Ⅱ. 피타고라스 정리
1 5 cm 5'2 cm 2 4'5 3'3 3 2'3 cm 2'6 cm
개념콕콕 46
1
"√4¤ +3¤ =5(cm) '2_5=5'2(cm) 2
x="√12¤ -8¤ =4'5 '2_x=3'6 x=3'3
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3
x cm
"√2¤ +x¤ =4 4+x¤ =16
x¤ =12 x=2'3 ( x>0) 2'3 cm x cm '2_x=4'3 x=2'6
2'6 cm
1
= _2='3(cm) = _2¤ ='3(cm¤ )
2
ABH A’H” ¤ =7¤ -x¤ y AHC A’H” ¤ =9¤ -(8-x)¤ y
7¤ -x¤ =9¤ -(8-x)¤
16x=32 x=2
ABH A’H”="√7¤ -2¤ =3'5(cm) ABC=;2!;_8_3'5=12'5(cm¤ )
'3 4 '3
2 x cm
"√x¤ +6¤ =2'∂13 x¤ +36=52 x¤ =16 x=4 ( x>0)
4_6=24(cm¤ ) 24 cm¤
-1
x cm
"√7¤ +x¤ =11 49+x¤ =121 x¤ =72 x=6'2 ( x>0)
7_6'2=42'2(cm¤ ) -2
BCD BD”="√6¤ +8¤ =10(cm) AB”_A’D”=A’H”_BD”
8_6=A’H”_10 A’H”=;;™5¢;;(cm) ;;™5¢;; cm
x cm '2x=4 x=2'2
ABCD
4_2'2=8'2(cm) 8'2 cm
-1
x cm '2x=10 x=5'2
ABCD
4_5'2=20'2(cm) 20'2 cm
-2
x cm '2x=20 x=10'2
10'2 cm 10'2 cm
대표 유형
24 cm¤ -1 -2 ;;™5¢;; cm
8'2 cm -120'2 cm -2 10'2 cm
47
삼각형의 높이와 넓이
개념
14
1 '3 cm '3 cm¤
2 2 3'5 cm 12'5 cm¤
개념콕콕 48
x cm
x=4'3 x=8
= _8¤ =16'3(cm¤ )
-1
x cm x=6'2 x=4'6
='3_(4'6)¤ =24'3(cm¤ ) 24'3 cm¤
4 '3
2
'3 4 '3
2
대표 유형
-1 24'3 cm¤ -22 cm 2'5 cm 8'5 cm¤
-1 8 cm 48 cm¤
-2 12 cm 84 cm¤
49
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본 교 재
-2
AD” ABC ABC
AD”= _2'3=3(cm)
G ABC
AG”=;3@;AD”=;3@;_3=2(cm) 2 cm
BH”=;2!;BC”=;2!;_8=4(cm)
ABH AH”="√6¤ -4¤ =2'5(cm) ABC=;2!;_8_2'5=8'5(cm¤ )
2'5 cm 8'5 cm¤
-1
A BC”
H
BH”=;2!;BC”=;2!;_12=6(cm) ABH AH”="√10¤ -6¤ =8(cm)
ABC=;2!;_12_8=48(cm¤ )
8 cm 48 cm¤¤
-2
A BC”
H BH”=x cm CH”=(14-x) cm ABH A’H” ¤ =15¤ -x¤ y AHC
A’H” ¤ =13¤ -(14-x)¤ y 15¤ -x¤ =13¤ -(14-x)¤
28x=252 x=9
ABH A’H”="√15¤ -9¤ =12(cm) ABC=;2!;_14_12=84(cm¤ )
12 cm 84 cm¤¤
A
B H C
15 cm
13 cm
x cm
14-x cm 6 cm 6 cm
A
B H C
10 cm 10 cm
'3 2
50
01(10+4'5) cm 02;;™5¡;; cm 03
0410 cm 05 06 cm¤
074'5 cm 08
27'3 4 배운대로해결하기
01
x cm
"√5¤ +x¤ =3'5 25+x¤ =45 x¤ =20 x=2'5 ( x>0)
ABCD
2_(5+2'5)=10+4'5(cm) (10+4'5) cm
02
ABD BD”=øπ9¤ +12¤ =15(cm) ABD AB” ¤ =BE”_DB”
9¤ =BE”_15 BE”=;;™5¶;;(cm) BCD CD” ¤ =DF”_DB”
