《2026年金考卷中考试题汇编45套数学山东专版》

17. (本小题满分8分)
数学综合实践小组进行了项目式学习的实践探究,请根据表格内容完成任务.

(结果精确到0.01m.参考数据:$\sin11^{\circ}≈0.191$,$\cos11^{\circ}≈0.982$,$\tan11^{\circ}≈0.194$,$\sin61^{\circ}≈0.875$,$\cos61^{\circ}≈0.485$,$\tan61^{\circ}≈1.804$)

答案：
17 任务1 如图,过点$B$分别作$BG \perp DF$于点$G$,$BH \perp MN$于点$H$.
$\because AB = CB = 4m$,$\angle ABC = 22^{\circ}$,$\therefore \angle GBC = 11^{\circ}$(依据:等腰三角形“三线合一”).在$Rt \triangle BGC$中,$\angle BGC = 90^{\circ}$,$\therefore GC = BC · \sin \angle GBC = 4 × \sin 11^{\circ} \approx 4 × 0.191 = 0.764(m)$,(2分)$\therefore GD = GC + CD = 0.764 + 2 = 2.764(m)$.$\because \angle BGD = \angle GDH = \angle BHD = 90^{\circ}$,$\therefore$四边形$BGDH$为矩形,$\therefore BH = GD = 2.764m \approx 2.76m$.
答:遮阳棚前端$D$到地面$MN$的距离约为$2.76m$.(4分)

任务2 如图,在$Rt \triangle BGC$中,$\angle BGC = 90^{\circ}$,$\angle GBC = 11^{\circ}$,$\therefore GB = BC · \cos \angle GBC = 4 × \cos 11^{\circ} \approx 4 × 0.982 = 3.928(m)$.(6分)$\because$四边形$BGDH$为矩形,$\therefore DH = GB = 3.928m$.在$Rt \triangle BHE$中,$\angle BHE = 90^{\circ}$,$\angle BEH = 61^{\circ}$,$\therefore EH = \frac{BH}{\tan \angle BEH} = \frac{2.76}{\tan 61^{\circ}} \approx \frac{2.76}{1.804} \approx 1.530(m)$,$\therefore DE = DH - EH = 3.928 - 1.530 = 2.398 \approx 2.40(m)$.
答:非机动车道有效遮阳宽度$DE$的长约为$2.40m$.(8分)

18. (本小题满分8分)
如图,$AB$为$\odot O$的直径,$AC$为$\odot O$的切线,连接$BC$交$\odot O$于点$D$,$D$为$\overset{\frown}{AE}$的中点,连接$AE$交$BD$于点$F$.
(1)求证:$AC = AF$;
(2)若$AC = \sqrt{5}$,$CF = 2$,求$BF$的长.

答案：
18
(1)证明:如图,连接$BE$.
$\because AB$为$\odot O$的直径,$\therefore \angle AEB = 90^{\circ}$(依据:直径所对的圆周角为$90^{\circ}$),$\therefore \angle EBF + \angle EFB = 90^{\circ}$.$\because AC$为$\odot O$的切线,$\therefore \angle BAC = 90^{\circ}$,$\therefore \angle C + \angle CBA = 90^{\circ}$.$\because D$为$\overset{\frown}{AE}$的中点,$\therefore \overset{\frown}{ED} = \overset{\frown}{AD}$,$\therefore \angle EBF = \angle CBA$(依据:等弧所对的圆周角相等),$\therefore \angle EFB = \angle C$.又$\because \angle AFC = \angle EFB$,$\therefore \angle C = \angle AFC$,$\therefore AC = AF$.(4分)

(2)如图,连接$AD$.
$\because AB$为$\odot O$的直径,$\therefore \angle ADB = 90^{\circ}$,即$AD \perp CF$.$\because AC = AF$,$DF = \frac{1}{2}CF = 1$(依据:等腰三角形“三线合一”),$\therefore \angle ADC = \angle BAC = 90^{\circ}$,$\angle C = \angle C$,$\therefore \triangle DAC \sim \triangle ABC$,$\therefore \frac{AC}{BC} = \frac{DC}{AC}$,即$\frac{\sqrt{5}}{BC} = \frac{1}{\sqrt{5}}$,$\therefore BC = 5$,$\therefore BF = BC - CF = 5 - 2 = 3$.(8分)