2026年北师大七下数学期末模拟卷（陕西专用七下全册）（参考答案）

-2026学年七年级数学下学期期末模拟卷参考答案一、选择题（本大题共8小题，每小题3分，满分24分．在每个小题给出的四个选项中，只有一项符合题目要求的）12345678CBCBABDA二、填空题（本大题共6小题，每小题3分，满分18分）9． 10．144 11．12．2 13．48 14．9.6/三、解答题（本大题共12小题，满分78分．解答应写出文字说明，证明过程或演算步骤）15．（5分）【答案】2【详解】解：····································3分.····································5分16.（5分）【答案】【详解】解：····································3分····································5分17．（5分）【答案】见解析【详解】如图，点即为所求，····································5分18．（5分）【答案】见解析【详解】证明：∵，，∴，····································2分在和中，∵∴．····································5分19．（5分）【答案】【详解】解：····································3分，····································4分当时，原式．····································5分20．（5分）【答案】(1)不可能(2)不公平，见解析【详解】（1）解：∵转盘被等分成6个扇形，分别标有数字1，2，3，4，5，6∴转出的数字大于6是不可能事件，故答案为：不可能····································1分（2）解：根据题意，转出的数字是2的倍数的可能性有3种，∴小明获胜的概率为，转出的数字是3的倍数的可能性有2种，∴小亮获胜的概率为．····································4分∵，∴游戏不公平．····································5分21．（6分）【答案】见解析【详解】解：如图所示，任意画出3个即可，画对一个给2分··········6分22．（7分）【答案】任务一：可行；任务二：见解析【详解】任务一：解：∵该方案可以证明，∴．故答案为：可行．····································1分任务二：解：理由如下，∵，，且，∴．····································2分∵，∴．∴．····································5分∵，∴．∴．····································7分故该方案可行．23．（7分）【答案】(1)离家的时间(2)1500，4(3)在整个上学的途中分钟时速度最快，在安全限度内【详解】（1）解：根据图象可得，横坐标为离家的时间，故图中自变量是离家的时间，故答案为：离家的时间；····································1分（2）解：轴表示路程，起点是家，终点是学校，小明家到学校的路程是米，由图象可知：小明在书店停留了（分钟），故答案为：；4；····································3分（3）解：由图象可知：分钟时，平均速度米/分，····································4分分钟时，平均速度米/分，····································5分分钟时，平均速度米/分，∴在整个上学的途中分钟时速度最快，在安全限度内．···································7分24．（8分）【答案】(1)见解析；(2)【详解】（1）证明：∵，∴，即，··································2分在和中，，∴；····································4分（2）解：由（1）可知，∴，∴，····································5分∵，∴，∴，····································7分又∵O点为中点，∴．····································8分25．（8分）【答案】(1)(2)①；②60【详解】（1）解：由题意得：阴影部分的面积，即；····································2分（2）解：①由（1）可得：，∵，，∴，解得：；····································5分②设，，∴，∵，∴，∴．····································8分26．（12分）【答案】（1）；（2）见解析；（3）【详解】解：（1）∵，∴，∴，又∵，∴，∴；····································2分（2）如图所示，过点C作于E，过点D作交延长线于F，∴；∵，∴；又∵，∴，又∵，∴，∴；∵，且，∴；····································6分（3）如图所示，过点D作交延长线