9¤ =DF”_15 DF”=;;™5¶;;(cm) EF”=BD”-BE”-DF”
=15-;;™5¶;;-;;™5¶;;=;;™5¡;;(cm) ;;™5¡;; cm
03
r cm 2r cm
'2_2r=4'2 r=2 p_2¤ =4p(cm¤ )
04
ABCD 8'2 cm
'2_BC”=8'2 BC”=8(cm)
ECFG 6'2 cm
'2_EC”=6'2 EC”=6(cm)
BCE BE”="√8¤ +6¤ =10(cm) 10 cm
05
x cm x¤ =18'3 x¤ =72
x=6'2 ( x>0)
_6'2=3'6(cm)
06
AD”= _6=3'3(cm)
ADE= _(3'3)¤ = (cm¤ ) 27'3 cm¤
4 27'3
4 '3
4 '3
2 '3
2 '3
4
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07
A BC”
H
;2!;_8_AH”=32 AH”=8(cm) BH”=;2!;BC”=;2!;_8=4(cm) ABH
AB”="√4¤ +8¤ =4'5(cm) 4'5 cm
08
A
BC” H
BH”=x cm CH”=(10-x) cm
ABH
A’H” ¤ =12¤ -x¤ y AHC
A’H” ¤ =8¤ -(10-x)¤ y 12¤ -x¤ =8¤ -(10-x)¤
20x=180 x=9
ABH A’H”="√12¤ -9¤ =3'7(cm) ABC=;2!;_10_3'7=15'7(cm¤ )
A
B H C
12 cm 8 cm
x cm 10-x cm H A
B 4 cm 4 cm C
특수한 직각삼각형의 세 변의 길이의 비
개념
15
1 '2 4'2 1 4 2 6 '3 3'3 2 x='2 y=2 x=3'2 y=3'2 3 x=5'3 y=5 x=3 y=6
개념콕콕 51
2
x : '2=1 : 1 x='2 y : '2='2 : 1 y=2
x : 6=1 : '2 '2x=6 x=3'2 y : 6=1 : '2 '2y=6 y=3'2
3
x : 10='3 : 2 2x=10'3 x=5'3 y : 10=1 : 2 2y=10 y=5 x : 3'3=1 : '3 '3x=3'3 x=3 y : 3'3=2 : '3 '3y=6'3 y=6
ABD AB” : AD”=2 : 1 4 : x=2 : 1 2x=4 x=2
ADC AD” : AC”=1 : '2
2 : y=1 : '2 y=2'2 x=2 y=2'2
-1
ADC AC” : AD”=2 : '3 4'3 : x=2 : '3 2x=12 x=6
ABD AB” : AD”='2 : 1
y : 6='2 : 1 y=6'2 x=6 y=6'2
-2
DBC BC” : CD”='3 : 1
BC” : 3'2='3 : 1 BC”=3'6(cm) ABC AB” : BC”=1 : '2
AB” : 3'6=1 : '2 '2 AB”=3'6 AB”=3'3(cm)
-3
BCD BC” : BD”='3 : 2 9 : BD”='3 : 2 '3 BD”=18
BD”=6'3(cm)
ABD AB” : BD”=1 : '2 AB” : 6'3=1 : '2 '2 AB”=6'3
AB”=3'6(cm) AB” : AD”=1 : 1 AD”=AB”=3'6(cm)
ABD=;2!;_3'6_3'6=27(cm¤ ) 27 cm¤
A
BC” H
ABH
AB” : AH”='2 : 1
8 : AH”='2 : 1 '2 AH”=8 AH”=4'2(cm)
ABCD=7'2_4'2=56(cm¤ )
45˘
A
B H C
D 8 cm
7'2 cm 대표 유형
x=2 y=2'2 -1 x=6 y=6'2
-2 -3 27 cm¤
-1 (9+3'3) cm¤
52
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본 교 좌표평면 위의 두 점 사이의 거리 재
개념
16
1 '∂10 2'∂17 5 6'2
2 '5 10 5'2 4'5
3 O’A”='∂∂13 OB”=2'5 AB”='∂29
개념콕콕 53
1
"√3¤ +1¤ ='∂10 øπ(-8)¤ +2¤ =2'∂17 øπ(-4)¤ +π(-3)¤ =5 øπ6¤ +(-6)¤ =6'2 2
