《2026年通城学典全国中考试题分类精粹中考数学》

24. (2025·苏州)两个智能机器人在如图所示的$Rt\triangle ABC$区域工作，$\angle ABC = 90^{\circ}$，$AB = 40m$，$BC = 30m$，直线$BD$为生产流水线，且$BD$平分$\triangle ABC$的面积(即$D$为$AC$的中点)。机器人甲从点$A$出发，沿$A\rightarrow B$的方向以$v_1m/min$的速度匀速运动，其所在位置用点$P$表示，机器人乙从点$B$出发，沿$B\rightarrow C\rightarrow D$的方向以$v_2m/min$的速度匀速运动，其所在位置用点$Q$表示。两个机器人同时出发，设机器人运动的时间为$t\ min$，记点$P$到$BD$的距离(即垂线段$PP'$的长)为$d_1m$，点$Q$到$BD$的距离(即垂线段$QQ'$的长)为$d_2m$。当机器人乙到达终点时，两个机器人立即同时停止运动，此时$d_1 = 7.5$。$d_2$与$t$的部分对应数值如下表($t_1 ＜ t_2$)：

(1)机器人乙运动的路线长为55$m$；
(2)求$t_2 - t_1$的值；
(3)当机器人甲、乙到生产流水线$BD$的距离相等(即$d_1 = d_2$)时，求$t$的值。

答案：
24.
(1) 55 解析：$\because \angle ABC = 90^{\circ}$，$AB = 40 m$，$BC = 30 m$，
$\therefore AC = \sqrt{30^{2} + 40^{2}} = 50 m$。$\because$D为AC的中点，$\therefore CD = \frac{1}{2}AC = 25 m$。$\because BC + CD = 30 + 25 = 55(m)$，$\therefore$机器人乙
运动的路线长为$55 m$。
(2) 根据题意，得$v_{2} = \frac{55}{5.5} = 10$。$\because$在$Rt \triangle ABC$中，
$\angle ABC = 90^{\circ}$，D为AC的中点，$\therefore BD = CD = AD = 25 m$。
$\therefore \angle ABD = \angle BAC$，$\angle DBC = \angle C$。$\therefore \sin \angle ABD = \sin \angle BAC = \frac{3}{5}$，$\sin \angle DBC = \sin C = \frac{4}{5}$。当点Q在BC上
时，$QQ' = BQ · \sin \angle DBC = 10t × \frac{4}{5} = 8t(m)$，即$d_{2} = 8t$。
$\therefore 8t_{1} = 16$，解得$t_{1} = 2$。当点Q在CD上时，过点A作
$AH \perp BD$，垂足为H（如图），则$AH = AB · \sin \angle ABD = 40 × \frac{3}{5} = 24(m)$。$\because \angle CDB = \angle ADH$，$\therefore \sin \angle CDB = \sin \angle ADH = \frac{24}{25}$。$\therefore QQ' = QD · \sin \angle CDB = (55 - 10t) × \frac{24}{25} = \frac{264}{5} - \frac{48}{5}t(m)$，即$d_{2} = \frac{264}{5} - \frac{48}{5}t$。$\therefore \frac{264}{5} - \frac{48}{5}t_{2} = 16$，
解得$t_{2} = \frac{23}{6}$。$\therefore t_{2} - t_{1} = \frac{11}{6}$
(3) 当$t = 5.5$时，
$d_{1} = 7.5$，此时$BP = \frac{PP'}{\sin \angle ABD} = \frac{7.5}{\frac{3}{5}} = 12.5(m)$，$\therefore AP = AB - BP = 40 - 12.5 = 27.5(m)$。$v_{1} = \frac{27.5}{5.5} = 5$。$\therefore PP' = BP · \sin \angle ABD = (24 - 3t) × \frac{3}{5} = \frac{72}{5} - \frac{9}{5}t(m)$，即$d_{1} = 24 - 3t$。当点Q在BC上时，由$d_{1} = d_{2}$，得$24 - 3t = 8t$，解得$t = \frac{24}{11}$。当点Q在CD上时，由$d_{1} = d_{2}$，得$24 - 3t = \frac{264}{5} - \frac{48}{5}t$，解得$t = \frac{48}{11}$。$\therefore t = \frac{24}{11}$或$t = \frac{48}{11}$

