home
***
CD-ROM
|
disk
|
FTP
|
other
***
search
/
Multimedia Geometry
/
geometry-3.5.iso
/
GEOMETRY
/
CHAPTER5.6Y
< prev
next >
Wrap
Text File
|
1995-04-22
|
5KB
|
196 lines
à 5.6èSimilar Triangles
äèPlease answer ê followïg questions about similar
triangles.
âè
èèèè If two angles ç one triangle are equal ë two anglesè
èèèè ç anoêr triangle, ên ê triangles are similar.
éS1
Defïition 5.6.1èSIMILAR TRIANGLES:èTwo triangles are similar if cor-
respondïg angles are equal, å correspondïg sides are proportional.
èèèèèèèèèèèèèèèèèèèèIn this figure ╬A ╧ ╬E, ╬B ╧ ╬H,
èèèèèèèèèèèèèèèèèèèè╬C ╧ ╬C, å
èèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèèABè BCè ACè
èèèèèèèèèèèèèèèèèèèè╓╓ = ╓╓ = ╓╓
èèèèèèèèèèèèèèèèèèèèEHè HCè EC.èThus, ΦABC is
@fig5601.BMP,55,130,147,74èèèèèèèèèèèèèèsimilar ë ΦEHC.
è We will use ê symbol "~" ë mean "is similar ë."èAccordïg ë
this defïiiën, ï order ë show that two triangles are similar, you
must show correspondïg angles are equal, å correspondïg sides are
proportional.èThe followïg axiom allows you ë show that two triangles
are similar by simply showïg ê angles are equal.
Axiom 20èTwo triangles are similar if three angles ç one triangle are
equal ë ê three correspondïg angles ç ê oêr triangle (AAA).
è The followïg êorem furêr reduces ê amount ç work required ë
show that two triangles are similar.
Theorem 5.6.1èTwo triangles are similar if two angles ç one triangle
are equal ë ê two correspondïg angles ç ê oêr triangle (AA).
Proç:èFor a proç please see Problem 1.
è The next two êorems provide two additional ways ë show that two
triangles are similar.
Theorem 5.6.2èTwo triangles are similar if one pair ç correspondïg
angles are equal, å ê ïcludïg sides are proportional (SAS).
Theorem 5.6.3èTwo triangles are similar if ê three sides ç one tri-
angle are proportional ë ê correspondïg three sides ç ê oêr
triangle (SSS).
Theorem 5.6.4èThe areas ç two similar triangles are proportional ë
ê squares ç two correspondïg sides.
Theorem 5.6.5èSimilarity ç triangles is reflexive, transitive, å
symmetric.
1èèè If ╬A ╧ ╬E å ╬B ╧ ╬H, can you show ΦABC ~ ΦEHP?
èèèèèèèèèèèèèèèèèèèèè A) Yesèèè B) No
@fig5602.BMP,35,40,147,74è
ü Show ΦABC ~ ΦEHP
Proç: StatementèèèèèèèèèèèReason
èèè 1. ╬A ╧ ╬E, ╬B ╧ ╬Hèèèèèè1. Given
èèè 2. ╬C ╧ ╬Pèèèèèèèèèè 2. Theorem 3.5.4
èèè 3. ΦABC ~ ΦEHPèèèèèèèè 3. (20)AAA axiom
Ç A
2èèèIs êre sufficient ïformation ë show ΦABC ~ ΦEHC?
èèèèèèèèèèèèèèèèèèèèè A) Yesèèè B) No
@fig5603.BMP,35,40,147,74è
ü Show ΦABC ~ ΦEHC
Proç: StatementèèèèèèèèèèèèèReason
èèè 1. ╬ABC, ╬EHC are right ╬sèèèè 1. Given
èèè 2. ╬ABC ╧ ╬EHCèèèèèèèèèè 2. (14)Right ╬s are congruent
èèè 3. ╬BCA ╧ ╬HCEèèèèèèèèèè 3. Vertical angles
èèè 4. ΦABC ~ ΦEHCèèèèèèèèèè 4. Similar by AA
Ç A
3èèèèIf AB = 8, EH = 6, å BC = 12, fïd HC.
èèèèèèèèèèèèèèèèèèèèA) 9èè B) 10èè C) 12
@fig5601.BMP,35,40,147,74è
üèèèèèèèèèèèèèHCè EH
èèèèèèèèèèèèèèèè╓╓ = ╓╓
èèèèèèèèèèèèèèèèBCè AB
èèèèèèèèèèèèèèèèHCè 6
èèèèèèèèèèèèèèèè╓╓ = ╓
èèèèèèèèèèèèèèèè12è 8
èèèèèèèèèèèèèèèèHC = 12·6/8
èèèèèèèèèèèèèèèèHC = 9
Ç A
4èèèèIf AB = 8, EH = 6, å EC = 12, fïd AC.
