第三章圆(B卷·培优卷·单元重点综合测试)(解析版)2024-2025学年九年级数学下册单元速记.巧练(北师大版，贵州专用)_初中全科知识点总结和试卷_赠送习题试卷_初中3年学习辅助资料_4-2、数学

      
      
第三章 圆（B 卷·培优卷）
班级___________ 姓名___________ 学号____________ 分数____________
考试范围：全章的内容；  考试时间：120 分钟；  总分：150 分
一、选择题（本大题共12 小题，每小题3 分，共36 分）
1⊙O 	
 3OP
2 P 
A
B
C
D
 P 	 ! 2"#	
 3
$ P 
%&C
2'(AB )⊙O 	*AD = CD = BC+BOC 	,-
A45°
B50°
C55°
D60°
AB )/O 	*
$+AOB
+AOD++COD++BOC
180°
AD = CD = BC
$+AOD
+COD
+BOC
$+BOC
60°
%&D
3012345	)
A67	*8*#79:67;<	=>?
B@A5BCD
CEF	G;<	?EF
D@GH	@GH	@DI !EF
AJ67K)*	*8*#79:67;<	=>?KLMNOP
BJKQC	@A5BCDKLMNOP
      
      
CJQFREF	G;<	?EFKLMNOP
DJ@GH	@GH	@DI !EFLMNOP
%&D
4'(AB )
 O 	* D )? AC 	R+BAC
44°+DAC F#
A22°
B44°
C23°
D46°
AB )
 O 	*
$+ACB
90°
+BAC
44°
$+B
90.S+BAC
46°
TUH ABCD )
/O 	VTUH
$+D
180.S+B
134°
 D )? AC 	R
$AD = DC
$AD
DC
$+DAC
+DCA =
1
2180.S+D
23°
%&C
5'(AB )⊙O 	*7 CDWAB # E AB
10AE
17 CD 	X)
A 7
B2 7
C6
D8
YV OC
AB )/O 	*7 CDWAB
$CD
2CE+OEC
90°
AB
10AE
1
      
      
$OC
5OE
5S1
4
 RtZCOE RCE =
OC2−OE2 = 3
$CD
2CE
6
%&C
6'(⊙O ZABC 	[+C
90°AO 	\X]^ BC # D AC
6DC
2⊙O 	

F#
A
3
2
B
4
3
C
5
4
D
5
3
_ O ` AC 	[ M	
 r
'(YV OM
+C
90°
$CM
r
+OAM
+OAM+C
+OMA
90°
$ZAOMaZADC
$OMCD
AMAC
b r2
6Sr6
cr =
3
2
      
      
%&A
7'(* AB
12 	
d B efghijkl 30°mg A n Ao	pq(Rrst
6	u)
A2π
B3π
C4π
D12π
* AB
12 	
d B fghkl 30°
$S rs
SA'B *	
+S vH ABA'SSAB *	

wAB
A'B
$SA'B *	

SAB *	

$S rs
S vH ABA'
AB
12+ABA’
30°
$S rs
S vH ABA’ = nπr2
360 = 30π×122
360
= 12y
%zD
8{6|	}~€)‚ƒ„…†‡ˆ‰Š‹	ŒŽ	K‘’“”{6|R	 6 D•–|`
6 D—–|˜QC:;™••š	šX›EF'( 1œnCDžŸ	 UH4 U
H( 2 )¡¢O(+1 	,-
A130°
B120°
C110°
D60°
'( 2 UH ABCDEF )4 UH
$AB
AF
EF+BAF =
(6−2)× 180°
6
= 120°
$+ABF
+AFB =
180°−120°
2
= 30°
      
      
Q£+EAF
30°
$+1
180.S30.S30.
120°
%&B
9¤¥¦§£¨©gª«¬­H®,>	H¯°¢¨©ž	±6²0)QCD¨©®³KQ
´µg®,>	¢O(
¶¨©ž	±6² x g]µ MN 	X,· dx01¸¹RCB45	)
A¶ x1ºx2 gdx1ºdx2
B¶ dx1ºdx2gx1ºx2
C¶ x1
2x2 gdx1
2dx2
D¶ x1+x2
1 gdx1
dx2
AJ¶ x1ºx2 gdx1` dx2»AEF»AKF¼&½KLMNO
BJ¶ dx1ºdx2gx1ºx2  x1¾x2¼&½KLMNO
AJ¶ x1
2x2 gdx1¾2 dx2¼&½KLMNO
DJ¶ x1+x2
1 gdx1
dx2¼&½LMNO
%&D
10'( E )ZABC 	AE 	\X]¿ZABC 	VE^# D` BC E^# G0
1À①+BAD
+CADP② G  BC 	R+BGD
90°P③YV BECE+BAC