AB”=øπ(2-1)¤ +π(5-7)¤ ='5 CD”=øπ{-7-(-1)}¤ +π(-4-4)¤ =10 EF”=øπ{5-(-2)}¤ +π(-1-0)¤ =5'2 GH”=øπ(-1-3)¤ +π{2-(-6)}¤ =4'5
3
y
O 2 4
2
-2 x
A B
O’A”=øπ3¤ +2¤ ='∂13 OB”=øπ(-2)¤ +4¤ =2'5
AB”=øπ(-2-3)¤ +π(4-2)¤ ='∂29 AB” ¤ <O’A” ¤ +OB” ¤ OAB -1
A BC”
H ABH
AB” : AH”=2 : '3
2'6 : AH”=2 : '3 2AH”=6'2 AH”=3'2(cm)
AB” : BH”=2 : 1
2'6 : BH”=2 : 1 2BH”=2'6 BH”='6(cm)
AHC AH” : CH”=1 : 1 CH”=AH”=3'2(cm)
BC”=BH”+CH”='6+3'2(cm)
ABC=;2!;_('6+3'2)_3'2=9+3'3(cm¤ )
(9+3'3) cm¤
60˘
30˘45˘
45˘
A
B H C
2'6 cm
대표 유형
6 -1 -1 -2
AB”=3'2 BC”=3'2 CA”=6 B=90˘
-1 AB”=5'2 BC”=4'5 C’A”='∂10 A>90˘
-2
54
AB”=øπ{6-(-2)}¤ +π(2-a)¤ =4'5 8¤ +(2-a)¤ =(4'5)¤ a¤ -4a-12=0
(a+2)(a-6)=0 a=6 ( a>0) 6
-1
AB”=øπ(-1-3)¤ +(a-5)¤ =2'∂13
(-4)¤ +(a-5)¤ =(2'∂13)¤ a¤ -10a-11=0
(a+1)(a-11)=0 a=-1 ( a<0) -1
-2
øπ(4-2)¤ +π(1-3)¤ =2'2 øπ{5-(-3)}¤ +π(0-1)¤ ='∂65 øπ(1-6)¤ +π{-1-(-2)}¤ ='∂26 øπ{0-(-2)}¤ +π{-6-(-2)}¤ =2'5 øπ(-2-4)¤ +π{5-(-1)}¤ =6'2
AB”=øπ(-1-2)¤ +π(0-3)¤ =3'2 BC”=øπ{2-(-1)}¤ +π(-3-0)¤ =3'2 CA”=øπ(2-2)¤ +π{3-(-3)}¤ =6
AB”=BC” AB”¤ +BC”¤ =CA”¤ ABC B=90˘
AB”=3'2 BC”=3'2 CA”=6 B=90˘
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-1
AB”=øπ(-4-1)¤ +π{3-(-2)}¤ =5'2 BC”=øπ{0-(-4)}¤ +π(-5-3)¤ =4'5 CA”=øπ(1-0)¤ +π{-2-(-5)}¤ ='∂10 AB” ¤ +CA” ¤ <BC” ¤ ABC A>90˘
AB”=5'2 BC”=4'5 CA”='∂10 A>90˘
-2
AB”=øπ(-1-3)¤ +π(2-0)¤ =2'5 BC”=øπ{-3-(-1)}¤ +π(-2-2)¤ =2'5 CA”=øπ{3-(-3)}¤ +π{0-(-2)}¤ =2'∂10
AB”=BC” AB” ¤ +BC” ¤ =CA” ¤ ABC B=90˘
ABC=;2!;_2'5_2'5=10
01
ABD AB” : BD”=1 : '2 6 : BD”=1 : '2 BD”=6'2(cm)
BCD BC” : BD”='3 : 2 BC” : 6'2='3 : 2 2BC”=6'6
BC”=3'6(cm) 3'6 cm
02
ABC AB” : AC”=1 : '2 5'3 : AC”=1 : '2 AC”=5'6
ABD AB” : AD”='3 : 2 5'3 : AD”='3 : 2 '3 AD”=10'3
AD”=10
ABD AB” : BD”='3 : 1
5'3 : BD”='3 : 1 '3 BD”=5'3 BD”=5 ABC AB” : BC”=1 : 1
BC”=AB”=5'3
CD”=BC”-BD”=5'3-5
03
A
BC” H
AHC
AC” : AH”=2 : 1
6'6 : AH”=2 : 1 2AH”=6'6 AH”=3'6(cm)
ABH AB” : AH”='2 : 1
AB” : 3'6='2 : 1 AB”=6'3(cm)
04
=45˘
ABC AB” : BC”=1 : '2
