삼각비의 활용

본 교 재 20

OB”=OD”-BD”=1-0.3982=0.6018 yy

cos x= = =O’B”=0.6018=cos 53˘

x=53˘ yy

tan 53˘= = =CD”

CD”=1.3270 ^yy

1.3270

21

ABH cos B= = BH”=3'3

AH”=øπ6¤ -(3'3)¤ ='9=3 AHC

sin C= =;5#; ;5#;

22 CDB

sin 30˘= =;2!; CD”=4

tan 30˘= = DB”=2'3

DCA= CAD=15˘ AD”=CD”=4 AB”=4+2'3

ABC

tan 15˘= = =2-'3 2-'3

23

A=180˘_ =30˘

=

={ _ +;2!;}÷{;2!;+ }

=1÷ =

='3-1

24

0˘<x<90˘ 0<sin x<1 sin x+1>0 sin x-1<0

øπ(sin x+1)¤ -øπ(sin x-1)¤ =sin x+1-{-(sin x-1)}

=sin x+1+sin x-1

=2 sin x 2 sin x 2

1+'3 1+'3

2

'3 2 '3

2 '3

3

tan 30˘_cos 30˘+sin 30˘

sin 30˘+cos 30˘

tan A_cos A+sin A sin A+cos A

1 1+2+3

2 4+2'3 CB”

AB”

'3 3 2 DB”

2 CD”

AH”

AC”

'3 2 BH”

6 CD”

1 CD”

OD”

OB”

1 OB”

OA” 개념

26

직각삼각형의 변의 길이

일반 삼각형의 변의 길이

개념

27

1 3'3 3 6 3'7 2 ^60˘ 4'2 8'6

3

개념콕콕 ⁹⁰

1

ABH A’H”=6 sin 60˘=6_ =3'3 ABH B’H”=6 cos 60˘=6_;2!;=3 CH”=BC”-BH”=9-3=6

AHC AC”=ø∑(3'3)¤ +∑6¤ ='∂63=3'7

2

A=180˘-(45˘+75˘)=60˘

BCH CH”=8 sin 45˘=8_ =4'2

AHC AC”= = =4'2_ =8'6

3 2 '3 4'2

sin 60˘

CH”

sin A

'2 2 '3

2

A BC”

H ABH

A’H”=4 sin 30˘=4_;2!;=2 B’H”=4 cos 30˘=4_ =2'3

CH”=BC”-BH”=3'3-2'3='3 AHC AC”=øπ2¤ +('3)¤ ='7

-1

A BC”

H AHC

A’H”=10 sin 60˘=10_

=5'3(km)

CH”=10 cos 60˘=10_;2!;=5(km) '3

2

60˘

H A

B C

16 km 10 km

'3 2

30˘

A

H C

B

3'3 4

BH”=BC”-CH”=16-5=11(km) ABH AB”=øπ11¤ +(5'3)¤ ='∂196=14(km)

A B 14 km 14 km

-2

A BC”

H ACH=180˘-120˘=60˘

ACH

A’H”=4 sin 60˘=4_ =2'3 CH”=4 cos 60˘=4_;2!;=2

BH”=BC”+CH”=3+2=5 ABH AB”=ø∑5¤ +∑(2'3)¤ ='∂37

B=180˘-(75˘+45˘)=60˘

A BC”

H AHC

A’H”=6 sin 45˘=6_ =3'2 ABH

AB”= = =3'2_ =2'6 2'6

-1

B AC”

H BHC

B’H”=12 sin 60˘=12_ =6'3(m) AHB

AB”= = =6'3_ =6'6(m)

A B 6'6 m

-2

A=180˘-(30˘+105˘)=45˘

C AB”

H BCH

C’H”=20 sin 30˘=20_;2!;=10

30˘ 105˘

45˘

A

C H

B 20

2 '2 6'3

sin 45˘

B’H”

sin A

'3 2

2 '3 3'2

sin 60˘

A’H”

sin B

'2 2

H 75˘

60˘ 45˘

6 A

B C

'3 2

H 120˘

A

B 3 C

4

A

C H 45˘ B

60˘

12 m 대표 유형

-114 km -2

2'6 -1 -2 10'2

http://zuaki.tistory.com

91

본 교 재 AHC

AC”= = =10_ 2 =10'2 10'2

'2 10

sin 45˘

C’H”

sin A

삼각형의 높이

개념

28

1tan 60˘ '3h tan 45˘ h '3h h '3 1 '3+1 5('3-1)

2 ^{BAH=60˘ CAH=30˘} '3h h

4'3

'3 3

개념콕콕 ⁹²

2

ABH BAH=180˘-(30˘+90˘)=60˘

ACH CAH=180˘-(60˘+90˘)=30˘

ABH BH”=h tan 60˘='3h ACH CH”=h tan 30˘= h BC”=BH”-CH” 8='3h- h 8= 2'3h h=4'3

3

'3 3 '3

3

A’H”=h BAH=30˘ CAH=45˘

ABH B’H”=h tan 30˘= h AHC CH”=h tan 45˘=h

BC”=BH”+CH” 12= h+h

h=12 h= =6(3-'3)

AH”=6(3-'3) 6(3-'3)

-1

AH”=h BAH=50˘ CAH=20˘

ABH B’H”=h tan 50˘=1.2h AHC C’H”=h tan 20˘=0.4h

36 '3+3 '3+3

3

'3 3 '3 3 대표 유형

6(3-'3) -1 -2 50('3-1) m

-16 -2 50'3 m

93

BC”=B’H”+CH” 32=1.2h+0.4h 1.6h=32 h=20

AH”=20

-2

A BC”

H

AH”=h m BAH=45˘

CAH=60˘

ABH B’H”=h tan 45˘=h(m) AHC CH”=h tan 60˘='3h(m)

BC”=B’H”+C’H” 100=h+'3h (1+'3)h=100 h= =50('3-1)

50('3-1) m

50('3-1) m

A’’H”=h BAH=60˘ CAH=45˘

ABH B’H”=h tan 60˘='3h ACH CH”=h tan 45˘=h

BC”=B’H”-C’H” 6='3h-h ('3-1)h=6 h= =3('3+1)

A’H”=3('3+1)

-1

A’’H”=h BAH=40˘ CAH=15˘

ABH B’H”=h tan 40˘=0.8h ACH CH”=h tan 15˘=0.3h

BC”=B’H”-C’H” 3=0.8h-0.3h 0.5h=3 h=6

A’’H”=6 6

-2

AD”=h m BAD=60˘ CAD=30˘

ABD BD”=h tan 60˘='3h(m) ACD CD”=h tan 30˘= h(m)

BC”=BD”-CD” 100='3h- h h=100 h=50'3

50'3 m 50'3 m

2'3 3

'3 3 '3

3 6 '3-1

100 1+'3

100 m

45˘ 30˘

B H

A

C h m

http://zuaki.tistory.com

삼각형의 넓이

개념

29

1 ^{12 cm¤} 15'3 cm¤ 6'2 cm¤ 14 cm¤

2 21'2 cm¤ cm¤ 10 cm¤ 35'3 cm¤

2 9'3

2

개념콕콕 ⁹⁵

1

ABC=;2!;_6_8_sin 30˘

=;2!;_6_8_;2!;=12(cm¤ ) ABC=;2!;_5_12_sin 60˘

=;2!;_5_12_ =15'3(cm¤ ) ABC=;2!;_6_4_sin 45˘

=;2!;_6_4_'2=6'2(cm¤ ) 2

'3 2 94

01 02⁷ 03

04(30+10'3) m 05 069'6

0730(3-'3) m 084(3+'3) cm¤

배운대로해결하기

01

tan A=;bA; a=b tan A sin B=;cB; c=

02

x=5 cos 36˘=5_0.81=4.05 y=5 sin 36˘=5_0.59=2.95

x+y=4.05+2.95=7 7

03

ACB BC”=10 tan 50˘=10_1.2=12(m) BD”=BC”+CD”=12+1.6=13.6(m)

13.6 m 04

CHD

D’H”=30 tan 30˘=30_ =10'3(m) CEH

EH”=30 tan 45˘=30_1=30(m) DE”=D’H”+EH”=10'3+30(m)

B (30+10'3) m (30+10'3) m 05

B=180˘-135˘=45˘

A

BC” H

ABH

AH”=3'2 sin 45˘=3'2_ =3 BH”=3'2 cos 45˘=3'2_ =3

CH”=BC”-BH”=7-3=4 AHC AC”="√3¤ +4¤ ='∂25=5

06

A=180˘-(75˘+60˘)=45˘

B AC”

H BCH

BH”=18 sin 60˘=18_ =9'3 ABH

'3

2 ^75˘ _60˘

45˘

H A

C

B 18

'2 2 '2

2

45˘ 135˘

C D

H A

B 3'2

7

'3

3 ^C _45˘^30˘

D

30 m A

B H

E

b sin B

AB”= = =9'3_ =9'6 9'6

07

C CH”=h m

ACH=45˘ BCH=30˘

CAH AH”=h tan 45˘=h(m) CHB BH”=h tan 30˘= h(m)

AB”=AH”+BH” 60=h+ h

h=60 h= =30(3-'3)

30(3-'3) m 30(3-'3) m 08

AH”=h cm BAH=45˘ CAH=30˘

ABH BH”=h tan 45˘=h(cm) ACH CH”=h tan 30˘= h(cm)

BC”=BH”-CH” 4=h- h

h=4 h= =2(3+'3)

ABC=;2!;_4_2(3+'3)=4(3+'3)(cm¤ )

4(3+'3) cm¤

12 3-'3 3-'3

3

'3 3 '3

3 180 3+'3 3+'3

3

'3 3 '3

3 2 '2 9'3

sin 45˘

BH”

sin A

45˘ 60˘

H C

A B

h m

60 m

http://zuaki.tistory.com

본 교 재 ABC=;2!;_8_7_sin 30˘

=;2!;_8_7_;2!;=14(cm¤ )

2

ABC=;2!;_14_6_sin (180˘-135˘)

=;2!;_14_6_ =21'2(cm¤ ) ABC=;2!;_6_3_sin (180˘-120˘)

=;2!;_6_3_ = (cm¤ ) ABC=;2!;_8_5_sin (180˘-150˘)

=;2!;_8_5_;2!;=10(cm¤ ) ABC=;2!;_7_10_sin (180˘-120˘)

