《2025年启东中学作业本九年级数学上册苏科版宿迁专版》

7.(2023·杭州)如图,在$\odot O$中,半径OA,OB互相垂直,点C在劣弧AB上.若$∠ABC= 19^{\circ }$,则$∠BAC=$(D)

A.$23^{\circ }$
B.$24^{\circ }$
C.$25^{\circ }$
D.$26^{\circ }$

答案： D

8.如图,正方形ABCD内接于$\odot O$,点P在$\widehat {AB}$上,则$∠BPC$的度数为(B)
A.$30^{\circ }$
B.$45^{\circ }$
C.$60^{\circ }$
D.$90^{\circ }$

答案： B

9.如图,A,B,C是半径为1的$\odot O$上的三个点,若$AB= \sqrt {2},∠CAB= 30^{\circ }$,则$∠ABC$的度数为(C)

A.$95^{\circ }$
B.$100^{\circ }$
C.$105^{\circ }$
D.$110^{\circ }$

答案： C

10.在半径等于4 cm的圆内有长为$4\sqrt {3}cm$的弦,则此弦所对的圆周角的度数为$60^{\circ }$或$120^{\circ }$.

答案： $60^{\circ }$或$120^{\circ }$

11.(2023·武汉)如图,OA,OB,OC都是$\odot O$的半径,$∠ACB= 2∠BAC.$
(1)求证:$∠AOB= 2∠BOC;$
(2)若$AB= 4,BC= \sqrt {5}$,求$\odot O$的半径.

答案：

(1)证明:$\because ∠ACB=\frac {1}{2}∠AOB,∠BAC=\frac {1}{2}∠BOC,∠ACB=2∠BAC,\therefore ∠AOB=2∠BOC;$
(2)解:如答图，过点O作半径$OD⊥AB$于点E，连接BD,
　　$\therefore AE=BE,∠DOB=\frac {1}{2}∠AOB,$
　又$\because ∠AOB=2∠BOC,$
　　$\therefore ∠DOB=∠BOC,\therefore BD=BC;$
　　$\because AB=4,BC=\sqrt {5},\therefore BE=2,DB=\sqrt {5}.$
　在$Rt△BDE$中,$∠DEB=90^{\circ },$
　　$\therefore DE=\sqrt {BD^{2}-BE^{2}}=1,$
　在$Rt△BOE$中,$∠OEB=90^{\circ },OB^{2}=(OB - 1)^{2}+2^{2}$,解得$OB=\frac {5}{2}$,即$\odot O$的半径是$\frac {5}{2}$.
　　　　　　　　

12.如图,已知点A,B,C为$\odot O$上的三个点,且$\triangle ABC$为等边三角形,P为$\widehat {BC}$上一点.求证:$PA= PB+PC.$

答案：
证明:如答图，在PA上截取$PD = PB$，连接BD.
　　　　　　　　　
　　$\because △ABC$是等边三角形，
$\therefore ∠ABC = ∠ACB = 60^{\circ },AB = BC,$
　$\therefore ∠BPA = ∠ACB = 60^{\circ },∠APC = ∠ABC = 60^{\circ },$
　$\therefore ∠BPC = 120^{\circ }.$
　又$\because PD = PB,\therefore △PBD$是等边三角形，
　$\therefore ∠BDP = 60^{\circ },BD = PB,\therefore ∠BDA = 120^{\circ }.$
　又$\because ∠BAP = ∠BCP,$
　$\therefore △ABD\cong △CBP(AAS),$
　$\therefore AD = PC,\therefore PA = PD + AD = PB + PC.$

13.如图,$\odot O$中两条互相垂直的弦AB,CD交于点P,AB经过点O,E是AC的中点,连接OE,EP,延长EP交BD于点F.
(1)若$AB= 10,OE= \sqrt {10}$,求AC的长;
(2)求证:$EF⊥BD.$

答案：
(1)解:$\because E$是$AC$的中点，
　$\therefore OE⊥AC,AC = 2AE,$
　$\because AB = 10,\therefore OA=\frac {1}{2}AB = 5,$
　在$Rt△AOE$中,$OE = \sqrt{10},$
　$\therefore AE=\sqrt{OA^{2}-OE^{2}}=\sqrt{5^{2}-(\sqrt{10})^{2}}=\sqrt{15}$
　$\therefore AC = 2AE = 2\sqrt{15},\therefore AC$的长为$2\sqrt{15}$
(2)证明:$\because AB⊥CD,\therefore ∠APC = ∠BPD = 90^{\circ },$
　$\therefore ∠DPF + ∠BPF = 90^{\circ },$
　$\because E$是$AC$的中点，
　$\therefore EP = EC=\frac {1}{2}AC,\therefore ∠EPC = ∠C;$
　$\because ∠EPC = ∠DPF,∠B = ∠C,\therefore ∠DPF = ∠B,$
　$\therefore ∠B + ∠BPF = 90^{\circ },$
　$\therefore ∠BFP = 180^{\circ }-(∠B + ∠BPF)= 90^{\circ },\therefore EF⊥BD.$