AB” : 4'2=1 : '2 '2 AB”=4'2 AB”=4(cm)
2_4+4'2=8+4'2(cm) (8+4'2) cm
05 (3 -1)
øπ{3-(-2)}¤ +π(-1-2)¤ ='∂34 øπ{3-(-1)}¤ +π{-1-(-4)}¤ =5 øπ{3-(-1)}¤ +π(-1-3)¤ =4'2 øπ(3-0)¤ +π(-1-2)¤ =3'2 øπ(3-2)¤ +π(-1-5)¤ ='∂37
(3 -1)
06
PQ”=øπ(a-2)¤ +(-1-2)¤ =3'5 (a-2)¤ +(-3)¤ =(3'5)¤
a¤ -4a-32=0 (a+4)(a-8)=0
a=-4 a=8
Q 3 a<0
a=-4 -4
07
y=x¤ -6x+7=(x-3)¤ -2 P(3 -2) y=x¤ -6x+7 x=0
y=7 Q(0 7)
PQ”=øπ(0-3)¤ +π{7-(-2)}¤ =3'∂10
45˘ 4'2 cm A B
C
360˘
8
45˘
45˘
30˘
60˘ 6'6 cm A
B H C
55
013'6 cm 02 03
04(8+4'2) cm 05 06-4
07 08
배운대로해결하기
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본 교 재 08
AB”=øπ{1-(-3)}¤ +π(1-3)¤ =2'5 BC”=øπ(2-1)¤ +(3-1)¤ ='5 CA”=øπ(-3-2)¤ +(3-3)¤ =5
AB”” ¤ +BC” ¤ =CA” ¤ ABC B=90˘
ABC=;2!;_'5_2'5=5
직육면체의 대각선의 길이
개념
17
1 5 cm 5'2 cm 2 3'∂10 cm 8'3 cm 3 7 cm '∂15 cm
개념콕콕 56
1
FGH FH”="√4¤ +3¤ =5(cm) DFH DF”="√5¤ +5¤ =5'2(cm)
2
"√7¤ +5¤ +4¤ =3'∂10(cm) '3_8=8'3(cm) 3
"√2¤ +3¤ +6¤ =7(cm) '3_'5='∂15(cm)
대표 유형
-1 -2(15+5'5) cm
'6 cm -13'3 cm -2
57
BF”=x cm øπ5¤ +3¤ +x¤ =7'2 34+x¤ =98 x¤ =64 x=8 ( x>0)
BF”=8(cm) -1
øπx¤ +x¤ +2¤ =6 2x¤ +4=36 x¤ =16 x=4 ( x>0)
-2
FGH FH”=øπ6¤ +8¤ =10(cm) DFH DF”=øπ10¤ +5¤ =5'5(cm)
DFH
5'5+10+5=15+5'5(cm) (15+5'5) cm
x cm '3x=3'2 x='6
'6 cm '6 cm
-1
x cm '3x=9 x=3'3
3'3 cm 3'3 cm
-2
x cm '3x=4'6 x=4'2
(4'2)‹ =128'2(cm‹ )
2
DM”= _3= (cm) D’H”=;3@;DM”=;3@;_ ='3(cm) A’H”=ø∑3¤ -('3)¤ ='6(cm)
BCD= _3¤ = (cm¤ )
=;3!;_ _'6=9'2(cm‹ ) 4
9'3 4
9'3 4 '3
4
3'3 2 3'3
2 '3
2
정사면체의 높이와 부피
개념
18
1 3'3 3'3 2'3 2'3 2'6 2'6 18'2
2 cm '3 cm '6 cm cm¤
9'2 cm‹
4
9'3 4 3'3
2 '3
2
개념콕콕 58
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정사각뿔의 높이와 부피
개념
19
1'2 6'2 6'2 3'2 3'2 3'7 3'7 36'7 2 2'2 cm '2 cm '6 cm 4'6 cm‹
3
개념콕콕 60
2
AC”='2_2=2'2(cm)
A’H”=;2!;AC”=;2!;_2'2='2(cm) O’H”=øπ(2'2)¤ -('2)¤ ='6(cm)
=;3!;_2¤ _'6=4'6(cm‹ ) 3
AC”='2_4=4'2(cm)
AH”=;2!; AC”=;2!;_4'2=2'2(cm) OAH OH”=øπ6¤ -(2'2)¤ =2'7(cm)
;3!;_4¤ _2'7= (cm‹ ) 2'7 cm cm‹
-1
AC”='2_6=6'2(cm)
AH”=;2!;AC”=;2!;_6'2=3'2(cm)