=;2!;_7_10_ =35'3 (cm¤ ) 2

'3 2

9'3 2 '3

2 '2

2

;2!;_4_7_sin B=7'3 sin B=

sin 60˘= B=60˘ 60˘

-1

;2!;_11_12_sin C=33 sin C=;2!;

sin 30˘=;2!; C=30˘ 30˘

-2

;2!;_6'2_BC”_sin 30˘=27 ;2!;_6'2_BC”_;2!;=27 BC”=27 BC”=9'2(cm)

;2!;_8_AC”_sin (180˘-135˘)=18'2

;2!;_8_AC”_ =18'2 2'2 AC”=18'2 AC”=9(cm)

'2 2 3'2

2

'3 2

'3 2 대표 유형

60˘ -130˘ -2

-16 cm -2 14'3 cm¤

96

-1

;2!;_AB”_10_sin (180˘-120˘)=15'3

;2!;_AB”_10_ =15'3 AB”=15'3

AB”=6(cm) 6 cm

-2

BD”

ABCD

= ABD+ BCD

=;2!;_2'3_4_sin (180˘-150˘) +;2!;_6_8_sin 60˘

=;2!;_2'3_4_;2!;+;2!;_6_8_

=2'3+12'3=14'3(cm¤ ) 14'3 cm¤

'3 2

60˘

150˘

C D A

B 8 cm

6 cm 4 cm

2'3 cm

5'3 2 '3

2

사각형의 넓이

개념

30

1 24'3 cm¤ 21'2 cm¤ 60'3 cm¤ 12 cm¤

2 20'3 cm¤ 63'2 cm¤

4

개념콕콕 ⁹⁷

1

ABCD=6_8_sin 60˘

=6_8_ =24'3(cm¤ ) ABCD=7_6_sin 45˘

=7_6_ =21'2(cm¤ ) ABCD=10_12_sin (180˘-120˘)

=10_12_ =60'3(cm¤ ) ABCD=4_6_sin (180˘-150˘)

=4_6_;2!;=12(cm¤ )

2

ABCD=;2!;_8_10_sin 60˘

=;2!;_8_10_ =20'3(cm¤ ) ABCD=;2!;_7_9_sin (180˘-135˘)

=;2!;_7_9_ = 63'2 (cm¤ ) 4

'2 2

'3 2 '3

2 '2

2 '3

2

http://zuaki.tistory.com

8_BC”_sin 30˘=40 8_BC”_;2!;=40 4BC”=40

BC”=10(cm) 10 cm

-1

ABCD x cm

x_x_sin (180˘-135˘)=16'2 x_x_ =16'2 x¤ =16'2 x¤ =32 x=4'2 ( x>0)

ABCD 4'2 cm 4'2 cm

-2

APD=;4!; ABCD=;4!;_(4_6_sin 60˘)

=;4!;_{4_6_ }=3'3(cm¤ )

ABCD

O OBC

AOB=24˘+36˘=60˘

ABCD=;2!;_10_14_sin 60˘

=;2!;_10_14_ =35'3(cm¤ ) 35'3 cm¤

-1

ABCD

O AOD

AOB=30˘+15˘=45˘

ABCD=;2!;_7_8_sin 45˘

=;2!;_7_8_ =14'2(cm¤ )

-2

;2!;_12_10_sin x=30 sin x=;2!;

sin 30˘=;2!; x=30˘ 30˘

'2 2

C B

30˘ 15˘ D A O

8 cm 7 cm

'3 2

24˘ 36˘

14 cm 10 cm A

B C

D O

'3 2 '2

2 '2

2 대표 유형

10 cm -14'2 cm -2

35'3 cm¤ -1 -2 30˘

98

01

C= B=75˘ A=180˘-(75˘+75˘)=30˘

ABC=;2!;_6_6_sin 30˘

=;2!;_6_6_;2!;=9(cm¤ )

02

;2!;_5'3_AC”_sin 60˘=30

;2!;_5'3_AC”_ =30 ;;¡4∞;;AC”=30 AC”=8(cm)

03

O

4 cm 45˘

8

=8_{;2!;_4_4_sin 45˘}

=8_{;2!;_4_4_ }=32'2(cm¤ ) 32'2 cm¤

04

;2!;_13_12_sin (180˘-B)=39'2 sin (180˘-B)=

sin 45˘=

180˘- B=45˘ B=135˘ 135˘

05

B’D”

ABCD

= ABD+ BCD

=;2!;_20_20_sin (180 -120 ) +;2!;_20'3_20'3_sin 60˘

=;2!;_20_20_ +;2!;_20'3_20'3_

=100'3+300'3=400'3(cm¤ ) 400'3 cm¤

'3 2 '3

2

60˘

20 cm

20 cm A

B C

D

20 3 cm 20 3 cm 120˘

'2 2

'2 2

'2 2

4 cm O

45˘

'3 2

99

01 02 0332'2 cm¤ 04^135˘

05400'3 cm¤ 06^30˘ 07

08

배운대로해결하기

http://zuaki.tistory.com

본 교 재 06

12_9_sin x=54 sin x=;2!;

sin 30˘=;2!; x=30˘ 30˘

07

AMC=;4!; ABCD

=;4!;_(14_16_sin 45˘)

=;4!;_{14_16_ }

=28'2(cm¤ )

08

BD”=x cm ABCD

AC”=BD”=x(cm)

;2!;_x_x_sin (180˘-120˘)=16'3

;2!;_x_x_ =16'3 x¤ =16'3 x¤ =64 x=8 ( x>0)

BD”=8(cm)

'3 4 '3

2

'2 2

100 102

01 02 ^pcm‹ 03

04(12-6'3 ) cm 0550'3 m 06

07 088'6 09 10

11^45˘ 129'3 cm¤ 134'2 cm¤ 14

15^{126 cm¤} 16 17^{16 cm} 188'2 cm¤

19 20^60˘ 21 ^cm¤

22 ^12'3₅ 23(12p-9'3) cm¤ 24 25'3

3 8'3

3 개념 넓히기로마무리

01

A=180˘-(90˘+51˘)=39˘

AB”=7 sin 51˘=7 cos 39˘

02 ABO

AO”=4 sin 60˘=4_ =2'3(cm) BO”=4 cos 60˘=4_;2!;=2(cm)

'3 2

=;3!;_p_2¤ _2'3

= p(cm‹ ) pcm‹

03

AB”=8 tan 32˘

=8_0.6=4.8(m) AC”=

AC”=8÷0.8=10(m)

=AB”+AC”

=4.8+10=14.8(m)

04

B O’A”

H OBH

OH”=12 cos 30˘

OH”=12_ =6'3(cm) A’H”=O’A”-O’H”

=12-6'3(cm)

B A (12-6'3) cm

(12-6'3 ) cm

05 ABH

A’H”=100 cos 30˘

A’H”=100_ =50'3(m) ^yy

AHC

C’H”=50'3 tan 45˘

C’H”=50'3_1=50'3(m)

50'3 m ^yy

50'3 m

06

A BC”

H ABH

AH”=15 sin B AH”=15_;5#;=9 BH”=15 cos B BH”=15_;5$;=12

CH”=BC”-BH”=18-12=6 AHC AC”="√9¤ +6¤ ='∂117=3'∂13

H A

C B

15

18

'3 2 '3

2

12 cm 30˘

A

B B'

O

H

8 cos 32˘

8'3 3 8'3

3

http://zuaki.tistory.com

07

A BC”

H ACH=180˘-135˘=45˘

ACH

A’H”=2'2 sin 45˘=2'2_ =2 CH”=2'2 cos 45˘=2'2_ =2

BH”=BC”+CH”=4+2=6 ABH AB”=ø∑6¤ +2¤ ='∂40=2'∂10

08

A=180˘-(60˘+75˘)=45˘

C AB”

H AHC

CH”=24 sin 45˘=24_ =12'2 BCH

BC”= = =12'2_ =8'6 8'6

09

A’H”=h BAH=40˘ CAH=50˘

ABH BH”=h tan 40˘

AHC CH”=h tan 50˘

BC”=BH”+CH” 12=h tan 40˘+h tan 50˘

h(tan 40˘+tan 50˘)=12 h=

AH”

10

A’H”=h m CAH=60˘-15˘=45˘

ABH BH”=h tan 60˘='3h(m) ACH CH”=h tan 45˘=h(m)

BC”=B’H”-CH” 10='3h-h

('3-1)h=10 h= =5('3+1)

5('3+1) m

11

;2!;_12_8'3_sin B=24'6 sin B=

sin 45˘='2 B=45˘ 45˘

2

'2 2 10

'3-1

12 tan 40˘+tan 50˘

2 '3 12'2

sin 60˘

CH”

sin B

'2 2

45˘

A

B C

H 60˘

75˘

24

'2 2 '2

2

H 135˘

A

B 4 C

2'2

12

AE” DC” AED= AEC

ABED= ABE+ AED

= ABE+ AEC

= ABC

=;2!;_4_9_sin 60˘

=;2!;_4_9_

=9'3(cm¤ ) 9'3 cm¤

13

G ABC

GBC=;3!; ABC

=;3!;_{;2!;_6_8_sin 45˘}

=;3!;_{;2!;_6_8_ }

=4'2(cm¤ ) 4'2 cm¤

14

ABD A’D”=AB”=3'2(cm)

BD”= =3'2_ =6(cm)

ABCD= ABD+ BCD

=;2!;_3'2_3'2+;2!;_6_5_sin 30˘

=;2!;_3'2_3'2+;2!;_6_5_;2!;

=9+;;¡2∞;;=;;£2£;;(cm¤ )

15

BD”

ABD

=;2!;_6_6'2_sin (180˘-135˘)

=;2!;_6_6'2_

=18(cm¤ ) ^yy

BCD=;2!;_12'2_18_sin 45˘

=;2!;_12'2_18_ =108(cm¤ ) yy

ABCD= ABD+ BCD

=18+108=126(cm¤ ) ^yy

126 cm¤

'2 2 '2

2

45˘

135˘

C D

A

B 6'2 cm

12'2 cm

18 cm 6 cm

2 '2 3'2

sin 45˘

'2 2 '3 2

http://zuaki.tistory.com

본 교 재 16

AB”=AD”=8(cm)

ADE AE”=8 sin 60˘=8_ =4'3(cm) BAE= BAD+ DAE=90˘+30˘=120˘

ABE=;2!;_8_4'3_sin (180˘-120˘)

=;2!;_8_4'3_

=24(cm¤ )

