《2026年金考卷中考试题汇编45套数学山东专版》

9. 如图,在$Rt△ABC$中,$∠C = 90^{\circ}$,以点$A$为圆心,适当长为半径作弧,分别交$AC$,$AB$于点$M$,$N$,再分别以点$M$,$N$为圆心,大于$\frac{1}{2}MN$的长为半径作弧,两弧交于点$G$,作射线$AG$交$BC$于点$D$,再分别以点$A$,$B$为圆心,大于$\frac{1}{2}AB$的长为半径作弧,两弧相交于点$P$,作直线$PQ$,分别交$AB$,$AD$于点$E$,$F$.若$AC = 6$,$BC = 8$,则$DF$的长为 (A)

A.$\frac{\sqrt{5}}{2}$
B.$\frac{4}{3}$
C.1
D.$\frac{3\sqrt{5}}{5}$

答案：
9 A
快招解题法 试题秒解 考场速用
如图,过点$D$作$DH \perp AB$于点$H$.在$Rt \triangle ABC$中,由勾股定理,得$AB = \sqrt{AC^{2} + BC^{2}} = \sqrt{6^{2} + 8^{2}} = 10$.由作图可知,$AD$平分$\angle CAB$,$PQ$垂直平分$AB$,$\therefore CD = DH$(依据:角平分线上的点到角两边的距离相等),$AE = BE = \frac{1}{2}AB = 5$.又$\because AD = AD$,$\therefore Rt \triangle ACD \cong Rt \triangle AHD(HL)$,$\therefore AH = AC = 6$,$\therefore HB = AB - AH = 4$.设$CD = HD = x$,则$DB = CB - CD = 8 - x$.

在$Rt \triangle DHB$中,由勾股定理,得$DH^{2} + BH^{2} = DB^{2}$,即$x^{2} + 4^{2} = (8 - x)^{2}$,解得$x = 3$,$\therefore CD = HD = 3$,$\therefore AD = \sqrt{AC^{2} + CD^{2}} = \sqrt{6^{2} + 3^{2}} = 3\sqrt{5}$.$\because DH \perp AB$,$EF \perp AB$,$\therefore \triangle AFE \sim \triangle ADH$(点拨:“A”字型相似模型),$\therefore \frac{AF}{AD} = \frac{AE}{AH}$,即$\frac{AF}{3\sqrt{5}} = \frac{5}{6}$,$\therefore AF = \frac{5\sqrt{5}}{2}$,$\therefore FD = AD - AF = 3\sqrt{5} - \frac{5\sqrt{5}}{2} = \frac{\sqrt{5}}{2}$.
巧作辅助线:见角平分线,常向两边作垂线
更多讲解详见《解题有招》折页“快招2”
一题多解
如图,过点$F$作$FJ \perp AC$于点$J$.在$Rt \triangle ABC$中,由勾股定理,得$AB = \sqrt{AC^{2} + BC^{2}} = \sqrt{6^{2} + 8^{2}} = 10$.由作图可知$AD$平分$\angle CAB$,$PQ$垂直平分$AB$,$\therefore FJ = EF$,$AE = \frac{1}{2}AB = 5$.又$\because AF = AF$,$\therefore Rt \triangle AFJ \cong Rt \triangle AFE$,$\therefore AJ = AE = 5$.易得$\frac{CD}{DB} = \frac{AC}{AB} = \frac{3}{5}$,$\therefore CD = \frac{3}{8}BC = 3$,$\therefore AD = \sqrt{AC^{2} + CD^{2}} = \sqrt{6^{2} + 3^{2}} = 3\sqrt{5}$.$\because FJ \perp AC$,$\angle C = 90^{\circ}$,$FJ // CD$,$\therefore \triangle AFJ \sim \triangle ADC$,$\therefore \frac{AF}{AD} = \frac{AJ}{AC}$,即$\frac{AF}{3\sqrt{5}} = \frac{5}{6}$,$\therefore AF = \frac{5\sqrt{5}}{2}$,$\therefore FD = AD - AF = \frac{\sqrt{5}}{2}$.