OAH O’H”=øπ(5'2)¤ -(3'2)¤ =4'2(cm)
;3!;_6¤ _4'2=48'2(cm‹ ) 4'2 cm 48'2 cm‹‹
-2
OAH AH”=øπ8¤ -(2'6)¤ =2'∂10(cm) AC”=2AH”=2_2'∂10=4'∂10(cm)
x cm '2x=4'∂10 x=4'5
;3!;_(4'5 )¤ _2'6=160'6(cm‹ ) 3
32'7 3 32'7
3 대표 유형
2'7 cm cm‹
-1 4'2 cm 48'2 cm‹ -2
-1 32'7
3
61
= _12=4'6(cm)
= _12‹ =144'2(cm‹ )
4'6 cm 144'2 cm‹
-1
= _2'3=2'2(cm)
= _(2'3)‹ =2'6(cm‹ )
2'2 cm 2'6 cm‹
-2
x cm
x=6 x=3'6
_(3'6)‹ =27'3(cm‹ )
BCD DM”= _3'2= (cm)
MH”=;3!;DM”=;3!;_ = (cm) AH”= _3'2=2'3(cm)
AMH=;2!;_ _2'3= (cm¤ )
-1
BCD BM”= _4'6=6'2(cm) BH”=;3@; BM”=;3@;_6'2=4'2(cm)
AH”= _4'6=8(cm)
ABH=;2!;_4'2_8=16'2(cm¤ ) 16'2 cm¤
'6 3
'3 2
3'2 2 '6
2 '6
3
'6 2 3'6
2
3'6 2 '3
2 '2
12 '6 3
'2 12 '6 3 '2 12 '6 3 대표 유형
4'6 cm 144'2 cm‹
-1 2'2 cm 2'6 cm‹ -2
-116'2 cm¤
59
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본 교 재 AC”='2_12=12'2(cm)
CH”=;2!; AC”=;2!;_12'2=6'2(cm) OHC O’H”=øπ11¤ -(6'2)¤ =7(cm) OHC=;2!;_6'2_7=21'2(cm¤ )
-1
AC”='2_3'2=6(cm) AH”=;2!;AC”=;2!;_6=3(cm)
OAH O’H”=øπ9¤ -3¤ =6'2(cm) OAH=;2!;_3_6'2=9'2(cm¤ )
01 AD”=x cm
øπx¤ +12¤ +3¤ =13 x¤ +153=169 x¤ =16 x=4 ( x>0)
A’D”=4(cm)
02 DH”=x cm
øπ8¤ +6¤ +x¤ =10'2 100+x¤ =200 x¤ =100 x=10 ( x>0)
FH”="√8¤ +6¤ =10(cm)
BFHD=10_10=100(cm¤ ) 100 cm¤
03
EG”
EG”='2_4=4'2(cm) AG”="√4¤ +4¤ +7¤ =9(cm)
AEG AE”_EG”=AG”_E’I’
7_4'2=9_E’I’’
E’I’= (cm)
28'2 cm 9 28'2
9
4 cm 4 cm
7 cm A
B
H C
D
E
F G
I 62
01 02100 cm¤ 03 cm04
0554'6 cm‹ 06 0732'2 cm¤ 0836'3 cm‹
28'2 9 배운대로해결하기
04
x cm '3x=3'6 x=3'2
FH”='2_3'2=6(cm) DFH=;2!;_6_3'2=9'2(cm¤ )
05
x cm x=6'2 x=6'3
_(6'3)‹ =54'6(cm‹ ) 54'6 cm‹
06
DM”= _9= (cm) AH”= _9=3'6(cm)
DH”=;3@;DM”=;3@;_ =3'3(cm) AHD=;2!;_3'3_3'6= (cm¤ )
=4 ABC=4_{ _9¤ }=81'3(cm¤ )
= _9‹ = (cm‹ )
07
AC”='2_8=8'2(cm)
AH”=;2!; AC”=;2!;_8'2=4'2(cm)
OAH OH”=øπ(4'6)¤ -(4'2)¤ =8(cm)
OAC=;2!;_8'2_8=32'2(cm¤ ) 32'2 cm¤
08
AC”='2_6=6'2(cm)
AH”=;2!;AC”=;2!;_6'2=3'2(cm) OAH
OH”=øπ(3'5)¤ -(3'2)¤ =3'3(cm)
;3!;_6¤ _3'3=36'3(cm‹ ) 36'3 cm‹
A B
H D C
O
6 cm
6 cm 3'5 cm
243'2 4 '2
12
'3 4
27'2 2 9'3