17

ABCD x cm

x_x_sin (180˘-120˘)=8'3 x_x_ =8'3 x¤ =8'3 x¤ =16 x=4 ( x>0)

ABCD

4_4=16(cm) 16 cm

18

= PAB+ PCD

=;2!; ABCD

=;2!;_(4_8_sin 45˘)

=;2!;_{4_8_ }

=8'2(cm¤ ) 8'2 cm¤

19

B’D”=x cm AC”=;2!;x(cm)

;2!;_;2!;x_x_sin 60˘=18'3

;2!;_;2!;x_x_ =18'3 x¤ =18'3 x¤ =144 x=12 ( x>0)

BD”=12(cm)

20

AOB= x

;2!;_8_9_sin x=18'3 sin x= yy

sin 60˘= x=60˘

AOB=60˘ yy

60˘

'3 2

'3 2 '3

8 '3

2

'2 2 '3

2 '3

2

'3 2

'3 2

21

BE”

BEA BEC'

BAE= BC'E=90˘ BE”

B’A”=BC'”

BEA BEC' RHS

ABE= C'BE=;2!;_(90˘-30˘)=30˘

BC'E EC'”=5 tan 30˘=5_ = (cm) ABC'E=2 BC'E

=2_{;2!;_5_ }

= (cm¤ ) cm¤

22

BAD= CAD=;2!; BAC=;2!;_60˘=30˘

ABC= ABD+ ADC

;2!;_4_6_sin 60˘=;2!;_4_A’D”_sin 30˘

+;2!;_A’D”_6_sin 30˘

;2!;_4_6_ =;2!;_4_A’D”_;2!;+;2!;_A’D”_6_;2!;

6'3=A’D”+;2#; A’D” ;2%; A’D”=6'3 A’D”=

23

OC”

OA”=OC”

OCA= OAC=30˘

AOC=180˘-(30˘+30˘)=120˘

= AOC - AOC

=p_6¤ _;3!6@0);-;2!;_6_6_sin (180˘-120˘)

=p_6¤ _;3!6@0);-;2!;_6_6_

=12p-9'3(cm¤ ) (12p-9'3 ) cm¤

24

x ABCD=;2!;_4_6_sin x=12 sin x

sin x 1 ABCD

12 cm¤

'3 2

30˘

C

A 6 cm O B

12'3 5 12'3

5 '3

2

25'3 3 25'3

3

5'3 3

5'3 3 '3

3

30˘ C C'

D D' A E A'

B 5 cm

http://zuaki.tistory.com

현의 수직이등분선

개념

31 1. 원과 직선

Ⅳ. 원의 성질

1 OMB O’M” RHS B’M”

2 ² '3 14 2'5 6 4'2

개념콕콕 ¹⁰⁴

2

x=2AM”=2_7=14 x=2MB”=2_'5=2'5 x=;2!;AB”=;2!;_12=6 x=;2!;AB”=;2!;_8'2=4'2

OAM A’M”="√5¤ -4¤ ='9=3(cm)

AB”=2A’M”=2_3=6(cm) 6 cm

-1

OMB B’M”=øπ(2'2)¤ -2¤ ='4=2(cm)

AB”=2B’M”=2_2=4(cm) 4 cm

-2

A’M”=;2!;AB”=;2!;_30=15(cm) OA”

OAM

OA”="√15¤ +8¤ ='∂289=17(cm) O

2p_17=34p(cm) -3

OC”=OB”=6(cm) O’M”=6-3=3(cm)

OMB MB’”="√6¤ -3¤ ='∂27=3'3(cm) AB”=2MB”=2_3'3=6'3(cm)

A B

M O

8 cm 30 cm

-4

O’A”

O’A”=x cm OC”=O’A”=x(cm) O’M”=x-6(cm)

OAM

x¤ =8¤ +(x-6)¤ x¤ =64+(x¤ -12x+36) 12x=100 x=;;™3∞;;

O ;;™3∞;; cm ;;™3∞;; cm

O r cm

OA”=r cm OD”=r-3(cm) AOD r¤ =9¤ +(r-3)¤

r¤ =81+(r¤ -6r+9) 6r=90 r=15

15 cm 15 cm

-1

O r cm

OA”=r cm OD”=r-4(cm) AD”=;2!; AB”=;2!;_16=8(cm)

AOD r¤ =8¤ +(r-4)¤

r¤ =64+(r¤ -8r+16) 8r=80 r=10

10 cm 10 cm

-2

O

A’D”=;2!; AB”=;2!;_24=12(cm)

AOD O’D”="√13¤ -12¤ ='∂25=5(cm) C’D”=OC”-O’D”=13-5=8(cm)

OA” O

AB” M

OA”=8 cm

OM”=;2!;OA”=;2!;_8=4(cm) ^{A M B}

O D 12 cm 13 cm

A B

C

O

A B

C

O D

4 cm 8 cm

r cm (r-4) cm

A B

C

D 3 cm 9 cm

O

r cm (r-3) cm O

A B

C M 8 cm 6 cm

대표 유형

6 cm -14 cm -2

-3 -4;;™3∞;; cm

15 cm -110 cm -2

8'3 cm -14'6 cm -2

105 106

http://zuaki.tistory.com

본 교 재

1

OM”=ON” AB”=CB”

x=;2!;CB”=;2!;AB”=;2!;_16=8 OM”=ON” AB”=CD”

x=;2!;CD”=;2!;AB”=;2!;_6=3

2

AB”=2MB”=2_4=8

AB”=CD” x=ON”=3 CD”=2DN”=2_7=14

AB”=CD” x=OM”=6

OAM A’M”="√8¤ -4¤ ='∂48=4'3(cm) AB”=2AM”=2_4'3=8'3(cm) 8'3 cm

-1

OA” O AB”

M OA”=4'2 cm

OM”=;2!;OA”=;2!;_4'2=2'2(cm) AOM

AM”=øπ(4'2)¤ -(2'2)¤ ='∂24=2'6(cm)

AB”=2AM”=2_2'6=4'6(cm) 4'6 cm -2

OA” O

AB” M

O r cm

OA”=r cm OM”=;2!;OA”=;2!;r(cm) AM”=;2!;AB”=;2!;_14'3=7'3(cm)

OAM r¤ =(7'3)¤ +{;2!;r}¤ r¤ =147+;4!;r¤

;4#;r¤ =147 r¤ =196 r=14 ( r>0)

O 14 cm

B

A 14'3 cm O

M

A M B

O

현의 길이

개념

32

1 ^{5 7 8 3} 2 ^{5 4 3 6}

개념콕콕 ¹⁰⁷

OAM A’M”=ø∑4¤ -('7)¤ ='9=3(cm) AB”=2 A’M”=2_3=6(cm)

O’M”=O’N” C’D”=AB”=6(cm) 6 cm -1

OM”=ON” AB”=CD”=8(cm)

AM”=;2!; AB”=;2!;_8=4(cm) OAM OA”="√4¤ +4¤ ='∂32=4'2(cm) 4'2 cm

-2

O CD”

N AB”=CD”

ON”=OM”=2'6(cm) OND

DN”=øπ7¤ -(2'6)¤ ='∂25=5(cm) CD”=2D’N”=2_5=10(cm) OCD=;2!;_10_2'6=10'6(cm¤ )

O’M”=O’N” AB”=AC”

ABC

B=;2!;_(180˘-50˘)=65˘ ^65˘

-1

O’M”=O’N” AB”=AC”

ABC

BAC=180˘-2_70˘=40˘ 40˘

-2

O’D”=O’E”=O’F” A’B”=BC”=CA”

ABC ABC

3_12=36(cm)

O

M N

A B

D C 7 cm

2'6 cm 대표 유형

6 cm -1 4'2 cm -2

65˘ -1 40˘ -2

108

109

01 02 03^{16p cm¤} 04^{13 cm}

05 06 0716'3 cm 08^50˘

배운대로해결하기

http://zuaki.tistory.com

01

O’M”=;2!;OC”=;2!;_10=5(cm) OMB MB”="√10¤ -5¤ ='∂75=5'3(cm)

A’M”=MB”=5'3(cm)

02

OC”

OC”=;2!;AB”=;2!;_30=15(cm) CM”=;2!;CD”=;2!;_24=12(cm)

OMC O’M”="√15¤ -12¤ ='∂81=9(cm)

03

MCB MB”=øπ(2'6)¤ -3¤ ='∂15(cm)

OB” O

r cm

OB”=r cm OM”=r-3(cm) OMB

r¤ =(r-3)¤ +('∂15)¤

r¤ =(r¤ -6r+9)+15 6r=24 r=4

O =p_4¤ =16p(cm¤ ) 16p cm¤

04

O r cm

OA”=r cm OD”=r-4(cm) AD”=;2!;AB”=;2!;_12=6(cm)

AOD

r¤ =(r-4)¤ +6¤ r¤ =(r¤ -8r+16)+36 8r=52 r=;;¡2£;;

:¡2£:_2=13(cm) ^{13 cm}

05

OA” O

AB” M

O r cm

OA”=r cm OM”=;2!;OA”=;2!;r(cm) AM”=;2!;AB”=;2!;_10'3=5'3(cm)

O

10'3 cm A

B M A

C

D B O 6 cm

(r-4) cm 4 cm

r cm O

A B

C 3 cmM

2'6 cm

A O B

C D

M

30 cm 24 cm

AMO

r¤ =(5'3)¤ +{;2!;r}¤ r¤ =75+;4!;r¤

;4#;r¤ =75 r¤ =100 r=10 ( r>0)

O 10 cm

06

CN”=;2!; CD”=;2!;_12=6(cm) OCN O’N”="√10¤ -6¤ ='∂64=8(cm)

AB”=2 A’M”=2_6=12(cm)

AB”=CD” O’M”=ON”=8(cm) 07

O’A”

O’A”=8 cm OAM

A’M”="√8¤ -4¤ ='∂48=4'3(cm) AB”=2 A’M”=2_4'3=8'3(cm)

O’M”=O’N”

CD”=AB”=8'3(cm)

AB”+C’D”=8'3+8'3=16'3(cm) 16"3 cm 08

AMON MAN=360˘-(90˘+100˘+90˘)=80˘

O’M”=O’N” A’B”=AC”

