江蘇省徐州市區(qū)2022—2023學年上學期九年級數(shù)學期中檢測試題

這是一份江蘇省徐州市區(qū)2022—2023學年上學期九年級數(shù)學期中檢測試題，共8頁。試卷主要包含了10 18，2分，， 326分，連接OD.，①證明等內(nèi)容，歡迎下載使用。
2022—2023學年度第一學期期中檢測九年級數(shù)學參考答案   9．4                10.-1        11.（-2，3）      12.       13. 2    14.   15. 5        16.           17．10       18．  19． 方程化一般形式且正確給1分（1）  ···························································3分 ··························································6分（2）法一：，····················································10分∴，．·····················································12分法二：，（10分）∴，．········································12分法三：，（10分）∴，．········································12分 20．（1）··························································2分（2）把（﹣1，0），（3，0），（0，3）代入拋物線中得中·····························································4分·································································7分（3）①···························································8分   ②-5≤y≤4·····················································10分    21．（1）見下圖（有字母和關(guān)鍵格點即可）， (4分)   32·······················6分（2）見下圖（有字母和關(guān)鍵格點即可），（10分）   ······················12分         22．（1）矩形ABCD， AB=xBC=·················1分由題意，得 ．·························3分解得 ，． 當． 當····························································5分答：此時x的值為6 m．············································6分（2）設矩形養(yǎng)殖場的面積為·········································7分由（1）得，···················································10分····························································11分答:當························································12分 23．(1)連接OD.  ∵OA= OD，∴∠OAD=∠ODA  ∵EF垂直平分線交BD，∴DE=BE ∴∠EDB=∠B ·····························3分∵∠C=90°，∴∠OAD +∠B= 90o ∴∠ODA+∠EDB =90°·················································4分∵∠ADB=180°∴∠ODE= 180°-90°=90°∴OD⊥DE ∵OD為半徑∴DE是⊙O的切線····················································6分(2) 連接OE，設DE=x由題意，DE=BE=x，∵AC=6，BC=8，OA=2∴OA=OD=2,OC=4，CE=8-x∵在中，∠OCE=90°，由勾股定理得 ∴， ∵在中，∠ODE=90°，由勾股定理得 ∴， ····························································10分∴，x=，答：DE的長為·····················································12分24．（1）（方法不唯一）設拋物線的表達式為，由題意得：，把點B的坐標為（-3，0），點C（0，﹣3）代入得：，(4分)解得：，(5分)∴拋物線的解析式為；·························6分（2）（方法不唯一）過點D作 DH⊥y軸由題意，DH=1,BO=3,OC=3,CH=1,OH=4,··········································8分 , ∴········································11分（3）·····································14分 25．（1）①證明：∵AB＝BC，∴∠ADB＝∠CDB，··················································2分∴DB平分圓周角∠ADC，∴圓中存在“爪形D”；···············································4分②延長DC至點E，使得CE＝AD，連接BE， 5分∵∠A＋∠DCB＝180°∠ECB＋∠DCB＝180°∴∠A＝∠ECB∵CE＝AD，AB＝BC∴△BAD≌△BCE∴∠E＝∠ADB， BE＝BD······················8分∵∠ADC＝90°，∠ADB＝∠CDB∴∠E＝∠ADB＝45° ∴△BDE為等腰直角三角形····················10分∴由勾股定理得，，即, ∴CE+CD=AD+CD=··························12分 （2）120°··································14分