2 '6
3
9'3 2 '3
2 '2
12 '6 3
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원뿔의 높이와 부피
개념
20
1 4'3 cm pcm‹
2 12 cm 18'2p cm‹
3 6 2 4'2 2 4'2 16'2p
3 64'3
3
개념콕콕 63
1
=ø∑8¤ -4¤ =4'3(cm)
=;3!;_p_4¤ _4'3= p(cm‹ ) 2
=ø∑13¤ -5¤ =12(cm)
=øπ6¤ -(3'2)¤ =3'2(cm)
=;3!;_p_(3'2)¤ _3'2=18'2p(cm‹ ) 3
6 cm
2 cm h cm
64'3 3
대표 유형
-118'2p cm‹ -2 pcm‹ -196p cm‹ -2 4'∂21
3
64
r cm
2pr=12p r=6 h cm h="√8¤ -6¤ =2'7
;3!;_p_6¤ _2'7=24'7p(cm‹ ) -1
r cm
pr¤ =9p r¤ =9 r=3 ( r>0) h cm
h="√9¤ -3¤ =6'2
h cm
r cm 9 cm h cm
r cm 8 cm
;3!;_p_3¤ _6'2=18'2p(cm‹ ) 18'2p cm‹
-2
AB” : OB”=2 : 1
6 : OB”=2 : 1 2O’B”=6 OB”=3(cm) O’A” : OB”='3 : 1
O’A” : 3='3 : 1 O’A”=3'3(cm)
;3!;_p_3¤ _3'3=9'3p(cm‹ )
AB” : OB”=2 : 1
6 : OB”=2 : 1 2OB”=6 OB”=3(cm) O’A”=øπ6¤ -3¤ =3'3(cm)
;3!;_p_3¤ _3'3=9'3p(cm‹ )
h cm h="√5¤ -2¤ ='∂21
;3!;_p_2¤ _'∂21= p(cm‹ )
pcm‹
-1
h cm h="√10¤ -6¤ =8
;3!;_p_6¤ _8=96p(cm‹ ) 96p cm‹
-2
r cm 2p_12_;3!6@0);=2pr r=4
h cm h="√12¤ -4¤ =8'2
;3!;_p_4¤ _8'2=128'2p(cm‹ ) 3
12 cm
4 cm h cm
6 cm 10 cm h cm
4'∂21 3 4'∂21
3
5 cm
2 cm h cm
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본 교 재 입체도형에서의 최단 거리
개념
21
1 6 5 5'5
2 2'5p
개념콕콕 65
1
2
A’B'”=ø∑(4p)¤ +∑(2p)¤ =2'5p
A B
A' B'
4p
2p A
B
H
C D
4 G
5
6
대표 유형
5'5 cm -14'5 cm -2
13p cm -1 -2 8'2 cm
66
BH”
BH”=øπ(7+3)¤ +5¤
=5'5(cm) 5'5 cm
-1
AH”
AH”=øπ4¤ +(2+4+2)¤
=4'5(cm)
4'5 cm 4 cm
4 cm 2 cm
2 cm A
B
H C D
E
F G
7 cm
5 cm
3 cm
B C D
F G H
-2
DC”
DC”=øπ(4+8)¤ +12¤ =12'2(cm)
2p_6=12p(cm)
A’B'”
A’B'”=øπ(12p)¤ +(5p)¤
=13p(cm) 13p cm
-1
r cm
pr¤ =50p r¤ =50 r=5'2 ( r>0)
2p_5'2=10'2p(cm)
A’B'”
A’B'”=øπ(10'2p)¤ +π(10p)¤
=10'3p(cm) -2
x˘
2p_8_ =2p_2 x=90
ABB' A=90˘
B’B'”
B’B'””='2_8=8'2(cm) 8'2 cm
x 360
x˘
A
B B'
8 cm
2 cm A
B
A' B'
10'2p cm
10p cm
A A'
B B'
5p cm
12p cm
A B C
D E F
12 cm
4 cm 8 cm
67
01 024'5 cm 03 04324p cm‹
05 066p cm 07 089'3 cm
배운대로해결하기
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01
r cm
2pr=12p r=6
"√12¤ -6¤ =6'3(cm)
02