ABC

ACB=;2!;_(180˘-80˘)=50˘ ^50˘

O A

B

C D

N

M

4 cm 4 cm

원의 접선

개념

33

1 ^{60 4} 2 ^{130˘ 65˘}

3 ^{9 70}

개념콕콕 ¹¹⁰

1

PTO=90˘ OPT

POT=180˘-(30˘+90˘)=60˘ x=60

PTO=90˘ POT

PT”="√5¤ -3¤ ='∂16=4(cm) x=4 2

PAO= PBO=90˘ APBO

x=360˘-(90˘+50˘+90˘)=130˘

http://zuaki.tistory.com

본 교 재

PAO= PBO=90˘ AOBP

x=360˘-(90˘+115˘+90˘)=65˘

3

PA”=PB” PAB

PBA= PAB=70˘ x=70

대표 유형

4 cm -13 cm -2

17 cm -12'1ß0 cm -2 44˘

5 cm -110 cm -2 5 cm

13 cm 12 cm -1 4'6 cm

-214p cm¤

111 112

PTO=90˘ POT

PO”="√8¤ +6¤ ='∂100=10(cm)

PA”=PO”-AO”=10-6=4(cm) 4 cm

-1

PTO=90˘ PTO

PO”=øπ(3'3)¤ +3¤ ='∂36=6(cm)

PA”=PO”-AO”=6-3=3(cm) 3 cm

-2

O r cm

OA”=OT”=r(cm) OP”=r+8(cm)

OTP=90˘ OTP

(r+8)¤ =r¤ +12¤ r¤ +16r+64=r¤ +144 16r=80 r=5

O =p_5¤ =25p(cm¤ )

PB”=PA”=15(cm) PBO=90˘ PBO

PO”="√15¤ +8¤ ='∂289=17(cm) ^{17 cm} -1

OQ”=OA”=3(cm) PAO=90˘ POA

P’A”="√(4+3)¤ -3¤ ='∂40=2'1ß0(cm)

PB”=PA”=2'1ß0(cm) 2'1ß0 cm

-2

PAO=90˘ PAB=90˘-22˘=68˘

PBA PA”=PB”

APB=180˘-2_68˘=44˘ 44˘

BE”=BD”=7-4=3(cm)

AF”=AD”=7(cm) CE”=CF”=7-5=2(cm)

BC”=BE”+CE”=3+2=5(cm) 5 cm

-1

AD”=AF”=20(cm) BE”=BD”=20-16=4(cm) CE”=CF”=20-14=6(cm)

BC”=BE”+CE”=4+6=10(cm) 10 cm -2

BD”=BE” CF”=CE”

AD”+AF”=(AB”+BD”)+(AC”+CF”)

=(AB”+BE”)+(AC”+CE”)

=AB”+(BE”+CE”)+AC”

=AB”+BC”+AC”=7+8+9=24(cm) AD”=AF” 2AD”=24 AD”=12(cm)

BD”=AD”-AB”=12-7=5(cm) 5 cm

CE”=CA”=4(cm) DE”=DB”=9(cm) CD”=4+9=13(cm)

C

BD” H

HB”=CA”=4(cm) D’H”=9-4=5(cm)

CHD CH”="√13¤ -5¤ ='∂144=12(cm)

AB”=CH”=12(cm) 13 cm 12 cm

-1

CE”=CA”=8(cm) DE”=DB”=12(cm) CD”=8+12=20(cm)

C BD”

H HB”=CA”=8(cm) D’H”=12-8=4(cm)

CDH

CH”="√20¤ -4¤ ='∂384=8'6(cm)

8 cmA

12 cm B C

O

D H E A O 4 cm

9 cm

B D

C E

H

http://zuaki.tistory.com

삼각형의 내접원

개념

34

1 x=3 y=8 z=7 x=5 y=4 z=6 2 ^{4 17}

3 ¹⁵ AD”=9-r BD”=12-r 3

개념콕콕 ¹¹³

AB”=CH”=8'6(cm) O

;2!;_8'6=4'6(cm) 4'6 cm

-2

CE”=CA”=7(cm) DB”=DE”=11-7=4(cm)

D AC”

H HA”=DB”=4(cm) C’H”=7-4=3(cm)

CHD

HD”="√11¤ -3¤ ='∂112=4'7(cm) AB”=HD”=4'7(cm) O

;2!;_4'7=2'7(cm)

O =;2!;_p_(2'7)¤

=14p(cm¤ ) 14p cm¤

H

O 7 cm

11 cm

A B

D C

E

1

x=AD”=5 z=CF”=11-5=6 y=BE”=10-6=4 2

BE”=BD”=5

CF”=CE”=9-5=4 x=4

BE”=BD”=6 AF”=AD”=10-6=4 CE”=CF”=15-4=11

BC”=BE”+CE”=6+11=17 x=17 3

AB”="√12¤ +9¤ ='∂225=15 EC”=FC”=OE”=r

AD”=AF”=9-r BD”=BE”=12-r AB”=AD”+BD”

15=(9-r)+(12-r)

2r=6 r=3

AD”=x cm AF”=AD”=x(cm) BE”=BD”=12-x(cm) CE”=CF”=8-x(cm)

BC”=BE”+CE”

14=(12-x)+(8-x) 2x=6 x=3

AD”=3(cm) 3 cm

-1

BD”=x cm BE”=BD”=x(cm)

AF”=AD”=22-x(cm) CF”=CE”=17-x(cm) AC”=AF”+CF”

15=(22-x)+(17-x) 2x=24 x=12

BD”=12(cm) 12 cm

-2 AF”=x cm

AD”=AF”=x(cm) BE”=BD”=6(cm) CF”=CE”=8(cm)

ABC =2(AF”+BD”+CE”)

36=2(x+6+8) x=4 AF”=4(cm)

OE” OF”

O r cm

OECF

CE”=CF”=OF”=r(cm) AD”=AF”=3-r(cm) BD”=BE”=4-r(cm)

ABC AB”="√4¤ +3¤ ='∂25=5(cm) AB”=AD”+B’D”

5=(3-r)+(4-r) 2r=2 r=1

O 1 cm 1 cm

-1

OD” OE”

O r cm

DBEO

BD”=BE”=OE”=r(cm)

AF”=AD”=5-r(cm) CF”=CE”=12-r(cm)

A

B C

D E

F 5 cm

12 cm O

O A

4 cm

3 cm

B C

D

E F 대표 유형

3 cm -1 12 cm -2

1 cm -1 2 cm -29p cm¤

114

http://zuaki.tistory.com

본 교 재 ABC AC”="√5¤ +12¤ ='∂169=13(cm)

AC”=AF”+CF”

13=(5-r)+(12-r) 2r=4 r=2

O 2 cm 2 cm

-2

OD” OF”

O r cm

ADOF

AD”=AF”=OF”=r(cm)

BD”=BE”=6(cm) CF”=CE”=9(cm) AB”=r+6(cm) AC”=r+9(cm)

ABC (6+9)¤ =(r+6)¤ +(r+9)¤

r¤ +15r-54=0 (r-3)(r+18)=0 r=3 ( r>0)

O =p_3¤ =9p(cm¤ ) 9p cm¤

A

B C

D

E F

6 cm 9 cm O

외접사각형

개념

35

1 ^{13 cm 16 cm} 2^{48 cm}

3 ^{3 8 7 16}

개념콕콕 ¹¹⁵

1

AD”+BC”=AB”+CD”=5+8=13(cm) AD”+BC”=AB”+CD”=10+6=16(cm)

2

AB”+CD”=AD”+BC”

ABCD =AB”+BC”+CD””+AD”

=2(AD”+BC”)

=2_(9+15)=48(cm) 3

AB”+CD”=AD”+BC”

6+10=x+13 x=3 AB”+CD”=AD”+BC”

7+5=4+x x=8

AB”+CD”=AD”+BC”

x+13=8+12 x=7 AB”+CD”=AD”+BC”

20+x=12+24 x=16

AB”+CD”=AD”+BC”

6+12=8+(4+CF”) CF”=6(cm) 6 cm

-1

AB”+CD”=AD”+BC”

9+(DG”+5)=6+11 DG”=3(cm) 3 cm

-2

AB”+CD”=AD”+BC”

10+15=12+(BE”+7) BE”=6(cm) O AB”

F OE”

OFBE

OF”=BE”=6(cm)

O =p_6¤ =36p(cm¤ ) -3

O 4 cm AB”=2_4=8(cm)

AD”+BC”=AB”+CD”=8+12=20(cm)

ABCD=;2!;_(AD”+BC”)_8=;2!;_20_8=80(cm¤ ) 80 cm¤

ABE AE”="√10¤ -8¤ ='∂36=6(cm) DE”=x cm BC”=AD”=6+x(cm)

EBCD O EB”+CD”=ED”+BC”

10+8=x+(6+x) 2x=12 x=6

DE”=6(cm) 6 cm

-1

DEC EC”="√13¤ -12¤ ='∂25=5(cm) AD”=x cm BC”=AD”=x(cm)

BE”=x-5(cm)

ABED O AB”+ED”=AD”+BE”

12+13=x+(x-5) 2x=30 x=15

AD”=15(cm) 15 cm

A

10 cm

7 cm 15 cm 12 cm

B C

D

E F O 대표 유형

6 cm -1 3 cm -2

-380 cm¤

6 cm -1 15 cm

116

http://zuaki.tistory.com

01

O r cm

OA”=OT”=r(cm) OP”=r+8(cm)

OTP=90˘ OPT

(r+8)¤ =r¤ +16¤ r¤ +16r+64=r¤ +256 16r=192 r=12

O 2p_12=24p(cm)

02

PB”=PA”=4(cm)

PAB= PBA=;2!;_(180˘-60˘)=60˘

APB

APB= _4¤ =4'3(cm¤ ) 4'3 cm¤

03

ODA=90˘ AOD

AD”="√6¤ -2¤ ='∂32=4'2(cm) AD”=AF” BE”=BD” CE”=CF”

ACB =AC”+BC”+AB”

=AC”+(BE”+CE”)+AB”

=AC”+(BD”+CF”)+AB”

=(AC”+CF”)+(AB”+BD”)

=AF”+AD”=2AD”

=2_4'2=8'2(cm)

04

BE”=BC”=5(cm) AD”=AE”=8-5=3(cm)

A BC”

H HC”=AD”=3(cm) B’H”=5-3=2(cm)

ABH

AH”="√8¤ -2¤ ='∂60=2'∂15(cm) ABCD=;2!;_(3+5)_2'∂15

=8'∂15(cm¤ ) 8'∂15 cm¤

O D

8 cm

5 cm C A

E

B H

'3 4

05

AF”=AD”=4(cm)