名师敲重点
知识积累
与角平分线相关的重要结论
如图,$AD$是$\triangle ABC$的角平分线,有如下常用结论:
1. 点$D$到$AB,AC$的距离相等.
2. $\frac{AB}{AC} = \frac{BD}{CD}$(证明过程:$\because AD$是$\triangle ABC$的角平分线,$\therefore$点$D$到$AB,AC$的距离为$h$,则$\frac{S_{\triangle ABD}}{S_{\triangle ACD}} = \frac{\frac{1}{2}AB · h}{\frac{1}{2}AC · h} = \frac{BD}{CD}$,即$\frac{AB}{AC} = \frac{BD}{CD}$)
3. 若$\angle BAC = 90^{\circ}$,则点$D$到$AB,AC$的距离$h = \frac{AB · AC}{AB + AC}$

10. 新考法 定义函数的“$a$级封闭定义域” 在平面直角坐标系中,定义:已知$y$是$x$的函数,如果对于任意两个不相等的自变量$x_{1}$,$x_{2}(x_{1} ＜ x_{2})$,当$x_{1} ≤ x ≤ x_{2}$时,$y$的取值范围是$ax_{1} ≤ y ≤ ax_{2}(a ＞ 0)$,那么将$x_{1} ≤ x ≤ x_{2}$称为这个函数的“$a$级封闭定义域”.例如:函数$y = 3x$,当$1 ≤ x ≤ 3$时,$3 ≤ y ≤ 9$,所以$1 ≤ x ≤ 3$是函数$y = 3x$的“3级封闭定义域”.下列结论:①$1 ≤ x ≤ 2$是函数$y = -x + 3$的“1级封闭定义域”;②若$0 ≤ x ≤ b(b ＞ 0)$是函数$y = x^{2}$的“2级封闭定义域”,则$b = 2$;③若函数$y = kx + 4$存在“3级封闭定义域”,则$k = -3$;④函数$y = -4x^{2} + 3x + 4$不存在“4级封闭定义域”.其中正确结论的个数是 (C)

A.1
B.2
C.3
D.4

答案： 10 C 逐个分析如下,故正确结论的个数为3,故选C.
序号 分析 正误
对于$y = -x + 3$,因为$-1 ＜ 0$,所以$y$随$x$的增大而减小,所以当$1 \leq x \leq 2$时,$1 \leq y \leq 2$(点拨:当$x = 1$时,$y$取得最大值,当$x = 2$时,$y$取得最小值),所以$1 \leq x \leq 2$是函数$y = -x + 3$的“1级封闭定义域”. $\surd$
当$x = 0$时,$y = 0$,当$x = b$时,$y = b^{2}$.因为$0 \leq x \leq b(b ＞ 0)$是函数$y = x^{2}$的“2级封闭定义域”,所以$0 \leq y \leq 2b$.对于函数$y = x^{2}$,当$x \geq 0$时,$y$随$x$的增大而增大,所以$b^{2} = 2b$,所以$b = 2$或$b = 0$.又$b ＞ 0$,所以$b = 2$. $\surd$
对于函数$y = kx + 4$,分两种情况讨论.(i)当$k ＞ 0$时,$y$随$x$的增大而增大.若此时函数存在“3级封闭定义域”,则存在满足题意的$x_{1},x_{2}$,使得$\begin{cases}kx_{1} + 4 = 3x_{1}, \\kx_{2} + 4 = 3x_{2}.\end{cases}$两式相减,得$k(x_{2} - x_{1}) = 3(x_{2} - x_{1})$.因为$x_{1} ＜ x_{2}$,所以$x_{2} - x_{1} \neq 0$,所以$k = 3$.当$k = 3$时,$3x + 4 = 3x$,此时$0 = 4$,故此种情况不存在.(ii)当$k ＜ 0$时,$y$随$x$的增大而减小,若此时函数存在“3级封闭定义域”,则存在满足题意的$x_{1},x_{2}$,使得$\begin{cases}kx_{1} + 4 = 3x_{2}, \\kx_{2} + 4 = 3x_{1}.\end{cases}$两式相减,得$k(x_{1} - x_{2}) = 3(x_{2} - x_{1})$.因为$x_{1} - x_{2} \neq 0$,所以$4(x_{1} + x_{2}) - 7 = 0$,所以$x_{1} + x_{2} = \frac{7}{4}$.两式相加,得$-4(x_{1}^{2} + x_{2}^{2}) + 3(x_{1} + x_{2}) + 8 = 4(x_{2} + x_{1})$.将$x_{1} + x_{2} = \frac{7}{4}$代入并整理,得$x_{1}^{2} + x_{2}^{2} = \frac{25}{16}$,所以$x_{1}x_{2} = \frac{1}{2}[(x_{1} + x_{2})^{2} - (x_{1}^{2} + x_{2}^{2})] = \frac{3}{4}$,所以$x_{1},x_{2}$是方程$x^{2} - \frac{7}{4}x + \frac{3}{4} = 0$的两个根,所以$x_{1} = \frac{3}{4},x_{2} = 1$,$\frac{3}{4} \leq x \leq 1$是函数$y = -4x^{2} + 3x + 4$的“4级封闭定义域”. $×$
续表
序号 分析 正误
(iii)当$x_{1} ＜ \frac{3}{8} ＜ x_{2}$时,$y$的最大值为$\frac{73}{16}$,所以$4x_{2} = \frac{73}{16}$,所以$x_{2} = \frac{73}{64}$.当$- \frac{25}{64} ＜ x_{1} ＜ \frac{3}{8}$时,$-4x_{1}^{2} + 3x_{1} + 4 = 4x_{1}$,所以$x_{1} = \frac{2 \pm 271}{4096}$(不合题意,舍去).当$x_{1} ＜ - \frac{25}{64}$时,$-4x_{1}^{2} + 3x_{1} + 4 = 4x_{1}$,所以$x_{1} = \frac{-1 - \sqrt{65}}{8}$(不合题意的值已舍),所以$\frac{-1 - \sqrt{65}}{8} \leq x \leq \frac{73}{64}$是函数$y = -4x^{2} + 3x + 4$的“4级封闭定义域”.综上,函数$y = -4x^{2} + 3x + 4$存在“4级封闭定义域”. $\surd$

