《2025年新课程学习与测评同步学习九年级数学全一册湘教版》

5. 如图14,PA,PB是⊙O的切线,A,B为切点,AC是⊙O的直径,AC,PB的延长线相交于点D,∠1= 20°.求∠APB的度数.

答案： 解:由PA,PB是$\odot O$的切线,得$∠PAO = ∠PBO = 90^{\circ},PA = PB$.又$∠1 = 20^{\circ}$,故$∠BAP = 90^{\circ} - ∠1 = 70^{\circ},∠ABP = ∠BAP = 70^{\circ}$.因此$∠APB = 180^{\circ} - 70^{\circ}×2 = 40^{\circ}$.

6. 将一把直尺,一副含60°角的直角三角尺和一个光盘按图15的方式摆放,直角三角尺的直角边AD与直尺的一边重合,光盘与直角三角尺的斜边AC相切,与直尺的一边相切于点B,且AB= 3 cm,则光盘的直径是( ).

A.3 cm
B.3√3 cm
C.6 cm
D.6√3 cm

答案：
D 提示:如图82,设光盘的圆心为O,连接OA,OB.由$∠CAD = 60^{\circ}$,得$∠CAB = 120^{\circ}$.由AB,AC与$\odot O$相切,得$∠OAB = ∠OAC = 60^{\circ}$.又$AB = 3cm$,则$OB = AB\cdot tan60^{\circ} = 3\sqrt{3}cm$.故光盘的直径为$6\sqrt{3}cm$.
　　　　

7. 如图16,已知PA,PB是⊙O的切线,切点分别是A,B,PA= 12 cm,∠P= 56°,Q为⌢AB上一点,过点Q作⊙O的切线,分别交PA,PB于点E,F.
(1)求△PEF的周长.
(2)求∠EOF的度数.

答案：
(1)
∵PA,PB,EF是$\odot O$的切线,
∴$PB = PA = 12cm,EA = EQ,FQ = FB$.
∴$△PEF$的周长$ = PE + EQ + FQ + PF = PE + EA + FB + PF = PA + PB = 24cm$.
(2)
∵PA,PB,EF是$\odot O$的切线,
∴$PA⊥OA,PB⊥OB,EF⊥OQ,∠AEO = ∠QEO,∠QFO = ∠BFO$.
∴$∠AOE = ∠EOQ,∠QOF = ∠BOF$.
∴$∠EOF = ∠EOQ + ∠QOF = \frac{1}{2}(∠AOQ + ∠BOQ) = \frac{1}{2}∠AOB$.又$∠P = 56^{\circ}$,
∴$∠AOB = 360^{\circ} - 90^{\circ} - 90^{\circ} - 56^{\circ} = 124^{\circ}$.
∴$∠EOF = \frac{1}{2}×124^{\circ} = 62^{\circ}$.

8. 如图17,AD//BC,AB⊥BC,以AB为直径的⊙O与DC相切于点E.已知AB= 8,BC - AD= 6.
(1)求AD,BC的长.
(2)在直径AB上是否存在一动点P,使以点A,D,P为顶点的三角形与△BCP相似？若存在,则求出AP的长；若不存在,则说明理由.

答案：

(1)如图83,过点D作$DF⊥BC$于点F.
∵$AD//BC,AB⊥BC$,
∴四边形ABFD为矩形,$AB⊥AD$.
∴$DF = AB = 8,AD = BF$.在$Rt△DFC$中,$DF = 8,FC = BC - BF = BC - AD = 6$.
∴$DC = \sqrt{DF^{2} + FC^{2}} = 10$.设$AD = x$.
∵$AB⊥AD,AB⊥BC$,AB为$\odot O$的直径,
∴$\odot O$与AD,BC相切.又$\odot O$与DC相切于点E,
∴$DE = AD = x,EC = BC = x + 6$.
∴$DC = x + (x + 6) = 10$.解得$x = 2$.
∴$AD = 2,BC = 2 + 6 = 8$.
(2)存在符合条件的点P.设$AP = y$,则$BP = 8 - y$.以A,D,P为顶点的三角形与$△BCP$相似,有两种情况:①$△ADP∽△BCP$,此时有$\frac{AD}{BC} = \frac{AP}{BP}$,即$\frac{2}{8} = \frac{y}{8 - y}$.解得$y = \frac{8}{5}$.②$△ADP∽△BPC$,此时有$\frac{AD}{BP} = \frac{AP}{BC}$,即$\frac{2}{8 - y} = \frac{y}{8}$.解得$y = 4$.故存在符合条件的点P,此时AP的长为$\frac{8}{5}$或4.