《2026年金考卷中考45套汇编数学河北专版》

9. 如图,点$P_{1}\sim P_{8}$是$\odot O$的八等分点.若$\triangle P_{1}P_{3}P_{7}$,四边形$P_{3}P_{4}P_{6}P_{7}$的周长分别为$a,b$,则下列正确的是 (A)

A.$a ＜ b$
B.$a = b$
C.$a ＞ b$
D.$a,b$大小无法比较

答案：
9 A $\triangle P_{1}P_{3}P_{7}$的周长$= P_{1}P_{3} + P_{3}P_{7} + P_{1}P_{7} = a$，四边形$P_{3}P_{4}P_{6}P_{7}$的周长$= P_{7}P_{3} + P_{4}P_{6} + P_{3}P_{4} + P_{6}P_{7} = b$.如图，连接$P_{1}P_{8}$，$P_{7}P_{8}$，则$P_{1}P_{3} + P_{7}P_{8} ＞ P_{1}P_{7}$.$\because$点$P_{1} \sim P_{8}$是$\odot O$的八等分点，$\therefore P_{1}P_{3} = P_{4}P_{6}$，$P_{7}P_{8} = P_{7}P_{8} = P_{3}P_{4} = P_{6}P_{7}$，$\therefore P_{3}P_{4} + P_{6}P_{7} ＞ P_{1}P_{7}$，$\therefore b ＞ a$.

10. 光年是天文学上的一种距离单位,一光年是指光在一年内走过的路程,约等于$9.46×10^{12}km$.下列正确的是 (D)

A.$9.46×10^{12}-10 = 9.46×10^{11}$
B.$9.46×10^{12}-0.46 = 9×10^{12}$
C.$9.46×10^{12}$是 个12位数
D.$9.46×10^{12}$是一个13位数

答案： 10 D

11. 如图,在$Rt\triangle ABC$中,$AB = 4$,点$M$是斜边$BC$的中点,以$AM$为边作正方形$AMEF$.若$S_{正方形AMEF}=16$,则$S_{\triangle ABC}=$ (B)

A.$4\sqrt{3}$
B.$8\sqrt{3}$
C.12
D.16

答案： 11 B 由$S_{正方形AMEF} = 16$，可知$AM = 4.\because$点$M$是斜边$BC$的中点，$\therefore BC = 2AM = 8$，$\therefore AC = \sqrt{8^{2} - 4^{2}} = 4\sqrt{3}$，$\therefore S_{\triangle ABC} = \frac{1}{2}AB· AC = \frac{1}{2} × 4 × 4\sqrt{3} = 8\sqrt{3}$.

12. 如图(1),一个$2×2$的平台上已经放了一个棱长为1的正方体,要得到一个几何体,其主视图和左视图如图(2),平台上至少还需再放这样的正方体 (B)

A.1个
B.2个
C.3个
D.4个

答案：
12 B 要使得到的几何体的主视图和左视图符合题意，平台上至少还需按如图所示再放2个正方体

13. 在$\triangle ABC$和$\triangle A'B'C'$中,$\angle B = \angle B' = 30^{\circ},AB = A'B' = 6,AC = A'C' = 4$,已知$\angle C = n^{\circ}$,则$\angle C'=$ (C)

A.$30^{\circ}$
B.$n^{\circ}$
C.$n^{\circ}$或$180^{\circ}-n^{\circ}$
D.$30^{\circ}$或$150^{\circ}$

答案：
13 C 如图，当$B^{\prime}C^{\prime} = BC$时，$\triangle ABC ≌ \triangle A^{\prime}B^{\prime}C^{\prime}$，$\therefore \angle C^{\prime} = \angle C = n^{\circ}$；当$BC \neq B^{\prime}C^{\prime}$时，$\triangle ABC$与$\triangle A^{\prime}B^{\prime}C^{\prime}$不全等.$\because A^{\prime}C_{1}^{\prime} = A^{\prime}C_{2}^{\prime}$，$\therefore \angle A^{\prime}C_{2}^{\prime}C_{1}^{\prime} = \angle C^{\prime} = n^{\circ}$，$\therefore \angle A^{\prime}C_{2}^{\prime}B^{\prime} = 180^{\circ} - n^{\circ}$，$\therefore \angle C^{\prime} = n^{\circ}$或$180^{\circ} - n^{\circ}$.

名师敲重点
避坑指南 ►►►
本题需分$B^{\prime}C^{\prime} = BC$和$B^{\prime}C^{\prime} \neq BC$两种情况讨论.学生很容易忽略$B^{\prime}C^{\prime} \neq BC$这种情况，从而导致漏解.

