《2026年金考卷中考试题汇编45套数学山东专版》

20. (本小题满分10分)
(2025南充)如图,$Rt\triangle ABC$中,$\angle ACB = 90^{\circ}$,$CD\perp AB$于点$D$,以$CD$为直径的$\odot O$交$BC$于点$E$,交$AC$于点$F$,$M$为线段$DB$上一点,$ME = MD$,连接$OM$.
(1)求证:$ME$是$\odot O$的切线.
(2)若$CF = 3$,$\sin B=\frac{4}{5}$,求$OM$的长.

答案：
20
(1)证明：如图，连接OE.
$\because CD\perp AB,\therefore \angle ODM = 90°$. (1分)
在$\triangle ODM$与$\triangle OEM$中，$OD = OE,OM = OM,DM = EM,\therefore \triangle ODM\cong\triangle OEM(SSS)$ (3分)
$\therefore \angle OEM = \angle ODM = 90°,\therefore ME$为$\odot O$的切线. (5分)

(2)如图，连接DF. (6分)
$\because \angle ACB = 90°,CD\perp AB,\therefore \angle A + \angle B = \angle A + \angle ACD = 90°,\therefore \angle B = \angle ACD,\therefore \sin\angle ACD = \sin B = \frac{4}{5}$. (7分)
$\because CD$为$\odot O$的直径，$\therefore \angle CFD = 90°,\therefore \sin\angle ACD = \frac{DF}{CD}=\frac{4}{5}$.
设$DF = 4x,CD = 5x$，
在$Rt\triangle CDF$中，根据勾股定理得$CD² = DF² + CF²$，
$\therefore (5x)² = (4x)² + 3²,\therefore x = 1$(负值已舍去)，$\therefore CD = 5,\therefore OD = \frac{5}{2}$.
$\because \triangle ODM\cong\triangle OEM,\therefore \angle1 = \angle2$.
$\because \angle1 + \angle2 = \angle3 + \angle4,\angle3 = \angle4$，
$\therefore \angle1 = \angle3,\therefore OM// CB$，
$\therefore \sin\angle OMD = \sin B = \frac{4}{5}\therefore OM = \frac{OD}{\sin\angle OMD}=\frac{25}{8}$. (10分)

21. (本小题满分9分)
(2025山西)项目学习
项目背景:
如图(1),"源池泉涌"为我省某景区的一个景点,主体设计包括外栏墙与内栏墙,外栏墙高于内栏墙,两栏中间为步道,内栏墙内为泉池,池内泉水清澈见底.从正上方看,外栏墙呈正八边形,内栏墙呈圆形.综合实践小组的同学围绕"景物的测量与计算"开展项目学习活动,形成了如下活动报告.
请根据上述数据,计算内栏墙围成泉池的直径$BC$的长(结果精确到1米.参考数据:$\sin8.5^{\circ}\approx0.15$,$\cos8.5^{\circ}\approx0.99$,$\tan8.5^{\circ}\approx0.15$,$\sin37^{\circ}\approx0.60$,$\cos37^{\circ}\approx0.80$,$\tan37^{\circ}\approx0.75$).

答案： 21 由题意得，$\angle AEF = 90°$，四边形AEFD为矩形， (1分)
$\therefore EF = AD = 26$米，$AD// EF$，
$\therefore \angle ABE = \angle DAB = 37°,\angle ACE = \angle DAC = 8.5°$. (3分)
设$BE = CF = x$米，则$CE = (26 - x)$米，$BC = (26 - 2x)$米.在$Rt\triangle ABE$中，$\angle AEB = 90°,\tan\angle ABE = \frac{AE}{BE}$，
$\therefore AE = BE·\tan\angle ABE = x·\tan37°$. (5分)
在$Rt\triangle ACE$中，$\angle AEC = 90°,\tan\angle ACE = \frac{AE}{CE}$，
$\therefore AE = CE·\tan\angle ACE = (26 - x)·\tan8.5°$，
$\therefore x·\tan37° = (26 - x)·\tan8.5°$. (7分)
解得$x\approx\frac{13}{3}$，
$\therefore BC = 26 - 2×\frac{13}{3}\approx17$(米).
答：内栏墙围成泉池的直径BC的长约为17米. (9分)