11. 如图,转盘中四个扇形的面积都相等,任意转动这个转盘1次,当转盘停止转动时,指针落在灰色区域的概率是$\frac{1}{2}$.

答案： 11 $\frac{1}{2}$

12. 计算$\frac{x - 2}{x - 4}·\frac{x^{2} - 4x}{x^{2} - 4x + 4}$的结果为$\frac{x}{x-2}$.

答案： 12 $\frac{x}{x-2}$
【解析】原式$= \frac{x-2}{x-4} · \frac{x(x-4)}{(x-2)^{2}} = \frac{x}{x-2}$.

13. 如图,正八边形$ABCDEFGH$和正六边形$GHIJKL$的边长均为6,以顶点$H$为圆心,$HG$的长为半径画圆,则阴影部分的面积为$\frac{21\pi}{2}$.(结果保留$π$)

答案：
13 $\frac{21\pi}{2}$
【解析】$\because$八边形$ABCDEFGH$是正八边形,六边形$GHIJKL$是正六边形,$\therefore \angle AHG = \frac{(8-2) × 180^{\circ}}{8} = 135^{\circ}$,$\angle IHG = \frac{(6-2) × 180^{\circ}}{6} = 120^{\circ}$,$\therefore \angle AHI = 360^{\circ} - 135^{\circ} - 120^{\circ} = 105^{\circ}$(关键点),$\therefore S_{阴影} = \frac{105\pi × 6^{2}}{360} = \frac{21\pi}{2}$.
一题多解
如图,延长$GH$交圆$H$于点$M$,则$\angle AHM = 360^{\circ} ÷ 8 = 45^{\circ}$,$\angle IHM = 360^{\circ} ÷ 6 = 60^{\circ}$(点拨:多边形的外角和为$360^{\circ}$,正多边形的每个外角都相等),$\therefore \angle AHI = \angle AHM + \angle IHM = 105^{\circ}$.

14. 某中学组织甲、乙两个生态兴趣小队在公园进行自然寻宝徒步,由出发点步行前往6km远的集合点.学校安排两队在不同时刻出发,已知乙队始终以5km/h的速度匀速前进,甲队匀速前进0.5h后,速度降低为原来的一半,最后两队恰好同时到达集合点.甲、乙两队前进的路程$y$(单位:km)与甲队出发的时间$x$(单位:h)的函数图象如图所示,当甲出发1h时,甲、乙两队相距1km.