40°+BEC
140°P④BD
DE¡RCB45	D-)
A4
B3
C2
D1
 E )ZABC 	
$AD 6+BAC
$+BAD
+CAD%Á45P
      
      
_ZABC V OYV OD OD 8*6 BC
 G  BC 	R
$ G  OD ` BC 	^b+BGD
90°%Â45P
+BAC
40°
$+ABC++ACB
180.S+BAC
140°
 E )ZABC 	
$∠EBC = ∠ABE =
1
2∠ABC∠ECB = ∠ACE =
1
2∠ACB
$+BEC
180.S+EBC++ECB= 180°−
1
2(∠ABC + ∠ACB) = 180°−
1
2 × 140° = 110°%ÃÄÅP
+BAD
+CAD
$BD = CD
$+DBC
+BAD
+DBE
+DBC++EBC+DEB
+BAD++ABE
$+DBE
+DEB
$BD
DE%Æ45
Ç45	™ 3 D
%&B
11'(AB )⊙O 	*7 CDWABDEÈAC AB
26AC
24 F )7 DE 	R OF
	É
A595 5
13
B595 10
169
C
2035
169
D
2045
169
      
      
'(· ABJCD 	^ MYV BC\X ED ^ AB 	\X]# N
AB )/O 	*
$+ACB
90°
ÊËB£cBC =
AB2−AC2 = 10
7 CDWAB
$CM
DM
S△ABC =
1
2AB × CM =
1
2AC × BC
$
1
2 × 26 × CM =
1
2 × 24 × 10
cCM =
120
13 
ÊËB£cBM =
BC2−CM2 =
50
13
$OM = OB−BM =
119
13 AM = OA + OM =
288
13 
DEÈAC
$+N
+A+NDM
+ACM
$ZNDMÌZACMAAS
$MN = AM =
288
13 ，DN = AC = 24
$ON = OM + MN =
407
13 
 F )7 DE 	R
$OFWDEb+OFN
90.
+DMN
w+ONF
+DNM
$ZONFaZDNM
$
OF
DM =
ON
DNb
OF
120
13 =
407
13
24
cOF =
2035
169 
      
      
%&C
12'(ÍH ABCD R E ) AB 	R F ) BC 	RYV EFG ) EF 	RYV
DGZBEF RBE
2+BFE
30°ÎZBEF d B Ïghklkl	ÐÑR]µ
DG X	Ò"É)
A 67
B2 17
C10
D12
YV BG
ZBEF RBE
2+BFE
30°
$EF
2BE
4BF =
3BE
2 3
G ) EF 	R
$BG =
1
2EF
2
$G /B :
 2
$¶ G  DB 	\X]gDG Ò"
BE
2BF
2 3 E ) AB 	R F ) BC 	R
$AB
4BC
4 3
$BD =
42 + (4 3)2 = 8
$DG 	Ò"É 8+2
10
%&C
二、填空题（本大题共4 小题，每小题4 分，共16 分）
13CDvH	G 120°u 3πmvH	?X 2π
_mvH	
 RmvH	?X l
S = 120°⋅π×R2
360°
= 3π
      
      
$πR2
3 = 3y
$R
3
$l =
120°⋅πR
180° =
120°⋅π×3
180°
= 2π
$mvH	?X 2y
%z2y
14'(AB )⊙O 	7: AB
6 C )? AB R D )Ó? AB 	C+ADC
30°
 O 7 AB 	 !F#  3
'(
YV OAJOCOC ^ AB # E
 C )? AB RAB
6
$OCWAB: AE
BE
3
+ADC
30°
$+AOC
2+ADC
60°
$+OAE
30°
$OE
AE•tan30.
3 ×
3
3 =
3
% O 7 AB 	 ! 3
%z 3
15ÕÖ AB )⊙O 	* P ) AB \X]	CD×Ð P Ø⊙O 	[][ C+APC 	
6]^ AC # D+CDP F# 45°
      
