ． ∴3=-CB a ． ··································· （1分） ∵23=BD BA ，∴23=BD BA ． ····················································· （1分） ∵=BA b ；∴23=BD b ． ························································· （1分） ∵=+CD CB BD ，∴233=-+
CD a b ． ······································ （1分） 21．解：（1）∵四边形ABCD 是平行四边形，
∴AD//BC ，AD =BC ． ································································· （1分）
∴=AD DE BG BE ，=DH DF BG FB
． ····················································· （2分） ∵BE=EF=FD ，AD=BC =8， ∴821
=BG ， ∴BG =4． ····························································· （1分） ∴
142=DH ，∴DH =2． ····························································· （1分） （2）∵AD ∥BC ，
∴△BEG ∽△DEA ，△HFD ∽△GFB ． ∴2⎛⎫= ⎪⎝⎭BEG DEA S BE S
ED ，2⎛⎫= ⎪⎝⎭HFD GFB S DF S FB ． ····································· （2分） ∵=BEG S a ，BE =EF ，∴2=BGF S a ．
∴14=DEA a
S ，124
=HFD S a ． ∴=4DEA S a ，12
=HFD S a ． ····················································· （2分） ∵=四边形-DEA HFD AEFH S S S ，∴17=422
四边形-=AEFH S a a a ． ··········· （1分）
22．解：该车不超速．
过点P 作PH ⊥AC ，垂足为点H ． ······················································ （1分） 由题意，得PH =50米．
在Rt △AHP 中，∵tan ∠=PH PAC AH ，∴50100tan 26.5=≈︒
AH ． ··········· （3分） 在Rt △BHP 中，∵tan ∠=
PH PBH BH ，∴5020tan 68.2=≈︒BH ． ············ （3分） ∵AB =AH -BH ，∴AB =100 -20=80（米）． ·········································· （1分） ∵这辆车通过AB 段的时间为9秒，∴这辆车通过AB 段的速度为809
米/秒．（1分） ∵809
米/秒=32千米/时 < 40千米/时，∴该车不超速． ··························· （1分） 23．证明：（1）∵AE CE DE BE ⋅=⋅， ∴=AE DE BE EC
． ····································································· （1分） 又∵∠AED =∠BEC ，∴△AED ∽△BEC ． ····································· （1分） ∴∠ADE =∠BCE ． ·································································· （1分） ∵AB =AD ，∴∠ABE =∠ADE ． ··················································· （1分） ∴∠ABE =∠BCE ． ··································································· （1分） 又∵∠BAE =∠CAB ，∴△ABE ∽△ACB ． ···································· （1分）
（2）∵2
DA DE DB =⋅，∴=DE DA DA DB ． ············································· （1分） 又∵∠EDA =∠ADB ，∴△EDA ∽△ADB ． ··································· （1分） ∴∠DAE =∠DBA ． ·································································· （1分） ∵∠ABE =∠BCE ，∴∠DAE =∠BCE ．
∴AD ∥BC ． ··········································································· （1分） ∴=AD AE BC EC
． ······································································· （1分） ∴⋅=⋅AD EC BC AE ．
∵AB =AD ， ∴⋅=⋅AB EC BC AE ． ·········································· （1分）
24．解：（1）将A （-4，0）、B （2，0）代入2
+4=-y ax bx ，得 164404240.，--=⎧⎨+-=⎩a b a b 解得：121.
，⎧=⎪⎨⎪=⎩a b ········································ （2分） 所以，2142
=+-y x x ． ························································· （1分） 当x =0时，4=-y ．∴点C 的坐标为（0，-4） ······························ （1分）
（2）过点A 作AH ⊥DC ，垂足为点H ．
∵D （-8，0）、C （0，-4），∴
= ······················ （1分） ∵1122
=⋅=⋅ADC S CD AH DA OC ， ·················································· （1分）
∴44=⨯AH
．∴=AH ． ··············································· （1分） ∵
=
5
=． ················ （1分） ∴tan ∠
ACD=13
==AH HC ． ············································· （1分） （2）由题意可知，点P 在第一象限．过点P 作PQ ⊥x 轴，垂足为点Q ．
∵A （-4，0）、C （0，-4），∴OA =OC ．∴∠OAC =∠OCA ．
∵∠OCD =∠OCA +∠ACD ，∠CAP =∠CAO +∠BAP ，∠OCD =∠CAP ，
∴∠ACD =∠BAP ．∴tan ∠BAP =tan ∠ACD=13
． ···························· （1分） 设PQ =a ，则AQ =3a ，OQ =3a -4．
∴P （3a -4，a ）． ····································································· （1分） 将P （3a -4，a ）代入2142
=+-y x x ，得()21343442-+--=a a a ． 解得120=
9a ，2=0a （舍）．∴P （83，209
）． ······························ （1分）
25．解：（1）∵∠C= 90°，AC =2，BC =
∴AB 4=． ···························································· （1分）
∴2
=BC AB ．
∵BQ =2BP ，∴=BQ BP ． ∴=BQ BC BP AB
． ········································································ （1分） 又∵∠B =∠B ，∴△BQP ∽△BCA ． ············································· （1分） ∴∠BQP =∠BCA ．∵∠C= 90°，∴∠BQP =90°．
即PQ ⊥AB ． ············································································ （1分）
（2）（i ）当∠PQD =90°时，
∵∠PQD < ∠PQA =90°，
∴此种情况不存在． ·································································· （1分） （ii ）当∠QPD =90°时，
∵∠PQB =∠QPD =90°，
∴AB ∥PD ，∴=CP CD BP DA
． ∵CD =DA ， ∴BP =CP ．
∵BC =BP ． ··························································· （2分） （iii ）当∠QDP =90°时，
过点Q 作QH ⊥AC ，垂足为点H ．
设BP =2x ，则BQ ，PC =2x ，QA =4．
∴AH =22-x ，QH =3
2x ，HD =12-x ．
∵∠QDC =∠CDP +90°，∠QDC =∠DQH +90°，
∴∠CDP =∠DQH ．
∴tan ∠CDP =tan ∠DQH ． ∴=CP HD DC QH
．
2
=x ．
解得1x ，2=x （舍）．
∴BP =6
． ································································· （3分）
综上所述，当△PQD 是直角三角形时，线段BP 6．
（3）
33<<BP ． ··························································· （2+2分）。