OHB OH”="√5¤ -4¤ =3(cm) O’A”=OB”=5(cm)
A’H”=5+3=8(cm)
AHB AB”="√8¤ +4¤ =4'5(cm)
4'5 cm 4'5 cm
03
l 1
h cm h=øπ(6'2)¤ -3¤ =3'7
;3!;_p_3¤ _3'7=9'7p(cm‹ )
04
r cm
2p_15_ =2pr r=9
h cm
h="√15¤ -9¤ =12
;3!;_p_9¤ _12=324p(cm‹ ) 324p cm‹
05
MH”=;2!; DH”=;2!;_12=6(cm)
FM”
FM”="√(7+2)¤ +6¤ =3'∂13(cm)
7 cm
6 cm
2 cm B
H C D
M
F G
9 cm 15 cm h cm
216 360
h cm l
3 cm 6'2 cm
12 cm
r cm
06
2p_4=8p(cm)
h cm h=øπ(10p)¤ -(8p)¤ =6p(cm)
6p cm
6p cm
07
x˘
2p_12_ =2p_3 x=90
ABM A=90˘
B’M”
B’M”=øπ12¤ +6¤ =6'5(cm) 08
ACDB AD” CB”
AD” CB” H
ACD CH”= _9= (cm)
CB”
CB”=2CH”=2_ 9'3 =9'3(cm) 9'3 cm 2
9'3 2 '3
2
A
H B
C
D 9 cm
x 360
x˘
A
3 cm
B B'
12 cm 6 cmM
A A'
B' B
10p cm
8p cm
h cm
68 70
01 024'∂10 03 0412'3 cm¤
05 06 07(6+2'3) cm
08 cm0914'3 cm¤ 10 11
12 1350'6 cm¤ 14 15
16(48+48'7) cm¤ 17
18 pcm‹ 19192p cm¤ 20
214'3 2210'2 23 2415p cm
250'2 3 10'3
3
개념 넓히기로마무리
01
3k cm 2k cm(k>0) øπ(3k)¤ +(2k)¤ =4'∂13 '∂13k=4'∂13 k=4
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본 교 재 3k=3_4=12(cm)
02
x
øπ(3x)¤ +x¤ =8'5 '∂10x=8'5 x=4'2
AC”=øπ(8'2)¤ +(4'2)¤ =4'∂10 4'∂10 03
5_2=10(cm) x cm
'2x=10 x=5'2 (5'2)¤ =50(cm¤ ) 04
G ABC
AD”=;2#; AG”=;2#;_4=6(cm) yy
ABC x cm
x=6 x=4'3 yy
ABC= _(4'3)¤ =12'3(cm¤ ) yy 12'3 cm¤
05
6 cm 6
6_{ _6¤ }=54'3(cm¤ )
06
A BC”
H BH”=x cm CH”=(7-x) cm
ABH
A’’H” ¤ =6¤ -x¤ y AHC
A’’H” ¤ =8¤ -(7-x)¤ y 6¤ -x¤ =8¤ -(7-x)¤
14x=21 x=;2#;
ABH A’H”=æ≠6¤ -{;2#;}¤ = (cm)
07
ABH AH” : BH”=1 : 1 BH”=AH”=6(cm)
3'∂15 2
A
B C
H 7 cm
6 cm 8 cm
x cm
'3 4
'3 4 '3
2
AHC A’H” : CH”='3 : 1
6 : CH”='3 : 1 '3 CH”=6 CH”=2'3(cm)
BC”=6+2'3(cm) (6+2'3) cm
08
ABC AB” : AC”=1 : 2
AB” : 10=1 : 2 2AB”=10 AB”=5(cm) BAC=180˘-(90˘+30˘)=60˘
BAD= DAC
BAD=;2!; BAC=;2!;_60˘=30˘
ABD AB” : AD”='3 : 2 5 : AD”='3 : 2 '3AD”=10
AD”= (cm) cm
09
D BC”
H DHC
DH” : CD”='3 : 2
DH” : 4='3 : 2 2DH”=4'3
DH”=2'3(cm) yy
CH” : CD”=1 : 2
CH” : 4=1 : 2 2 CH”=4 CH”=2(cm)
AD”=BH”=8-2=6(cm)ˇ yy
ABCD=;2!;_(6+8)_2'3=14'3(cm¤ )ˇ yy 14'3 cm¤
10
P(-2 1) Q(3 4) x
