《2026年1号卷中考试题精编九年级数学安徽专版》

7. 在$Rt\triangle ABC$中,$\angle ACB=90^{\circ}$,$CD$是$AB$边上的中线,$CE\perp AB$于点$E$,若$AC=2BC$,则$\cos\angle DCE$的值为 (D)


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

答案： 7 D 【解析】设$BC = x$，则$AC = 2x$，则$AB = \sqrt{AC^2 + BC^2} = \sqrt{5}x$，由条件可知$\sin \angle CAE = \frac{CE}{AC} = \frac{BC}{AB} = \frac{x}{\sqrt{5}x} = \frac{\sqrt{5}}{5}$，$\therefore CE = \frac{\sqrt{5}}{5}AC = \frac{2\sqrt{5}x}{5}$，由条件可知$AD = CD = \frac{1}{2}AB = \frac{\sqrt{5}}{2}x$，$\therefore \cos \angle DCE = \frac{CE}{CD} = \frac{\frac{2\sqrt{5}x}{5}}{\frac{\sqrt{5}x}{2}} = \frac{4}{5}$.故选D.

8. 新趋势·代数推理 已知实数$x,y$满足$x+3y-6=0$,$y＞3$,$x＞-5$,下列结论正确的是 (C)

A.$x＞-3$
B.$y＞\frac{11}{3}$
C.$-\frac{4}{3}＜x+y＜0$
D.$8＜y-x＜\frac{20}{3}$

答案： 8 C 【解析】逐项分析如下.故选C.
选项 分析 结论是否正确
A $\because x + 3y - 6 = 0$，$\therefore y = \frac{6 - x}{3}$，$\because y ＞ 3$，$\therefore \frac{6 - x}{3} ＞ 3$，解得$x ＜ -3$，又$\because x ＞ -5$，$\therefore -5 ＜ x ＜ -3$. 否
B $\because x + 3y - 6 = 0$，$\therefore x = 6 - 3y$，$\because x ＞ -5$，$\therefore 6 - 3y ＞ -5$，解得$y ＜ \frac{11}{3}$，$\therefore 3 ＜ y ＜ \frac{11}{3}$. 否
C $\because x + 3y - 6 = 0$，$\therefore x + y = 6 - 2y$，$\because 3 ＜ y ＜ \frac{11}{3}$，$\therefore \frac{-22}{3} ＜ -2y ＜ -6$，$\therefore -\frac{4}{3} ＜ 6 - 2y ＜ 0$，$\therefore -\frac{4}{3} ＜ x + y ＜ 0$. 是
D $\because x + 3y - 6 = 0$，$\therefore y - x = 4y - 6$，$\because 3 ＜ y ＜ \frac{11}{3}$，$\therefore 12 ＜ 4y ＜ \frac{44}{3}$，$\therefore 6 ＜ 4y - 6 ＜ \frac{26}{3}$，$\therefore 6 ＜ y - x ＜ \frac{26}{3}$. 否

9. 如图1,平行四边形$ABCD$中,$AB=3$,$BD\perp AB$,动点$F$从点$A$出发,沿折线$ADB$以每秒1个单位长度的速度运动到点$B$.图2是点$F$运动时,$\triangle FBC$的面积$y$随时间$x$变化的图象,则$m$的值为 (A)

A.$6$
B.$10$
C.$12$
D.$20$

答案： 9 A 【解析】由图可知，$AD = a$，$AD + BD = 9$，则$BD = 9 - a$，由$BD \perp AB$，可得$\triangle ABD$是直角三角形，由勾股定理可得$a^2 = (9 - a)^2 + 3^2$，解得$a = 5$，即$AD = 5$，$\therefore BD = 4$，$\therefore m = S_{\triangle BDC} = \frac{1}{2} × 3 × 4 = 6$.故选A.