BE”=BD”=11-4=7(cm) CE”=CF”=10-4=6(cm) BC”=BE”+CE”=7+6=13(cm)

06

OE” DBEO

BE”=BD”=DO”=3(cm) CF”=CE”=8-3=5(cm)

AF”=x cm AD”=AF”=x(cm)

AB”=x+3(cm) AC”=x+5(cm)

ABC (x+5)¤ =(x+3)¤ +8¤

x¤ +10x+25=(x¤ +6x+9)+64 4x=48 x=12 AF”=12(cm)

AC”=AF”+CF”=12+5=17(cm) 17 cm

07

BCD CD”="√13¤ -12¤ ='∂25=5(cm) AB”+CD”=AD”+BC”

11+5=AD”+12 AD”=4(cm)

08

AE”=x cm AECD O

AE”+CD”=AD”+EC” x+6=9+EC” EC”=x-3(cm)

BE”=9-(x-3)=12-x(cm) ABE

x¤ =6¤ +(12-x)¤ x¤ =36+(144-24x+x¤ ) 24x=180 x=;;¡2∞;;

AE”=;;¡2∞;;(cm) ;;¡2∞;; cm

A

O

B C

D E 3 cm F

8 cm 117

01 024'3 cm¤ 03 048'∂15 cm¤

05 06^{17 cm} 07 08;;¡2∞;; cm

배운대로해결하기

118 120

01 02 03;2(; cm 04^{30 cm}

05 06 078'5 cm¤ 08

0916'3 cm¤ 10 116'3 cm 12 13^{2 cm} 14^{6 cm} 15 16

17 18 19^{3 cm} 20^{13 cm}

21^{25p cm¤} 226'3 cm 2316'∂15 cm¤

244'2 cm

개념 넓히기로마무리

http://zuaki.tistory.com

본 교 재 01

OC”

OC”=;2!;AB”=;2!;_20=10 OM”=10-4=6

OMC CM”="√10¤ -6¤ ='∂64=8

CD”=2 CM”=2_8=16

02

O A’M”=;2!;AB”=;2!;_6=3(cm)

OAM O’M”="√4¤ -3¤ ='7(cm)

'7 cm 03

BMC

B’M”=ø∑(3'3)¤ -3¤ =3'2(cm) ^yy

OB” O

r cm

OB”=r cm O’M”=r-3(cm) ^yy OMB

r¤ =(r-3)¤ +(3'2)¤

r¤ =(r¤ -6r+9)+18 6r=27 r=;2(;

O ;2(; cm ^yy

;2(; cm

04

O r cm

OA”=r cm OD”=r-12(cm)

AOD r¤ =24¤ +(r-12)¤

r¤ =576+(r¤ -24r+144) 24r=720 r=30

30 cm 30 cm

05

OA” O

AB” M

OA”=10(cm)

OM”=;2!;OA”=;2!;_10=5(cm) ^M

O

A B

A B

C

D

O 12 cm

(r-12) cm r cm

24 cm

O A

C

M B

3 cm 3'3 cm O

A M B

4 cm 6 cm

A B

M C

D O 20

4

OAM

AM”="√10¤ -5¤ ='∂75=5'3(cm) AB”=2 AM”=2_5'3=10'3(cm)

06

OND D’N”=ø∑(2'5)¤ -2¤ ='∂16=4(cm) CD”=2 D’N”=2_4=8(cm)

O’M”=ON” AB”=CD”=8(cm)

07

O CD”

N

AB”=CD” ON”=OM”=4(cm) OND

DN”="√6¤ -4¤ ='∂20=2'5(cm) CD”=2DN”=2_2'5=4'5(cm)

OCD=;2!;_4'5_4=8'5(cm¤ ) 8'5 cm¤

08

O’D”=OF” AB”=AC”

ABC

ABC=;2!;_(180˘-56˘)=62˘

DBEO

DOE=360˘-(90˘+62˘+90˘)=118˘

09

MBNO MBN=360˘-(90˘+120˘+90˘)=60˘^yy OM”=ON” AB”=BC”

BAC= BCA=;2!;_(180˘-60˘)=60˘

ABC yy

AB”=2BM”=2_4=8(cm) ^yy

ABC= _8¤ =16'3(cm¤ ) ^yy

16'3 cm¤

10

APBO AOB=360˘-(90˘+45˘+90˘)=135˘

PAO PBO

PAO= PBO=90˘ PO” OA”=OB”

PAO PBO (RHS )

=p_4¤ _;3@6@0%;=10p(cm¤ ) '3

4

A

B M

N

C D

O 6 cm 4 cm

http://zuaki.tistory.com

11

PAO=90˘ AOP=60˘ AOP

AO” : AP”=1 : '3 6 : AP”=1 : '3 AP”=6'3(cm)

AOBP APB=360˘-(90˘+120˘+90˘)=60˘

PA”=PB” ABP

AB”=AP”=6'3(cm) 6'3 cm

12

BD”=BE” CF”=CE”

AD”+AF”=(AB”+BD”)+(AC”+CF”)

=(AB”+BE”)+(AC”+CE”)

=AB”+(BE”+CE”)+AC”

=AB”+BC”+AC”

=9+7+10=26(cm)

AD”=AF” 2AF”=26 AF”=13(cm) CF”=AF”-AC”=13-10=3(cm)

13

E CD”

H EH”=BC”=8(cm)

EB”=x cm HC”=EB”=x(cm) D’H”=8-x(cm)

EP”=EB”=x(cm) DP”=DC”=8(cm) ED”=x+8(cm)

DEH (x+8)¤ =8¤ +(8-x)¤

x¤ +16x+64=64+(64-16x+x¤ ) 32x=64 x=2

EB”=2(cm) 2 cm

14

BE”=x cm BD”=BE”=x(cm) ^yy

AF”=AD”=9-x(cm) CF”=CE”=10-x(cm) ^yy AC”=AF”+CF”

7=(9-x)+(10-x) 2x=12 x=6

BE”=6(cm) yy

6 cm

15

ABC C=180˘-(50˘+66˘)=64˘

CE”=CF” CFE

x=;2!;_(180˘-64˘)=58˘

A

B E

P

O C

H D

8 cm

8 cm

16

OD” OF”

O r cm

ADOF

AD”=AF”=OF”=r(cm)

BD”=BE”=6(cm) CF”=CE”=4(cm) AB”=r+6(cm) AC”=r+4(cm)

ABC

(6+4)¤ =(r+6)¤ +(r+4)¤ r¤ +10r-24=0 (r-2)(r+12)=0 r=2 ( r>0)

O 2p_2=4p(cm)

17

AB”+CD”=AD”+BC” 2x+(x+4)=(x+2)+(3x-4) 3x+4=4x-2 x=6

18

AB”+CD”=AD”+BC”=18+12=30(cm) AB” : CD”=2 : 3

CD”=30_ =18(cm)

19

OE”

OEBF

BF”=OF”=4(cm) ^yy

CG”=CF”=11-4=7(cm) ^yy D’H”=DG”=10-7=3(cm)yy

3 cm

20

BE”=x cm CE”=15-x(cm)

ABED O AB”+DE”=AD”+BE”

12+DE”=15+x DE”=x+3(cm) DEC (x+3)¤ =(15-x)¤ +12¤

x¤ +6x+9=(225-30x+x¤ )+144 36x=360 x=10

DE”=10+3=13(cm) 13 cm

21

O AB”

H

AH”=;2!;AB”=;2!;_10=5(cm) R cm r cm

OAH R¤ =5¤ +r¤ R¤ -r¤ =25

O

A H B

10 cm A

4 cm

11 cm 10 cm

B C

D

E

F G H

O

3 2+3

A

B C

D

E F O 6 cm 4 cm

http://zuaki.tistory.com

본 교 재

=p_R¤ -p_r¤ =p(R¤ -r¤ )

=25p(cm¤ ) 25p cm¤

22

ADO AEO RHS DAO= EAO=30˘

AOD AD” : AO”='3 : 2 AD” : 6='3 : 2

2AD”=6'3 AD”=3'3(cm)

BC” O F

AD”=AE” BF”=BD” CF”=CE”

ACB =AC”+CB”+BA”

=AC”+(CF”+BF”)+B’A”

=AC”+(CE”+BD”)+BA”

=(AC”+CE”)+(BD”+BA”)

=AE”+AD”=2AD”

=2_3'3=6'3(cm) 6'3 cm 23

CE”=CA”=6(cm) DE”=DB”=10(cm) CD”=6+10=16(cm)

C

BD” H

HB”=CA”=6(cm) DH”=10-6=4(cm)

CHD CH”="√16¤ -4¤ ='∂240=4'∂15(cm) AB”=CH”=4'∂15(cm) O

;2!;_4'∂15=2'∂15(cm)

OE” OE” CD” OE”=2'∂15 cm

COD=;2!;_16_2'∂15=16'∂15(cm¤ ) 16'∂15 cm¤

24

AB”+CD”=AD”+BC”=4+8=12(cm) AB”=CD” AB”=;2!;_12=6(cm)

A D

BC” H I

HI”=AD”=4(cm) BH”=;2!;_(8-4)=2(cm)

ABH

A’H”="√6¤ -2¤ ='∂32=4'2(cm) O

4'2 cm 4'2 cm

A

O 4 cm

8 cm

B H I C

D

A B

C

D E

O H

6 cm 10 cm

A 60˘

B F D

C E 6 cm O

대표 유형

-1 -2120˘

70˘ -1 44˘ -2118˘

45˘ -1 80˘ -2

70˘ -1 58˘ -2

123 124

μAPB 130˘ μAB

360˘-130˘=230˘

x=;2!;_230˘=115˘

-1

μAB 2 APB=2_110˘=220˘

x=360˘-220˘=140˘

-2

OB”

AOB=2 APB=2_25˘=50˘

BOC=2 BQC=2_35˘=70˘

x= AOB+ BOC

=50˘+70˘=120˘

120˘

25˘35˘

x A

B

P Q

C O

원주각과 중심각

개념

36 2. 원주각

Ⅳ. 원의 성질

1 ^{40˘ 50˘ 60˘}

2 ^{45˘ 30˘ 35˘}

개념콕콕 ¹²²

1

x=;2!; AOB=;2!;_80˘=40˘

x=2 APB=2_25˘=50˘

x=;2!; AOB=;2!;_120˘=60˘

2

ACB=90˘

x=180˘-(90˘+55˘)=35˘

http://zuaki.tistory.com

원주각의 크기와 호의 길이

개념

37

1 ^{26 30 6 3} 2 ^{36 21 9 15}

개념콕콕 ¹²⁵

2

12˘ : x˘=4 : 12 12 : x=1 : 3 x=36

x˘ : 42˘=7 : 14 x : 42=1 : 2 2x=42 x=21

45˘ : 30˘=x : 6 3 : 2=x : 6

2x=18 x=9

40˘ : 24˘=x : 9 5 : 3=x : 9 3x=45 x=15

OA” OB”

