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青浦区2022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
Q2023.2
一、选择题：（本大题共6题，每小题4分，满分24分）
1．D； 2．C； 3．C； 4．B； 5．B； 6．A．
二、填空题：（本大题共12题，每小题4分，满分48分）
7． ； 8． ； 9． ； 10． ； 11．上升； 12． ；
13． ； 14． ； 15． ； 16． ； 17． ； 18． ．
三、解答题：
19．解：原式= ．···········································（4分）
= ．································································（4分）
= ．·······················································································（2分）
20．解：（1）∵四边形ABCD是平行四边形，
∴AB//CD，AD//BC，AB=CD．·······················································（1分）
∴ ．···································································（2分）
∵DF=2AF， ∴ ．····························································（1分）
∴ ．·············································································（1分）
（2）∵四边形ABCD是平行四边形，∴AD∥BC，AD=BC．················（1分）
∵DF=2AF，∴ ．······················································（1分）
∵ ， ，∴
，
．···························（2分）∴ ．·························································（1分）
21．解：（1）∵AD⊥BC，AD=4，sin∠C= ，
∴ ∴ ．·················································（2分）
，
∴在Rt△ACD中， ．
∵BC=5， ∴BD=BC–CD=5–2=3．······················································（1分）
∵在Rt△ABD中， ，·········································（1分）
∴sin∠BAD= ．····································································（1分）
（2）∵AB=BC=5，BF平分∠ABC，
∴BF⊥AC， ．··························································（2分）
∴∠AFE=∠ADC，又∵∠EAF=∠CAD，∴ ∽ ，·····················（1分）
∴ ．即 ．∴ ．······································（2分）
22．解：设OH的长为x米．··········································································（1分）
在Rt△OBH中，∵ ，∴ ．····················（3分）
在Rt△AOH中，∵ ，∴ ．·········（3分）
∵AB =AH -BH，AB=13，
∴ ． 解得x= （米）．·····································（2分）
∴路灯顶端O到地面的距离OH的长为8.7米．······································（1分）23．证明：（1）∵ ，∴ ．·············································（1分）
又∵∠BAE=∠CAB，∴△ABE∽△ACB．··············································（1分）
∴∠ABF=∠C，∠ABC=∠AEB．·························································（1分）
∵∠ABC=∠AFE，∴∠AFE=∠AEB．···················································（1分）
∴180°–∠AFE=180°–∠AEB，即∠AFB=∠BEC．····································（1分）
∴△ABF ∽△BCE．·········································································（1分）
（2）∵△ABF ∽△BCE，∴ ，∠CBE=∠BAF．··························（2分）
又∵∠BDF=∠ADB，∴△DBF ∽△DAB．············································（1分）
∴ ∴ ．······························································（2分）
，
∴ ． ·····································································（1分）
24．解：（1）将A（-1，0）、B（2，0）代入 ，得
解得： ··············································（2分）
所以， ． ·······························································（1分）
当x=0时， ．∴点C的坐标为（0，2）······································（1分）
（2）①过点P作PH⊥BC，垂足为点H．
∵P（1，m）在 上，
∴ ，P（1，2） ．················································（1分）
∵C（0，2），B（2，0） ，
∴ ．PC⊥OC，∠BCO=45°，∠PCH=45°．····························（1分）
∴ ．BH=BC–CH= ．·····················（1分）
∴tan∠PBC= ．····················································（1分）
②由题意可知，点Q在第二象限．过点Q作QD⊥x轴，垂足为点D．∵∠QBP=∠CBA=45°，∴∠QBD=∠CBP．
∵tan∠PBC= ．∴tan∠QBD = ．···········································（1分）
设DQ=a，则BD=3a，OD=3a-2．∴Q（2-3a，a）．····························（1分）
将Q（2-3a，a）代入 ，得 ．
解得 ， （舍）．∴P（ ， ）．··································（2分）
25．解：（1）过D作DH⊥BC，垂足为点H．···················································（1分）
∵∠C= 90° ，∴DH∥AC．∴ ．··································（1分）
∵BD=DE=5t，∴BH=EH=4t．··························································（1分）
又∵BC=8，CE=4t，∴12t=8，t= ．····················································（1分）
（2）当t= 时，得BD=2，CE= ，BE= ．
∵BE>BD，∴点F是射线ED与直线BF的交点···································（1分）
过E作EG∥AC，交AB于点G，则BF∥GE∥AC．
∴ ， ．∴ ．·····························（1分）
∴ ．········································（1分）
，
∴ ．····················································（1分）
（3）（i）当点F是射线ED与BF的交点时，
∵∠BDE>∠F，∠BDE>∠FBD，又∵△BDE与△BDF相似，
∴∠BDE=∠BDF=90°．∵∠BDE=∠C，∠DBE=∠CBA，
∴ ∽ ．·································································（1分）
∴ ．即 ．解得 ． ∴ ．·········（1分）∵∠F=∠DBE，∴sin∠F=sin∠DBE．∴ ．解得 ．（1分）
（ii）当点F是射线DE与BF的交点时，
∵△BDE与△BDF相似，又∵∠BDE=∠BDF， ∴∠DBE=∠F，即∠ABC=∠F，
又∵∠EBF=∠C，∴ ∽ ．
，即 ．解得 ．·················（1分）
∴
过D作DM⊥BC，垂足为点M．由BD=5t，得DM=3t，BM=4t，EM==8t–8．
∵BF∥DM，∴ ∠EDM=∠F=∠ABC．∴tan∠EDM=tan∠ABC．
DM= ．∴ ．解得 ．························（1分）
∴
．··························································（1分）
∴
综上所述，当△BDE与△BDF相似时， BF的长为 或 ．