(a 0)
øπ{a-(-2)}¤ +π(0-1)¤ =øπ(a-3)¤ +(0-4)¤
(a+2)¤ +1=(a-3)¤ +16 10a=20 a=2
P Q x x 2
11
AB”=øπ{2-(-1)}¤ +π(1-3)¤ ='∂13 CA”=øπ(-1-4)¤ +π(3-4)¤ ='∂26
BC”=øπ(4-2)¤ +π(4-1)¤ ='∂13 AB”=BC”
AB”=BC” AB”¤ +BC”¤ =CA”¤ ABC B=90˘
ABC=;2!;_'∂13_'∂13=;;¡2£;;
60˘
8 cm H
4 cm A
B C
D 10'3
3 10'3
3
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12
2k 4k 5k(k>0) øπ(2k)¤ +(4k)¤ π+(5k)¤ =6'5 45k¤ =180
k¤ =4 k=2 ( k>0) 4 8 10 4_8_10=320
13
FN”=ND”=D’M”=MF”
MFND
DF”='3_10=10'3(cm) MN”=EG”='2_10=10'2(cm)
MFND=;2!;_10'3_10'2
=50'6(cm¤ ) 50'6 cm¤
NFG NDC
FG”=DC” NG”=NC” FGN= DCN NFG NDC (SAS )
NFG NDC MDA MFE
FN”=DN”=DM”=FM”
MFND 14
x
AF”='2x AFH
_('2x)¤ =24'3 x¤ =48 x=4'3 ( x>0)
'3_4'3=12 15
DM”=;2#; DH”=;2#;_2'2=3'2(cm) x cm x=3'2 x=2'6
_(2'6)‹ =8'3(cm‹ )
16
O BC”
E
HE”=;2!; AB”=;2!;_4'3=2'3(cm) OHE
OE”=øπ(6'2)¤ +(2'3)¤ =2'∂21(cm) 4'3 cm
6'2 cm
E
A B
H D C
O
4'3 cm
'2 12 '3 2 '3
4
A B
N H
M C
D
E F 10 cm G
(4'3)¤ +4_{;2!;_4'3_2'∂21}=48+48'7(cm¤ )
(48+48'7) cm¤
17
A
BCDE H
BD”='2_5'2=10(cm) BH”=;2!;BD”=;2!;_10=5(cm)
ABH
AH”=øπ(5'2)¤ -5¤ =5(cm)
2_[;3!;_(5'2)¤ _5]=;;∞;3);º;;(cm‹ )
=2_
18
r cm
2pr=10p r=5 yy
l cm
p_5_l=75p l=15 yy
h cm h="√15¤ -5¤
=10'2 yy
;3!;_p_5¤ _10'2= p(cm‹ ) yy
pcm‹
19
r cm r="√16¤ -8¤ =8'3
p_(8'3)¤ =192p(cm¤ ) 192p cm¤
R O
d
r r="√R¤ -d¤
R O d r 250'2
3 250'2
3
15 cm
5 cm h cm
5'2 cm A
B H
C D E
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본 교 재 20
BD”
BD”=øπ6¤ +(3+4+3)¤
=2'∂34(cm)
21
ABC= ABP+ APC _8¤ =;2!;_8_PQ”+;2!;_8_PR”
16'3=4PQ”+4PR” 16'3=4(PQ”+PR”)
PQ”+PR”=4'3 4'3
22
D BC”
D' AP”+PD”=AP”+P’D'”
æA’D'”
A’’D'” AP”+PD”
AA'D'
A’’D'”=øπ(3+7)¤ +10¤ =10'2
AP”+PD” 10'2 10'2
23
C’M” DM” ABC ABD
C’M”=D’M”= _12=6'3(cm)
M CD”
H CHM
MH”=øπ(6'3)¤ -6¤ =6'2(cm) CDM=;2!;_12_6'2
=36'2(cm¤ )
24
2p_3=6p(cm)
A’B"”
A’B"”=øπ(6p+6p)¤ +(9p)¤
=15p(cm)
15p cm 6p cm 6p cm
9p cm
A A' A"
B B' B"
H M
A
B
C D
12 cm
'3 2
A 3
7
7 7 B
10 10
P
A' D'
C D
'3 4
H G
A
B C
D E
F 6 cm
3 cm
3 cm
4 cm 개념