PAO= PBO=90˘

APBO

AOB=360˘-(90˘+40˘+90˘)

=140˘

ACB=;2!; AOB=;2!;_140˘=70˘ ^70˘

-1

OA” OB”

AOB=2 ACB

=2_68˘=136˘

OAP= OBP=90˘

AOBP

APB=360˘-(90˘+136˘+90˘)=44˘ 44˘

-2

O’A” OB”

PAO= PBO=90˘

APBO

AOB=360˘-(90˘+56˘+90˘)

=124˘

x=;2!;_(360˘-124˘)=118˘ ^118˘

ABD= ACD=40˘

ABP

85˘=40˘+ x x=45˘ 45˘

-1

ABD= ACD=45˘

ABP

x= ABP+ BAP=45˘+35˘=80˘ 80˘

-2

QB”

AQB= APB=20˘

BQC= BRC=30˘

x= AQB+ BQC

=20˘+30˘=50˘

20˘ x 30˘

A B P

Q R

C 56˘ x

A

B

P C O

68˘

A

B P

C O 40˘

A

B

P O C

AB” O ACB=90˘

ACB ABC=180˘-(20˘+90˘)=70˘

x= ABC=70˘ 70˘

-1

AB” O BCA=90˘

DCB= DAB=32˘

x= BCA- DCB=90˘-32˘=58˘ 58˘

-2

AQ” AC”

O AQC=90˘

AQB=90˘-47˘=43˘

x= AQB=43˘

x 47˘

A C

B

P Q

O

대표 유형

-1 -25 cm

80˘ -1 60˘ -2

126

μAC=μ BD DCB= ABC=24˘

PCB

x= PCB+ PBC=24˘+24˘=48˘

http://zuaki.tistory.com

본 교 재

네 점이 한 원 위에 있을 조건 - 원주각

개념

38

1

2 ^{60˘ 36˘ 68˘ 100˘}

개념콕콕 ¹²⁷

1

BAC= BDC=35˘ A B C D

ABD+ ACD A B C D

BAC=85˘-25˘=60˘ BAC= BDC=60˘

A B C D 2

DBC BDC=180˘-(40˘+72˘)=68˘

x= BDC=68˘

BDC= BAC=55˘

PCD

x= PDC+ PCD=55˘+45˘=100˘

대표 유형

-1 -2

-33˘

128 -1

μAC=μ BD BCD= ABC= x

PBC

30˘= x+ x 2 x=30˘ x=15˘

-2

ABP 60˘= BAP+20˘ BAP=40˘

ABD : BAC=μAD : μ BC 20˘ : 40˘=μAD : 10 1 : 2=μAD : 10

2μAD=10 μAD=5(cm) 5 cm

BCA : BAC : ABC=μAB : μ BC : μ CA=2 : 3 : 4

x=180˘_ =80˘ 80˘

-1

BCA : BAC : ABC=μAB : μ BC : μ CA=4 : 5 : 6

x=180˘_ =60˘ 60˘

-2

BC” μAB

;5!;

ACB=180˘_;5!;=36˘

μCD ;9!;

CBD=180˘_;9!;=20˘

PBC

x= PBC+ PCB=20˘+36˘=56˘

x A

B P

C D

5 4+5+6

4 2+3+4

BAC=180˘-(90˘+35˘)=55˘

BAC+ BDC A B C D

-1

BAC= BDC A B C D

DBC=180˘-(45˘+85˘)=50˘

DAC+ DBC A B C D

ABD+ ACD A B C D

ABD=180˘-(60˘+80˘)=40˘

ABD= ACD A B C D

A B C D

-2

A B C D ADB= ACB=25˘

DPB 65˘= x+25˘ x=40˘

-3

A B C D DAC= DBC=55˘

APD 86˘=55˘+ x x=31˘

PCD PCD=180˘-(60˘+86˘)=34˘

y= ACD=34˘

y- x=34˘-31˘=3˘ 3˘

http://zuaki.tistory.com

08

AB” O ACB=90˘

ACB BAC=180˘-(90˘+36˘)=54˘

DAB=82˘-54˘=28˘

ADC= ABC=36˘ APD

DPB= DAP+ ADP=28˘+36˘=64˘

09

AD” AB”

O ADB=90˘

ADE

EAD=180˘-(90˘+50˘)=40 x=2 CAD=2_40˘=80˘

80˘

10

BDC= BAC=44˘

μBC=μ CD DBC= BAC=44˘

DBC BCD=180˘-(44˘+44˘)=92˘

11

AB” O ACB=90˘

ABC= ADC=30˘ ACB

BAC=180˘-(90˘+30˘)=60˘

ADC : BAC=μAC : μ BC 30˘ : 60˘=4 : μ BC 1 : 2=4 : μ BC μBC=8(cm)

12

ADB : CBD=μAB : μ CD x : CBD=3 : 1 3 CBD= x CBD=;3!; x

DBE x=;3!; x+32˘

;3@; x=32˘ x=48˘

13

ACB : CAB : ABC=μAB : μ BC : μ CA=3 : 4 : 5

ACB=180˘_ =45˘ 45˘

14

BC” μAB

;9@;

BCA=180˘_;9@;=40˘

BCA : CBD=μAB : μCD

40˘ : CBD=2 : 3 2 CBD=120˘ CBD=60˘

A B

P

C

D

3 3+4+5

50˘

x

A B

C

D E

O 129 130

01 02 03^80˘ 04

05^52˘ 06^40˘ 07 08

09^80˘ 10 11 12

13^45˘ 14^100˘ 15 16^60˘

배운대로해결하기

01

AOB=2 APB=2_46˘=92˘

OAB OA”=OB”

x=;2!;_(180˘-92˘)=44˘

02

AOB=2 APB=2_60˘=120˘

μAB=2p_9_;3!6@0);=6p(cm)

03

x=;2!;_260˘=130˘ y=;2!;_(360˘-260˘)=;2!;_100˘=50˘

x- y=130˘-50˘=80˘ 80˘

04

OB”

AOB=2 APB=2_20˘=40˘

BOC=110˘-40˘=70˘

x=;2!; BOC=;2!;_70˘=35˘

05

OA” OB”

AOB=2 ACB=2_64˘=128˘

PAO= PBO=90˘

APBO

x=360˘-(90˘+128˘+90˘)=52˘

52˘

06

ACB= ADB=20˘ APC

60˘= x+20˘ x=40˘ 40˘

07

PB”

APB=;2!; AOB=;2!;_70˘=35˘

BPC=85˘-35˘=50˘

x= BPC=50˘

85˘

70˘

x A

B

P Q

C O

64˘

x A

P B

C O

110˘

20˘ x

A C

P Q

B O

http://zuaki.tistory.com

본 교 재 BCP

CPD= BCP+ CBP=40˘+60˘=100˘ 100˘

15

ADB+ ACB A B C D

ACB=65˘-35˘=30˘

ACB= ADB A B C D

BAC=90˘-35˘=55˘

BAC= BDC A B C D

ADB=180˘-(30˘+110˘)=40˘

ADB+ ACB A B C D

BDC=180˘-(43˘+77˘)=60˘

BAC= BDC A B C D

16

A B C D DAC= DBC=35˘

BAC=95˘-35˘=60˘

x= BAC=60˘ 60˘

원에 내접하는 사각형의 성질

개념

39

1 ^{x=110˘ y=80˘ x=130˘ y=75˘}

x=65˘ y=60˘ x=102˘ y=110˘

2

개념콕콕 ¹³¹

1

x=180˘-70˘=110˘

y=180˘-100˘=80˘

x=180˘-50˘=130˘

y=180˘-105˘=75˘

x= DAB=65˘

y=180˘-120˘=60˘

x=180˘-78˘=102˘

y= BCD=110˘

2

B+ D+180˘ ABCD

대표 유형

-1 122˘ -2108˘

-1 -2140˘

51˘ -1 63˘

-1 -298˘

132 133

BCD BCD=180˘-(55˘+65˘)=60˘

ABCD

x=180˘- BCD=180˘-60˘=120˘

-1

BC” O BDC=90˘

DBC BCD=180˘-(90˘+32˘)=58˘

ABCD O

x=180˘- BCD=180˘-58˘=122˘ 122˘

-2

BC”=BD” BCD

BCD=;2!;_(180˘-36˘)=72˘

ABCD

x=180˘- BCD=180˘-72˘=108˘ 108˘

BAD=;2!; BOD=;2!;_140˘=70˘

ABCD O

x= BAD=70˘

-1

BAD=;2!; BOD=;2!;_114˘=57˘

ABCD O

x= BAD=57˘

-2 ABCD

40˘+ x=100˘ x=60˘

CBD= CAD=40˘

y=180˘-(60˘+40˘)=80˘

x+ y=60˘+80˘=140˘ 140˘

ABCD

CBQ= PBA= ADC= x

http://zuaki.tistory.com

BQC BCD= x+43˘

APB BAD=35˘+ x

ABCD BAD+ BCD=180˘

(35˘+ x)+( x+43˘)=180˘

2 x=102˘ x=51˘ 51˘

ABCD

ABP= ADC= x

AQD BAP=43˘+ x

PBA 35˘+ x+(43˘+ x)=180˘

2 x=102˘ x=51˘

-1 ABCD

CBQ= PBA= ADC= x BQC BCD= x+24˘

APB BAD=30˘+ x

ABCD BAD+ BCD=180˘

(30˘+ x)+( x+24˘)=180˘

2 x=126˘ x=63˘ 63˘

ABCD

ABP= ADC= x

AQD BAP=24˘+ x

PBA 30˘+ x+(24˘+ x)=180˘

2 x=126˘ x=63˘

B=180˘-(52˘+68˘)=60˘ B+ D+180˘

ABCD

BAC+ BDC ABCD

A+ C+180˘ ABCD

-1

A+ C 180˘ ABCD

BAD=180˘-100˘=80˘ BAD+100˘

ABCD ABCD

-2

BAC= BDC ABCD

ABC=180˘- ADC=180˘-(50˘+32˘)=98˘ 98˘

접선과 현이 이루는 각

개념

40

1 ^{100˘ 50˘ 65˘ 55˘}

2 ^{25˘ 32˘ 80˘ 96˘}

개념콕콕 ¹³⁴

1

x= CAT=180˘-(60˘+55˘)=65˘

BCA= BAT=80˘ ABC

x=180˘-(45˘+80˘)=55˘

2

CB” O CAB=90˘

x= BCA=180˘-(90˘+65˘)=25˘

CB” O CAB=90˘

x= CBA=180˘-(90˘+58˘)=32˘

CAB= CBA=50˘

x= BCA=180˘-(50˘+50˘)=80˘

BCA= BAC=42˘

x= ABC=180˘-(42˘+42˘)=96˘

대표 유형

35˘ -1 64˘ -2

40˘ -1 62˘ -2

135

ABCD BCD=180˘-80˘=100˘

BCD DBC=180˘-(100˘+45˘)=35˘

x= DBC=35˘ 35˘

-1

ABCD BCD=180˘-102˘=78˘

BDC= BCT=38˘ BCD

x=180˘-(78˘+38˘)=64˘ 64˘

-2

BCA= BAT=56˘

BOA=2 BCA=2_56˘=112˘

OA”=OB”

