《2025年实验班提优大考卷九年级数学上册苏科版》

15. 在$Rt\triangle ABC$中$,\angle C = 90^{\circ},\tan A = \frac{4}{3},AB = 10\ cm$,则$S_{\triangle ABC} =$24$ cm^2$.

答案： 15.24 [解析]在Rt△ABC中,∠C = 90°,$\tan A = \frac{BC}{AC} = \frac{4}{3}$,
∴设BC = 4x cm,则AC = 3x cm,
∴AB = $\sqrt{AC^2 + BC^2} = \sqrt{(3x)^2 + (4x)^2} = 5x$(cm).
∵AB = 10 cm,
∴5x = 10,解得x = 2,
∴BC = 8 cm,AC = 6 cm,
∴$S_{\triangle ABC} = \frac{1}{2} AC · BC = \frac{1}{2} × 6 × 8 = 24(cm^2)$.

16. (2023·常州中考)如图,在$Rt\triangle ABC$中$,\angle A = 90^{\circ}$,点$D$在边$AB$上,连接$CD$.若$BD = CD$,$\frac{AD}{BD} = \frac{1}{3}$,则$\tan B =$$\frac{\sqrt{2}}{2}$.

答案： 16.$\frac{\sqrt{2}}{2}$ [解析]设AD = t.
∵BD = CD,$\frac{AD}{BD} = \frac{1}{3}$,
∴BD = CD = 3t,
∴AC = $\sqrt{CD^2 - AD^2} = 2\sqrt{2}t$,AB = AD + BD = 4t,
∴$\tan B = \frac{AC}{AB} = \frac{2\sqrt{2}t}{4t} = \frac{\sqrt{2}}{2}$.

17. 如图是一张宽为$m$的矩形台球桌$ABCD$,一球从点$M$(点$M$在长边$CD$上)出发沿虚线$MN$射向边$BC$,然后反弹到边$AB$上的点$P$.如果$MC = n$,$\angle CMN = \alpha$,那么点$P$与点$B$的距离为$\frac{m - n · \tan \alpha}{\tan \alpha}$.

答案：
17.$\frac{m - n · \tan \alpha}{\tan \alpha}$ [解析]如图,由题意知,∠NPB =
∠α.在Rt△MNC中,MC = n,∠NMC = α,
∴NC = MC·$\tan \alpha$ = n·$\tan \alpha$,
∴BN = BC - NC = m - n·$\tan \alpha$.
在Rt△BPN中,∠BPN = α,$\tan \alpha = \frac{BN}{PB}$,
∴PB = $\frac{BN}{\tan \alpha} = \frac{m - n · \tan \alpha}{\tan \alpha}$.

18. (2025·苏州姑苏区立达中学期中)如图,在$Rt\triangle ABC$中$,\angle ACB = 90^{\circ},\cos A = \frac{3}{5}$,$D$是边$AB$的中点$,BE\perp CD$,垂足为$E$,$DE = \frac{7}{2}$,则$CD$的长为$\frac{25}{2}$.

答案：
18.$\frac{25}{2}$ [解析]过点C作CF⊥AB于点F,如图所示.

在Rt△ABC中,$\cos \angle BAC = \frac{AC}{AB} = \frac{3}{5}$,
∴设AC = 3k,AB = 5k,
由勾股定理,得BC = $\sqrt{AB^2 - AC^2} = 4k$.
由三角形的面积公式,得$S_{\triangle ABC} = \frac{1}{2} AB · CF = \frac{1}{2} AC ·BC$,
∴CF = $\frac{AC · BC}{AB} = \frac{3k × 4k}{5k} = \frac{12}{5}k$.
∵点D是△ABC斜边AB的中点,
∴CD = AD = BD = $\frac{1}{2} AB = \frac{5}{2}k$,$S_{\triangle ACD} = S_{\triangle BCD}$,
∴$\frac{1}{2} AD · CF = \frac{1}{2} CD · BE$,
∴CF = BE = $\frac{12}{5}k$.
在Rt△BDE中,BD = $\frac{5}{2}k$,BE = $\frac{12}{5}k$,DE = $\frac{7}{2}$,
由勾股定理,得$BD^2 - BE^2 = DE^2$,
∴$(\frac{5}{2}k)^2 - (\frac{12}{5}k)^2 = (\frac{7}{2})^2$,
解得k = 5或k = -5(不合题意,舍去),
∴CD = $\frac{5}{2}k = \frac{25}{2}$.

19. (8分)(2024·黑龙江中考)先化简,再求值:$\frac{m^2 - 2m + 1}{m^2 - 1} ÷ (\frac{m^2}{m^2 + m} - 1)$,其中$m = \cos 60^{\circ}$.

答案： 19.原式 = $\frac{(m - 1)^2}{(m + 1)(m - 1)} ÷ \frac{-m}{m^2 + m} = \frac{m - 1}{m + 1} · \frac{m(m + 1)}{-m} = 1 - m$,
当m = $\cos 60° = \frac{1}{2}$时,原式 = 1 - $\frac{1}{2} = \frac{1}{2}$.

20. (8分)在一次数学活动课上,数学老师在同一平面内将一副直角三角板按如图所示位置摆放,点$C$在$FD$的延长线上$,AB// CF$,$\angle F = \angle ACB = 90^{\circ}$,$\angle E = 45^{\circ}$,$\angle A = 60^{\circ},AC = 10$,求$CD$的长.

答案：
20.如图,过点B作BM⊥FD于点M.

在△ACB中,∠ACB = 90°,∠A = 60°,AC = 10,
∴∠ABC = 30°,BC = AC·$\tan 60° = 10\sqrt{3}$.
∵AB//CF,
∴∠BCM = ∠ABC = 30°,
∴BM = BC·$\sin 30° = 10\sqrt{3} × \frac{1}{2} = 5\sqrt{3}$,
CM = BC·$\cos 30° = 10\sqrt{3} × \frac{\sqrt{3}}{2} = 15$.
在△EFD中,∠F = 90°,∠E = 45°,
∴∠EDF = 45°,
∴MD = BM = 5$\sqrt{3}$,
∴CD = CM - MD = 15 - 5$\sqrt{3}$.

21. (8分)(2025·南京秦淮区钟英中学模拟)如图,平地上有一竖直旗杆$AB$,从地面上的点$C$、$D$、$E$处测得旗杆顶端$A$的仰角分别是$56^{\circ}19'$、$26^{\circ}34'$、$45^{\circ}$.若点$C$、$D$之间的距离为$m$,用含$m$的代数式表示旗杆$AB$的高和点$C$、$E$之间的距离.
(图中所有点均在同一平面内,参考数据:$\tan 56^{\circ}19'\approx1.5,\tan 26^{\circ}34'\approx0.5$)

答案： 21.
∵AB⊥DE,
∴∠ABD = ∠ABE = 90°.
∵∠ADB = 26°34′,∠ACB = 56°19′,
∴$\tan \angle ADB = \frac{AB}{BD} \approx 0.5$,$\tan \angle ACB = \frac{AB}{BC} \approx 1.5$,
∴BD = 2AB,BC = $\frac{2}{3}$AB.
∵CD = m,
∴BD - BC = 2AB - $\frac{2}{3} AB$ = CD = m,
∴AB = $\frac{3}{4}m$.
∵∠AEB = 45°,
∴BE = AB = $\frac{3}{4}m$,
∴CE = BC + BE = $\frac{2}{3} × \frac{3}{4}m + \frac{3}{4}m = \frac{5}{4}m$.
故旗杆AB的高为$\frac{3}{4}m$,点C、E之间的距离为$\frac{5}{4}m$.