10. 如图,动点$P$在等边$\triangle ABC$的边$AC$上,$AB=2$,连接$PB$,$AD\perp PB$于点$D$,以$AD$为边在其右侧作等边$\triangle ADE$,$ED$的延长线交$BC$于点$F$,连接$PF$,则下列结论错误的是 (C)


A.$PB$的最小值是$\sqrt{3}$
B.$CD$的最小值是$\sqrt{3}-1$
C.$PF$的最小值是$1$
D.$EF$的最大值是$2$

答案：
10 C 【解析】逐项分析如下.故选C.
选项 分析 结论是否正确
A 由题意得，$AB = AC = BC = 2$，$\angle ABC = \angle C = 60°$，$\therefore$当$BP \perp AC$时，$BP$最小，此时$CP = AP = \frac{1}{2}AC = 1$，$\therefore PB$的最小值为$\sqrt{BC^2 - CP^2} = \sqrt{3}$. 是
B 如图，作$CM \perp AB$于$M$，连接$CD$，$DM$，则$AM = BM = \frac{1}{2}AB = 1$，$CM = \sqrt{BC^2 - BM^2} = \sqrt{3}$，$\because CD \geq CM - DM$，$\therefore$当$C$，$D$，$M$在同一直线上时，$CD$的值最小，$\because AD \perp PB$，$\therefore DM = \frac{1}{2}AB = 1$，$\therefore CD$的最小值为$CM - DM = \sqrt{3} - 1$. 是
C 如图，连接$AF$，$EC$，作$CG // BD$，交$EF$的延长线于$G$，由题意得，$AB = AC$，$AD = AE$，$\angle BAC = \angle DAE = 60°$，$\therefore \angle BAC - \angle DAC = \angle DAE - \angle DAC$，即$\angle BAD = \angle CAE$，$\therefore \triangle BAD \cong \triangle CAE (SAS)$，$\therefore BD = CE$，$\angle AEC = \angle ADB = 90°$，$\because \angle ADP = 180° - \angle ADB = 90°$，$\therefore \angle EDP = \angle ADP - \angle ADE = 30°$，$\angle CEG = \angle AEC - \angle AED = 30°$，$\therefore \angle BDG = \angle EDP = 30°$，$\because CG // BD$，由平行线的性质可得$\angle BDG = \angle G = 30°$，$\angle CEG = \angle G = 30°$，$\therefore CG = CE = BD$，$\because \angle CFG = \angle BFD$，$\therefore \triangle BDF \cong \triangle CGF (AAS)$，$\therefore BF = CF = \frac{1}{2}BC = 1$，$\therefore AF \perp BC$，即$\angle AFC = 90°$，$\therefore \angle AFC + \angle AEC = 180°$，$\therefore A$，$F$，$C$，$E$四点共圆，$\therefore$当$EF$取最大值时，$EF$等于直径$AC$，即为2. 是
D 当$FP \perp AC$时，$FP$最小，此时$FP = CF · \sin 60° = \frac{\sqrt{3}}{2}$. 否

思路点拨
由垂线段最短可得，当$BP \perp AC$时，$BP$最小，此时$CP = AP = 1$，再由勾股定理计算即可判断选项A；作$CM \perp AB$于$M$，连接$CD$，$DM$，由$CD \geq CM - DM$，可得当$C$，$D$，$M$在同一直线上时，$CD$的值最小，即可判断选项B；证明$A$，$F$，$C$，$E$四点共圆，得出当$EF$取最大值时，$EF$等于直径$AC$，即可判断选项D；再由垂线段最短结合解直角三角形即可判断选项C.

11. 因式分解:$3a^{2}-18a+27=$$3(a - 3)^2$.

答案： 11 $3(a - 3)^2$

12. 某校九年级共有1,2,3,4四个班,现从这四个班中随机抽取两个班进行一场篮球比赛,则恰好抽到1班和2班的概率是$\frac{1}{6}$.

答案： 12 $\frac{1}{6}$

13. 如图,$AE$是直径,点$B,C,D$在半圆上,若$\angle B=120^{\circ}$,则$\angle D=$150$^{\circ}$.

答案： 13 150 【解析】如图，连接$BE$，由条件可知$\angle ABE = 90°$，$\therefore \angle CBE = \angle ABC - \angle ABE = 120° - 90° = 30°$，$\therefore \angle D + \angle CBE = 180°$，$\because \angle D = 180° - \angle CBE = 150°$.