25. (2025·内江)如图，在$\triangle ABC$中，$\angle C = 90^{\circ}$，$\angle ABC$的平分线$BD$交$AC$于点$D$，$O$是边$AB$上一点，以点$O$为圆心、$OB$长为半径作圆，$\odot O$恰好经过点$D$，交$AB$于点$E$。
(1)求证：直线$AC$是$\odot O$的切线。
(2)若$E$为$AO$的中点，$AD = 3$，求涂色部分的面积。
(3)连接$DE$。若$\sin\angle DBA = \frac{\sqrt{5}}{5}$，求$\cos A$的值。

答案： 25.
(1) 连接OD。$\because \angle C = 90^{\circ}$，$\therefore BC \perp AC$。$\because$BD是
$\angle ABC$的平分线，$\therefore \angle OBD = \angle CBD$。$\because OB$是$\odot O$的半
径，$\odot O$恰好经过点D，交AB于点E，$\therefore OE = OD = OB$。
$\therefore \angle ODB = \angle OBD$。$\therefore \angle ODB = \angle CBD$。$\therefore OD // BC$。
$\therefore OD \perp AC$。又$\because OD$是$\odot O$的半径，$\therefore$直线AC是$\odot O$的
切线
(2) 设$\odot O$的半径为R，则$OD = OE = R$。$\because$E
为AO的中点，$\therefore AE = OE = R$。$\therefore AO = 2R$。由
(1) 可知，
$OD \perp AC$，$\therefore$在$Rt \triangle AOD$中，$\sin A = \frac{OD}{AO} = \frac{R}{2R} = \frac{1}{2}$。
$\therefore \angle A = 30^{\circ}$。$\therefore \angle AOD = 60^{\circ}$。$\because AD = 3$，$\tan A = \frac{OD}{AD}$，
$\therefore OD = AD · \tan A = 3 × \tan 30^{\circ} = \sqrt{3}$。$\therefore S_{扇形EOD} = \frac{60\pi × (\sqrt{3})^{2}}{360} = \frac{\pi}{2}$。$\therefore$涂
色部分的面积为$S_{\triangle AOD} - S_{扇形EOD} = \frac{1}{2} × 3 × \sqrt{3} - \frac{\pi}{2} = \frac{3\sqrt{3} - \pi}{2}$
(3) $\because BE$是
$\odot O$的直径，$\therefore \angle BDE = 90^{\circ}$。$\because$在$Rt \triangle BDE$中，
$\sin \angle DBA = \frac{DE}{BE} = \frac{\sqrt{5}}{5}$，$\therefore$设$DE = \sqrt{5}a$，$BE = 5a$。由勾股定
理，得$BD = \sqrt{BE^{2} - DE^{2}} = \sqrt{(5a)^{2} - (\sqrt{5}a)^{2}} = 2\sqrt{5}a$。
$\because \angle OBD = \angle CBD$，$\angle BDE = \angle C = 90^{\circ}$，$\therefore \triangle BDE \sim \triangle BCD$。$\therefore \frac{DE}{CD} = \frac{BD}{BC} = \frac{BE}{BD}$。
$\therefore \frac{\sqrt{5}a}{CD} = \frac{2\sqrt{5}a}{BC} = \frac{5a}{2\sqrt{5}a} = \frac{\sqrt{5}}{2}$。$\therefore CD = 2a$，$BC = 4a$。由
(1) 可知，
$OD // BC$，$\therefore \triangle AOD \sim \triangle ABC$。$\therefore \frac{AD}{AC} = \frac{OD}{BC}$。$\therefore \frac{AD}{AD + 2a} = \frac{2.5a}{4a} = \frac{5}{8}$。
在$Rt \triangle AOD$中，由勾股定理，得$AO = \sqrt{AD^{2} + OD^{2}} = \sqrt{(\frac{10a}{3})^{2} + (2.5a)^{2}} = \frac{25a}{6}$，$\therefore \cos A = \frac{AD}{AO} = \frac{\frac{10a}{3}}{\frac{25a}{6}} = \frac{4}{5}$。