      
'(YV OC
OC
OAPD 6+APC
$+CPD
+DPA+A
+ACO
PC /O 	[]
$OCWPC
+CPO++COP
90°
$+CPD++DPA++A++ACO
90°
$+DPA++A
45°
b+CDP
45°
%z45°
16'(AB )⊙O 	*AB
8 M ⊙O +MAB
20°N )? MB 	RP )* AB
	C× MN
1ZPMN ÙX	ÒÚÉ 5
Ø N Û# AB 	<Ü NoYV OMJONJONoJMNo
 MNo` AB 	^b PM+PN 	ÒÚg	PM+PN 	ÒÚÉ
MNo
+MAB
20°
$+MOB
2+MAB
2×20.
40°
N )? MB 	R
      
      
$+BON =
1
2+MOB =
1
2 × 40.
20°
<ÜÞ+NoOB
+BON
20°
$+MONo
+MOB++NoOB
40°+20.
60°
$ZMONo)FU@GH
$MNo
OM
OB =
1
2AB =
1
2 × 8
4
$ZPMN ÙX	ÒÚÉ
1+4
5
%z5
@JN¼"Nß 9 ÚNß 98 6 
17'(AB )⊙O 	* C ⊙O D )BC	R+BAC
70°à+C 	,-
D )BC	R
$CD = BD
$∠CAD = ∠BAD =
1
2∠BAC =
1
2 × 70° = 35°
AB )/O 	*
$+ADB
90°
$+B
90.S35.
55°
AJBJCJD TD›/O 
$+B++C
180°
$+C
125°
18Cáâãääå)?Hâ,AB 16 æåçCN 4 æ"èÐéä0ãê,
DE 12 æàëìçníîæï
      
      
YV OD
NOcOCWAB
$AN
NB =
1
2AB
8æ
Q£»cDM
ME =
1
2DE
6æ
_?H;	
 R æ ON
RS4æ
 RtZAON ROA2
AN2+ON2b R2
82+RS42
cR
10
$OM =
OD2−DM2 =
102−62 = 8æ
 MN
OMSON
2æ
ëìçn 2 æ
19'(ZABC V#⊙OAC ⊙O 	*BD 6+ABC^⊙O # DYV ADCDÕÖ
AD = 2 2AB = 2 3à BC 	X
AC /O 	*
$+ADC
+ABC
90°
BD 6+ABC
$+CBD
+ABD
      
      
$CD = AD = 2 2
$AC2
AD2+CD2
16
$BC =
AC2−AB2 =
16−(2 3)2 = 2
20'(;¢	åäðABñ¢äå
1AB;	 OEF )7 CD 	8*6]òóôðõöØ(÷‹ OKøØ3
ùúûüØ(
ýþ
2åä	â,7 AB 	X 16måçAB	R7 AB 	 ! 4màåä	
 R
1Ø7 AB 	8*6]^# G^ AB # H^ CD 	8*6] EF # O
 O b;àØ	'( 12 6
2YV OA'( 2
1R	Ø(»ÖZAOH *G@GHH ) AB 	RGH
4
$AH =
1
2AB
83 6
GH
4
$OH
RS4
 RtZAOH RÊËB£cOA2
AH2+OH2
$R2
82+RS424 6
cR
105 6
$åä	
 R  10m
      
      
21'(AD )⊙O 	*7 BCWAD # PY ABBDOB
1+A
25°à+CBD 	,-
2 BC
16PD
4à⊙O 	

1AD )/O 	*7 BCWAD
$BD = CD
+A
25°
$+CBD
+A
25°
2_/O 	
 r OB
OD
r
AD )/O 	*7 BCWADBC
16
$BP =
1
2BC =
1
2 × 16
8
PD
4
$OP
ODSPD
rS4
 RtZBPO RôðÊËB£c BP2+OP2
OB2
$82+rS42
r2
$r
10
$/O 	
) 10
22TUH ABCD V#⊙OAC *E  DA 	\X]: BE `⊙O E[AB 6+EAC
1ÿ BO ` CD 	pqÛĀ92ā£P
      
      
2 BE
4AD
3AEà⊙O 	

1BOWCD£'0
'(YV ODBD
TUH ABCD V#/O
$+BAD++BCD
180°
+BAE++BAD
180°
$+BAE
+BCD
AB 6+EAC
$+BAE
+BAC
$+BCD
+BAC
+BAC
+BDC
$+BDC
+BCD
$BD
BC
$ B ]µ CD 	8*6]
OD
OC
$ O ]µ CD 	8*6]
$BO 8*6]µ CD
$BOWCDP
2'(YV ODBD
      