14. 如图是一种轨道示意图,其中$\overgroup{ADC}$和$\overgroup{ABC}$均为半圆,点$M,A,C,N$依次在同一直线上,且$AM = CN$.现有两个机器人(看成点)分别从$M,N$两点同时出发,沿着轨道以大小相同的速度匀速移动,其路线分别为$M\rightarrow A\rightarrow D\rightarrow C\rightarrow N$和$N\rightarrow C\rightarrow B\rightarrow A\rightarrow M$.若移动时间为$x$,两个机器人之间距离为$y$,则$y$与$x$关系的图象大致是 (D)

答案： 14 D 当两个机器人分别沿$M \to A$和沿$N \to C$运动时，$y$随$x$的增大而减小；当两个机器人分别沿$A \to D \to C$和沿$C \to B \to A$运动时，$y$等于圆的直径；当两个机器人分别沿$C \to N$和沿$A \to M$运动时，$y$随$x$的增大而增大.故选 D.

15. 如图,直线$l_{1}// l_{2}$,菱形$ABCD$和等边三角形$EFG$在$l_{1},l_{2}$之间,点$A,F$分别在$l_{1},l_{2}$上,点$B,D,E,G$在同一直线上.若$\angle\alpha = 50^{\circ},\angle ADE = 146^{\circ}$,则$\angle\beta=$ (C)

A.$42^{\circ}$
B.$43^{\circ}$
C.$44^{\circ}$
D.$45^{\circ}$

答案：
15 C $\because \triangle EFG$是等边三角形，$\therefore \angle FEG = \angle EFG = 60^{\circ}$.如图，延长$FE$交$l_{1}$于点$H$.在四边形$ADEH$中，$\angle \alpha = 50^{\circ}$，$\angle ADE = 146^{\circ}$，$\angle DEH = \angle FEG = 60^{\circ}$，$\therefore \angle AHE = 360^{\circ} - (50^{\circ} + 146^{\circ} + 60^{\circ}) = 104^{\circ}.\because l_{1} // l_{2}$，$\therefore \angle EFG + \angle \beta = \angle AHE = 104^{\circ}$，$\therefore \angle \beta = 104^{\circ} - 60^{\circ} = 44^{\circ}$.

一题多解
如图，设直线$BD$分别交$l_{1}$，$l_{2}$于点$H$，$G$，$\because \angle ADE = 146^{\circ}$，$\therefore \angle ADB = 180^{\circ} - \angle ADE = 34^{\circ}.\because \angle \alpha = \angle ADB + \angle AHD$，$\therefore \angle AHD = \angle \alpha - \angle ADB = 50^{\circ} - 34^{\circ} = 16^{\circ}.\because l_{1} // l_{2}$，$\therefore \angle GIF = \angle AHD = 16^{\circ}.\because \angle EGF = \angle \beta + \angle GIF$，$\therefore \angle \beta = \angle EGF - \angle GIF = 60^{\circ} - 16^{\circ} = 44^{\circ}$.

16. 已知二次函数$y = -x^{2}+m^{2}x$和$y = x^{2}-m^{2}$($m$是常数)的图象与$x$轴都有两个交点,且这四个交点中每相邻两点间的距离都相等,则这两个函数图象对称轴之间的距离为 (A)

A.2
B.$m^{2}$
C.4
D.$2m^{2}$

答案： 16 A 对于函数$y = -x^{2} + m^{2}x = -x(x - m^{2})$，当$y = 0$时，$x_{1} = 0$，$x_{2} = m^{2}$，$\therefore$抛物线$y = -x^{2} + m^{2}x$与$x$轴的交点坐标分别为$(0,0)$，$(m^{2},0)$，$\therefore$其对称轴为直线$x = \frac{m^{2}}{2}$.对于函数$y = x^{2} - m^{2} = (x + m)(x - m)$，当$y = 0$时，$x_{1} = m$，$x_{2} = -m$，$\therefore$抛物线$y = x^{2} - m^{2}$与$x$轴的交点坐标分别为$(m,0)$，$(-m,0)$，$\therefore$其对称轴为直线$x = 0.\because$这两个函数的图象与$x$轴都有两个交点，$\therefore m \neq 0$.当$m ＞ 0$时，$\because$这四个交点中每相邻两点间的距离都相等，$\therefore m^{2} - m = 0 - (-m)$，解得$m_{1} = 0$(舍去)，$m_{2} = 2$，$\therefore$这两个函数图象对称轴之间的距离为$2$.当$m ＜ 0$时，同理可求得这两个函数图象对称轴之间的距离也为$2$.