OAB x=;2!;_(180˘-112˘)=34˘

http://zuaki.tistory.com

본 교 재 AT”

BAT= BTC=65˘

ATB=90˘

ATP=180˘-(90˘+65˘)=25˘

APT

65˘= x+25˘ x=40˘ 40˘

-1

AT”

BAT= BTC= x ATB=90˘

ATP=180˘-(90˘+ x)

=90˘- x

APT x=34˘+(90˘- x)

2 x=124˘ x=62˘ 62˘

-2

BE”=BD” BED=;2!;_(180˘-30˘)=75˘

DFE= BED=75˘ DEF

DEF=180˘-(55˘+75˘)=50˘

34˘

T C

B

A P

O

x x 65˘

T C

B

P A

O

136 137

01 02 03^15˘ 04

05 06^73˘ 07^94˘ 08^36˘

09 10 11 12^15˘

13 14^33˘ 15^42˘

배운대로해결하기

01

x=;2!; AOC=;2!;_160˘=80˘

ABCD O y=180˘-80˘=100˘

y- x=100˘-80˘=20˘

02

AB”=AC” ABC

ABC=;2!;_(180˘-40˘)=70˘

ABCD x=180˘-70˘=110˘

03 ABCE

( x+30˘)+80˘=180˘ x=70˘

ABCD y=180˘-95˘=85˘

y- x=85˘-70˘=15˘ 15˘

04

AB” O ACB=90˘

μBC=μ CD DAC= CAB=20˘

ABCD O

(20˘+20˘)+( x+90˘)=180˘ x=50˘

05

AC”

BAC=;2!; BOC=;2!;_60˘=30˘

CAE=110˘-30˘=80˘

ACDE O

x=180˘- CAE=180˘-80˘=100˘

06

ABC ABC=180˘-(47˘+60˘)=73˘

ABCD

CDE= ABC=73˘ 73˘

07

PQ”

ABQP O

APQ=180˘-86˘=94˘

PQCD O'

DCQ= APQ=94˘ 94˘

08

ABQ PAD=52˘+40˘=92˘

ABCD

ADP= ABC=52˘

ADP

x=180˘-(92˘+52˘)=36˘ 36˘

ABCD

CDQ= ABC=52˘

PBC PCQ= x+52˘

DCQ 52˘+( x+52˘)+40˘=180˘

x=36˘

09

BAC+ BDC ABCD

B+ D+180˘ ABCD

78˘

86˘

A

B

P

Q C D

O O'

60˘

110˘

x A

B

C D

E O

http://zuaki.tistory.com

ACD ADC=180˘-(60˘+50˘)=70˘

B+ D=180˘ ABCD

ADB= ACB ABCD

A+ DCE ABCD

10

BCA= BAT=65˘

BOA=2 BCA=2_65˘=130˘

11 ABCD

DAB=180˘-116˘=64˘

ABD ADB=180˘-(64˘+42˘)=74˘

ABE= ADB=74˘

12

ACB : CAB : ABC=μAB : μ BC : μ CA=3 : 4 : 5

ACB=180˘_ =45 x= ACB=45˘

CAB=180˘_ =60˘ y= CAB=60˘

y- x=60˘-45˘=15˘ 15˘

13

AT”

BTA=90˘ BTA

BAT=180˘-(34˘+90˘)=56˘

ATP= ABT=34˘

ATP

56˘=34˘+ x x=22˘

14

BC”

ABC=90˘

ACB= ABT=57˘

ABC

BAC=180˘-(90˘+57˘)=33˘

BDC= BAC=33˘ 33˘

15

APE= ABP=75˘

DPE= DCP=63˘

x=180˘-(75˘+63˘)=42˘ 42˘

57˘

A

T B

C D

O 34˘

x A

T B

P O

4 3+4+5

3 3+4+5

원에서의 선분의 길이 사이의 관계

개념

41

1 ^{9 5 8} 5'2 2 ⁶ ;2(; 4 ;;¡3¢;;

개념콕콕 ¹³⁸

1

x_12=6_18 x=9 8_x=10_4 x=5 2_12=x_3 x=8

5_10=x¤ x=5'2 ( x>0)

2

3_8=4_x x=6

x_16=6_12 x=;2(;

3_(3+9)=x_9 x=4 2_(2+5)=3_x x=;;¡3¢;;

대표 유형

8 -1 2 -2

13 -1 6 -25

139

PD”=x PC”=14-x

P’A”¥PB”=PC”¥PD” 4_12=(14-x)_x x¤ -14x+48=0 (x-6)(x-8)=0

x=8 ( PC”<PD”)

PD”=8 8

-1

P’A”=x PB”=16-x

P’A”¥PB”=PC”¥PD” x_(16-x)=4_7 x¤ -16x+28=0 (x-2)(x-14)=0

x=2 ( P’A”<PB”)

P’A”=2 2

-2

PC” : PD”=2 : 3 PC”=2x P’D”=3x P’A”¥PB”=PC”¥PD” 8_3=2x_3x 6x¤ =24 x¤ =4 x=2 ( x>0)

PD”=3_2=6

http://zuaki.tistory.com

본 교 재

원에서의 선분의 길이 사이의 관계의 응용

개념

42

1 ^{PC” 4 4 8} 9 9 8 9 56 25 5 8 6 6 24 6 12 2'3

2

개념콕콕 ¹⁴⁰

2

6_4+9_3

2_(2+7)=3_(3+3)

AB”=x PA”¥P’B”=PC”¥PD”

5_(5+x)=6_(6+9) 25+5x=90 5x=65 x=13

AB”=13 13

-1

CD”=x PA”¥PB”=PC”¥PD”

8_(8+12)=10_(10+x) 160=100+10x 10x=60 x=6

CD”=6 6

-2

PA”=x P’A”¥PB”=PC”¥PD”

x_(x+7)=4_(4+11) x¤ +7x-60=0 (x+12)(x-5)=0 x=5 ( x>0)

PA”=5 5

PC”=x AB” CD” PD”=PC”=x P’A”¥PB”=PC”¥PD”

16_4=x¤ x¤ =64 x=8 ( x>0)

PC”=8 8

대표 유형

8 -12'∂21 -2 ;;¡2∞;; cm

3 -18 -2 36p cm¤

;2(; cm -115 cm -2 40p cm¤

-1

141 142

-1

PA”=x AB” CD” PB”=PA”=x P’A”¥PB”=PC”¥PD” x¤ =6_14

x¤ =84 x=2'∂21 ( x>0)

PA”=2'∂21 2'∂21

-2

CD” O

E

AB” CE” DE”=CD”=6(cm)

O r cm

DB”=2r-3(cm)

D’’A”¥DB”=DC”¥DE” 3_(2r-3)=6¤

6r-9=36 6r=45 r=;;¡2∞;;

O ;;¡2∞;; cm ;;¡2∞;; cm

OP”=x PA”¥PB”=PC”¥PD”

(7-x)(7+x)=8_5 49-x¤ =40 x¤ =9 x=3 ( x>0)

OP”=3 3

-1

OP”=x PA”¥PB”=PC”¥PD”

(10+x)(10-x)=4_9 100-x¤ =36 x¤ =64 x=8 ( x>0)

OP”=8 8

-2

O r cm

PA”¥PB”=PC”¥PD” 4_5=(r+4)(r-4) 20=r¤ -16 r¤ =36 r=6 ( r>0)

O

p_6¤ =36p(cm¤ ) 36p cm¤

O r cm

P’A”¥PB”=PC”¥PD” 4_(4+5)=3_(3+2r) 36=9+6r 6r=27 r=;2(;

O ;2(; cm ;2(; cm

6 cm

A D B

C

E O 3 cm

http://zuaki.tistory.com

원의 할선과 접선의 길이 사이의 관계

개념

43

1 ^{12 5 16 9}

2 ⁴ 2'∂14

개념콕콕 ¹⁴³

1

x¤ =9_(9+7)=144 x=12 ( x>0) (5'3)¤ =x_15 15x=75 x=5 8¤ =4_x 64=4x x=16 6¤ =3_(3+x) 36=9+3x

3x=27 x=9

2

PA”=PO”-O’A”=9-5=4

PT” ¤ =P’A”¥PB” PT” ¤ =4_(9+5)=56 PT”=2'∂14 ( PT”>0)

대표 유형

4 -1 2 -24'3

;;™3º;; cm -1 4 cm -2

-1 -23'6

25 -1 10'6 -2

144 145

PA”=x PT” ¤ =P’A”¥PB”

6¤ =x_(x+5) x¤ +5x-36=0

(x+9)(x-4)=0 x=4 ( x>0)

PA”=4 4

-1

PA”=x PT” ¤ =P’A”¥PB”

4¤ =x_(x+6) x¤ +6x-16=0

(x+8)(x-2)=0 x=2 ( x>0)

PT”=2 2

-2

ATP= ABT APT= ABT

ATP= APT

APT A’P”=A’T”=4

PT”=x PT” ¤ =P’A”¥PB”

x¤ =4_(4+8) x¤ =48 x=4'3 ( x>0)

PA”=4'3 4'3

O r cm O’A”=OB”=r(cm)

PT” ¤ =P’A”¥P’B’ 7¤ =3_(3+2r) 49=9+6r 6r=40 r=;;™3º;;