答案：
14 1
名师教解题

由题图可知,甲、乙两队从出发点到集合点所用时间分别为$1.5h,6 ÷ 5 = 1.2(h)$,故乙队比甲队晚出发$1.5 - 1.2 = 0.3(h)$.设当$0.5 \leq x \leq 1.5$时,$y_{甲} = kx + b$.将$(0.5,3),(1.5,6)$分别代入,得$\begin{cases}0.5k + b = 3, \\1.5k + b = 6,\end{cases}$解得$\begin{cases}k = 3, \\b = 1.5,\end{cases}\therefore y_{甲} = 3x + 1.5$.设当$0.3 \leq x \leq 1.5$时,$y_{乙} = mx + n$.将$(0.3,0),(1.5,6)$分别代入,得$\begin{cases}0.3m + n = 0, \\1.5m + n = 6,\end{cases}$解得$\begin{cases}m = 5, \\n = -1.5,\end{cases}\therefore y_{乙} = 5x - 1.5$.当$x = 1$时,$y_{甲} = 3 × 1 + 1.5 = 4.5,y_{乙} = 5 × 1 - 1.5 = 3.5$,$\therefore$当甲出发$1h$时,甲、乙两队相距$4.5 - 3.5 = 1(km)$.

15. 如图,在平行四边形$ABCD$中,$∠B = 45^{\circ}$,$AB = 8$,$AD = 10$,$G$为边$BC$上一点,连接$AG$,将$△ABG$沿$AG$翻折,点$B$的对应点为$B'$,$E$为$AD$的中点,$F$为边$CD$上一点,连接$EF$,将$△DEF$沿$EF$翻折,点$D$的对应点恰巧也为$B'$,则$DF =$$\frac{15\sqrt{2}}{7}$.

答案：
15 $\frac{15\sqrt{2}}{7}$
【解析】如图,连接$B'D$,过点$F$作$FH \perp AD$于点$H$,则$\angle EHF = \angle DHF = 90^{\circ}$.$\because$四边形$ABCD$是平行四边形,$\angle B = 45^{\circ}$,$\therefore \angle ADC = \angle B = 45^{\circ}$(依据:平行四边形的对角相等),$\angle HFD = \angle HDF = 45^{\circ}$,$\therefore FH = DH$.$\because E$为$AD$的中点,$AD = 10$,$\therefore DE = AE = \frac{1}{2}AD = 5$.由折叠的性质,得直线$EF$是线段$B'D$的垂直平分线,$AB' = AB = 8$,$B'E = DE = AE$,$\therefore \angle EB'D = \angle EDB'$,$\angle EB'A = \angle EAB'$（依据:等边对等角),$\therefore \angle AB'D = \angle EB'D + \angle EB'A = \angle EDB' + \angle EAB' = \frac{1}{2} × 180^{\circ} = 90^{\circ}$,$\therefore B'D = \sqrt{AD^{2} - AB'^{2}} = \sqrt{10^{2} - 8^{2}} = 6$.$\because EF \perp B'D$,$AB' \perp B'D$,$\therefore EF // AB'$,$\angle DEF = \angle DAB'$,$\therefore \tan \angle DEF = \tan \angle DAB'$,即$\frac{FH}{EH} = \frac{B'D}{AB'} = \frac{3}{4}$,$\therefore EH = \frac{4}{3}FH = \frac{4}{3}DH$.$\because EH + DH = DE = 5$,$\therefore \frac{4}{3}DH + DH = 5$,$\therefore DH = \frac{15}{7}$,$\therefore DF = \sqrt{DH^{2} + FH^{2}} = \sqrt{2}DH = \frac{15\sqrt{2}}{7}$.
巧作辅助线:见$45^{\circ}$角构造等腰直角三角形

16. (本小题满分8分)
(1)计算:$\sqrt{9} + |1 - \sqrt{2}| - (2 - π)^{0} - 2\cos45^{\circ} + (\frac{1}{2})^{-1}$.
(2)解不等式组:$\begin{cases}3(x - 2) ≤ x - 4,①\frac{x - 1}{2} ＜ \frac{1 + 2x}{3},②\end{cases}$并写出它的所有整数解.

答案： 16
(1)原式$= 3 + \sqrt{2} - 1 - 1 - 2 × \frac{\sqrt{2}}{2} + 2 = 3 + \sqrt{2} - 1 - 1 - \sqrt{2} + 2 = 3$.(3分)
(2)解不等式①,得$x \leq 1$,解不等式②,得$x ＞ -5$,(2分)$\therefore$不等式组的解集为$-5 ＜ x \leq 1$,(3分)$\therefore$不等式组的所有整数解为$-4,-3,-2,-1,0,1$.(4分)