      
BE `/O E[
$+OBE
90°
b+ABE++ABO
90°
AC /O 	*
$+ABC
90°
$+BAC++ACB
90°
OB
OA
$+ABO
+BAC
$+ABO++ACB
90°
$+ABE
+ACB
AB = AB
$+ACB
+BDE
$+BDE
+ABE
+E
+E
$ZEBAaZEDB
$
EB
ED =
AE
BE+DBE
+BAE
+BAC
BE
4AD
3AE+BAC++ACB
90°
$
4
4AE =
AE
4 +DBE++BDE
+BAC++ACB
90°
$AE
2AD
6+BED
90°
$AB =
42 + 22 = 2 5BC = BD =
EB2 + ED2 =
42 + 82 = 4 5
$AC =
AB2 + BC2 =
(2 5)2 + (4 5)2 = 10
$/O 	

1
2AC = 5
23'(ÕÖ AB )
	* C 
CEWAB8Ă E D )BC	RAD ^ CE
# F^ BC # G
      
      
1ÿZFGC 	H~92ā£P
2CD 60°AB
16àrst6	u
1ZCFG )Fă@GH£'0
AB )/O 	*
$+ACB
90°*;<	ÙG)*G
$+CAD++AGC
90°
CEWAB
+AFE++BAD
90°
D BC	R
$CD = BD
$+CAD
+BADQFRQ?F?;<	ÙGEF
$+AGC
+AFE
+AFE
+CFG
$+CGF
+CFG
$CF
CG
$ZCFG )Fă@GHP
2YV OCJOD
CD 60°
$+COD
60°
D BC	R
$+BOD
+COD
60°QFRQ?F?;<	ÙGEF
$+AOC
180.S60.S60.
60°
OA
OC
$ZAOC FU@GH™CDGF# 60.	Fă@GH)FU@GH
$OA = AC = OC =
1
2AB = 8
CEWAB
      
      
$AE =
1
2OA = 4
$CE =
AC2−AE2 =
82−42 = 4 3
$Srs = S扇形AOC−S△AOC = 60×π×82
360
−1
2 × 8 × 4 3 =
32
3 π−16 3
24'( 1Ą™ë	ëąĆqëć¡Ĉĉ)­ AB *	
 OAB
52cmMN ë
ĉ]MN
48cmGH ćĉ]MNÈGH
1Ø OCWMN # Cà OC 	XP
2Î(R	ëĊ‹Ct6c( 2“”ëç,0ċn 14cmàmgëĉ]Čîníî
1YV OM
O OCWMNMN
48cm
$MC =
1
2MN = 24cm
AB
52cm
$OM =
1
2AB = 26cm
 RtZOMC ROC =
OM2−MC2 =
262−242 = 10cm
$OC 	X 10cmP
2Ð O Ø ODWEFYV OE
NcOD
10+14
24cm
 RtZOED RED =
OE2−OD2 =
262−242 = 10cm
$EF
2ED
20cm
$48S20
28cm
$ëĉ]Čîn 28cm
      
      
25čĎď( 1ĐÜ“Ēēď”wĕĖJėďFĘęgÕCĚěĜĝĞ‹”ğĠg4¯‹”
čĎďĕÜğiġĢ`ģĤĥQĦ	ħĨØð'( 2 ;¢ĩčĎďRĪī‹Ĭ	ġĭĮį
ZABC R­ZABC 	U AB *Ø⊙O^ AC # PPD )⊙O 	[]: PDWBC8Ă
D
1àİ+A
+CP
2 PD
2BD
4à⊙O 	

1İāYV OP'( 2
PD )⊙O 	[]
$OPWPD
PDWBC
$OPÈBC
$+OPA
+C
OA
OP
$+OPA
+A
$+A
+CP
2YV PB'( 2
 RtZPBD RPD
2BD
4
$PB =
22 + 42 = 2 5
AB *
$+APB
90°
      
      
+BDP
+BPC+DBP
+PBC
$ZBDPaZBPC
$BPBC
BDBPb 2 5BC
22 5
c BC
10
+A
+C
$BA
BC
10
$⊙O 	
 5