O ;;™3º;; cm ;;™3º;; cm

-1

O r cm O’A”=OB”=r(cm)

PT” ¤ =P’A”¥P’B’ (2'5)¤ =2_(2+2r) 20=4+4r 4r=16 r=4

O 4 cm 4 cm

-1

O r cm

P’A”¥PB”=PC”¥PD” 6_(6+2r)=9_(9+15) 36+12r=216 12r=180 r=15

O 15 cm 15 cm

-2

O r cm

P’A”¥PB”=PC”¥PD” 5_(5+7)=(10-r)(10+r) 60=100-r¤ r¤ =40 r=2'∂10 ( r>0)

O p_(2'∂10)¤ =40p(cm¤ ) ^{40p cm¤}

4_8+9_5 8_15=12_10

4_(4+3)+3_(3+5) 3_(3+10)+5_(5+4) 3_(3+4)+4_(4+5)

A B C D

-1

2_6=4_3 2_3=6_1

2_(2+5)+5_(5+2) 2_8=4_4 3_(3+5)=2_(2+10)

A B C D

http://zuaki.tistory.com

본 교 재

-2

O r cm

OA”=OB”=r(cm) PA”=8-2r(cm) PT” ¤ =PA”¥PB” 4¤ =(8-2r)_8 16=64-16r 16r=48 r=3

O p_3¤ =9p(cm¤ )

E’A”¥EB”=EC”¥ED” E’A”_4=2_6 E’A”=3 PT” ¤ =PA”¥PB” PT” ¤ =3_(3+3+4)=30

PT”='∂30 ( PT”>0)

-1

EA”¥EB”=EC”¥ED” E’A”_4=8_3 E’A”=6 PT” ¤ =PA”¥PB” PT” ¤ =2_(2+6+4)=24

PT”=2'6 ( PT”>0)

-2

DA”¥DB”=DC”¥DT” D’A” ¤ =3_6=18 DA”=3'2 ( DA”>0)

PT” ¤ =PA”¥PB” PT” ¤ =3'2_(3'2+3'2+3'2)=54

PT”=3'6 ( PT”>0) 3'6

P’T'” ¤ =PA”¥PB”=PT” ¤ PT'”=PT” x=10 PT” ¤ =PA”¥PB” 10¤ =5_(5+y)

100=25+5y 5y=75 y=15

x+y=10+15=25 25

-1

P’T” ¤ =PA”¥PB”=P’T'” ¤ PT”=PT'” x=2'6 PT” ¤ =P’A”¥PB” (2'6)¤ =3_(3+y)

24=9+3y 3y=15 y=5

xy=2'6_5=10'6 10'6

-2

P’T” ¤ =PA”¥PB”=P’T'” ¤

PT”=PT'”=;2!;T’T'”=;2!;_8=4 PA”=x PT” ¤ =PA”¥PB”

4¤ =x_(x+6) x¤ +6x-16=0

(x-2)(x+8)=0 x=2 ( x>0) PA”=2

01

P’A”=PB”=x cm P’A”¥PB”=PC”¥PD”

x¤ =2_8 x¤ =16 x=4 ( x>0) PA”=4(cm)

02

PC”=x PD”=10-x

P’A”¥PB”=PC”¥PD” (11-3)_3=x_(10-x) x¤ -10x+24=0 (x-4)(x-6)=0

x=6 ( PC”>PD”)

PC”=6 6

03

PA” : AB”=2 : 1 PA”=2x cm AB”=x cm(x>0) P’A”¥PB”=PC”¥PD”

2x_(2x+x)=6_(6+10) 6x¤ =96 x¤ =16 x=4 ( x>0)

PA”=2_4=8(cm) 8 cm

04

O PA”¥PB”=PE”¥PF”

O' PC”¥PD”=PE”¥PF”

PA”¥PB”=PC”¥PD” AB”=x 4_(4+x)=3_(3+5) 16+4x=24

4x=8 x=2

AB”=2

05

AB” CD” PC”=PD”

PC”=PD”=x cm PA”¥PB”=PC”¥PD”

4_(28-4)=x¤ x¤ =96 x=4'6 ( x>0) CD”=2PC”=2_4'6=8'6(cm)

06

O r cm

PA”=r+;2!;r=;2#;r(cm) PB”=;2!;r(cm)

146 147

01 02⁶ 03^{8 cm} 04

05 0612'2p cm 07

08^{3 cm} 09 10^{8 cm} 11^{6 cm} 12^{4 cm} 13 ^cm¤ 14 152'∂10 16

27'2 2 배운대로해결하기

http://zuaki.tistory.com

PA”¥PB”=PC”¥PD” ;2#;r_;2!;r=9_6

;4#;r¤ =54 r¤ =72 r=6'2 ( r>0) O

2p_6'2=12'2p(cm) 12'2p cm

07

PO”=x P’A”¥PB”=PC”¥PD”

(x-9)(x+9)=4_(4+6) x¤ -81=40 x¤ =121 x=11 ( x>0)

PO”=11 08

BAC= BDC A B C D

CD”=x cm PA”¥PB”=PD”¥PC”

5_(5+9)=7_(7+x) 70=49+7x

7x=21 x=3

CD”=3(cm) 3 cm

09

AB”=BP”=x cm PT” ¤ =PB”¥P’A”

8¤ =x_(x+x) x¤ =32 x=4'2 ( x>0) AB”=4'2(cm)

10

PT” PQ” PQ”=PT”=12(cm)

PA”=x cm PT” ¤ =PA”¥PB”

12¤ =x_(12+6) 18x=144 x=8

PA”=8(cm) 8 cm

11

PT” ¤ =PA”¥PB”=4_(4+12)=64 PT”=8(cm) ( PT”>0)

PTA PBT

PTA= PBT P

PTAª PBT AA

PA” : PT”=A’T” : T’B” 4 : 8=AT” : 12

8AT”=48 AT”=6(cm) 6 cm

12

AO” O

B PA”=x cm PT” ¤ =PA”¥ PB”

(2'∂14)¤ =x_(x+5+5)

x¤ +10x-56=0 (x-4)(x+14)=0

O

P T A

B

5 cm

2 14 cm

x=4 ( x>0)

PA”=4(cm) 4 cm

13

PT” ¤ =PB”¥PA” PT” ¤ =3_(3+6)=27 PT”=3'3(cm) ( PT”>0)

PT” O ATP=90˘

ATP

AT”=øπ9¤ -(3'3)¤ ='∂54=3'6(cm)

ATP=;2!;_3'3_3'6= (cm¤ ) cm¤

14

QA”¥QB”=QC”¥QT”

QA”_6=4_12 QA”=8 PA”=x PT” ¤ =PA”¥PB”

(6'2)¤ =x_(x+8+6) x¤ +14x-72=0 (x-4)(x+18)=0 x=4 ( x>0)

PA”=4 15

PT” ¤ =PA”¥PB” x¤ =2_(2+3)=10 x='∂10 ( x>0)

PT'” ¤ =PA”¥PB” PT'”=PT” y='∂10

x+y='∂10+'∂10=2'∂10 2'∂10

16

O PT” ¤ =P’A”¥PB”

O' PT” ¤ =PC”¥PD”

P’A”¥PB”=PC”¥PD” AB”=x 4_(4+x)=3_(3+7) 16+4x=30 4x=14 x=;2&;

AB”=;2&;

27'2 2 27'2

2

148 150

01 02^{20 m} 03^216˘ 04

05^69˘ 06 07 08^{27 cm}

09^128˘ 10^135˘ 11 12

13^32˘ 1418'3 cm¤ 15⁵ 16

17⁸ 18 194'3 cm 2080'6

21 ^'5₃ 22 233'6 24^{6 cm¤}

개념 넓히기로마무리

http://zuaki.tistory.com

본 교 재 01

OB”

BOC=2 BQC=2_25˘=50˘

AOB=124˘-50˘=74˘

APB=;2!; AOB=;2!;_74˘=37˘

02

A B O

AOB=2 APB=2_30˘=60˘

OA”=OB”

OAB= OBA=;2!;_(180˘-60˘)=60˘

AOB

10 m

2_10=20(m) 20 m

03

PAO= PBO=90˘ APBO

y=360˘-(90˘+36˘+90˘)=144˘

x=;2!; AOB=;2!;_144˘=72˘

x+ y=72˘+144˘=216˘ 216˘

04

DPB 62˘= ADB+42˘ ADB=20˘

ACB= ADB=20˘

05

AD”

CAD=;2!; COD

=;2!;_42˘=21˘ ^yy

AB” O

ADB=90˘ yy

PAD

x=180˘-(21˘+90˘)=69˘ yy

69˘

06

BAC= BDC=40˘

μAB=μ`BC ACB= BDC=40˘

ABC ABC=180˘-(40˘+40˘)=100˘

x= ABC- DBC=100˘-78˘=22˘

42˘

x

A B

P

C D

O 30˘

10 m

O

A B

P

O 25˘

124˘

A

B

P Q

C

07

μAC=2μ`BD μAC : μ`BD=2 : 1 ABC : DCB=μAC : μ BD

ABC : 15˘=2 : 1 ABC=30˘

PCB

x= PCB+ PBC=15˘+30˘=45˘

08

APD 75˘= PAD+35˘ PAD=40˘

l cm

CAD : 180˘=μ CD : l 40˘ : 180˘=6 : l 2 : 9=6 : l 2l=54 l=27

27 cm 27 cm

09

BOD=2 BAD=2_52˘=104˘

ABCD O BCD=180˘-52˘=128˘

OBCD 104˘+ x+128˘+ y=360˘

x+ y+232˘=360˘ x+ y=128˘ 128˘

10

CF” ABCF

AFC=180˘-105˘=75˘

CDEF

CFE=180˘-120˘=60˘

AFE= AFC+ CFE

=75˘+60˘=135˘ 135˘

11

PQCD O' PQB= PDC=115˘

ABQP O BAP=180˘-115˘=65˘

x=2 BAP=2_65˘=130˘

12

μBC=μ CA ABC= BAC=;2!;_(180˘-110˘)=35˘

x= ABC=35˘

13 ABCD

ABC=180˘-108˘=72˘ yy

BCP= BAC=40˘ ^yy

BPC

72˘= BPC+40˘ BPC=32˘ ^yy

32˘

105˘

120˘

A

B

C D

E

http://zuaki.tistory.com

F

문서에서 본교재 (페이지 55-87)