èèèèèèèèèèèèèèèèèèèèA) 18èè B) 14èè C) 16
@fig5601.BMP,35,40,147,74è
üèèèèèèèèèèèèèACè AB
èèèèèèèèèèèèèèèè╓╓ = ╓╓
èèèèèèèèèèèèèèèèECè EH
èèèèèèèèèèèèèèèèACè 8
èèèèèèèèèèèèèèèè╓╓ = ╓
èèèèèèèèèèèèèèèè12è 6
èèèèèèèèèèèèèèèèAC = 12·8/6
èèèèèèèèèèèèèèèèAC = 16
Ç C
5èèèèèè If AB = 8, EH = 6, å ê area çΦABC is 56,
èèèèèèèèèèèfïd ê area ç ΦEHC.
èèèèèèèèèèèèèèèèèèèA) 34èè B) 63/2èè C) 32
@fig5601.BMP,35,40,147,74è
üèèèèèèèèèèèarea ΦEHCè (EH)ìèèèèè
èèèèèèèèèèèèèè╓╓╓╓╓╓╓╓╓ = ╓╓╓╓╓
èèèèèèèèèèèèèèarea ΦABCè (AB)ì
èèèèèèèèèèèèèèarea ΦEHCè (6)ì
èèèèèèèèèèèèèè╓╓╓╓╓╓╓╓╓ = ╓╓╓╓
èèèèèèèèèèèèèèèè56èèè(8)ì
èèèèèèèèèèèèèèarea ΦEHC = 56·(36)/64
èèèèèèèèèèèèèèarea ΦEHC = 63/2
Ç B
6èèèèèè If m╬B = 105 å m╬H = 105, is êre sufficient
èèèèèèèèèèèèèè ïformation ë show ΦABC ~ ΦEHC?
èèèèèèèèèèèèèèèèèèèèA) YesèèèèB) No
@fig5601.BMP,35,40,147,74è
ü Show ΦABC ~ ΦEHCèèèèèèèèè
Proç: Statementèèèèèèèèèèèè Reason
èèè 1. m╬B = 105, m╬H = 105èèèèè 1. Given
èèè 2. ╬B ╧ ╬Hèèèèèèèèèèèè2. Defïition ç congruence
èèè 3. ╬C ╧ ╬Cèèèèèèèèèèèè3. Congruence is reflexive
èèè 4. ΦABC ~ ΦEHCèèèèèèèèèè4. Similar by AA
Ç A
7èèèèèèIf BC = 10, HC = 6, AC = 15, å EC = 9, is êreè
èèèèèèèèèèèèè sufficient ïformation ë show ΦABC ~ ΦEHC?
èèèèèèèèèèèèèèèèèèèèA) YesèèèèB) No
@fig5601.BMP,35,40,147,74è
ü Show ΦABC ~ ΦEHCèèèèèèèèè
Proç: Statementèèèèèèèèè Reason
èèè 1. BCè 10è 5èèèèèèè1. Given
èèèèè╓╓ = ╓╓ = ╓
èèèèèHCèè6è 3
èèè 2. ACè 15è 5èèèèèèè2. Given
èèèèè╓╓ = ╓╓ = ╓
èèèèèECèè9è 3
èèè 3. BCè ACèèèèèèèèè3. Transitive axiom
èèèèè╓╓ = ╓╓
èèèèèHCè EC
èèè 4. ╬C ╧ ╬Cèèèèèèèèè4. Congruence is reflexive
èèè 5. ΦABC ~ ΦEHCèèèèèèè5. Similar by SAS
Ç A
8èèèèèèèèIf AB = 6, BC = 8, AC = 9, EH = 12, HP = 16,è
èèèèèèèèèèèè å EP = 18, can you show ΦABC ~ ΦEHP?
èèèèèèèèèèèèèèèèèèèèA) YesèèèèB) No
@fig5602.BMP,35,40,147,74è
ü Show ΦABC ~ ΦEHPèèèèèèèèè
Proç: Statementèèèèèèèèè Reason
èèè 1. ABèè6è 1èèèèèèè1. Given
èèèèè╓╓ = ╓╓ = ╓
èèèèèEHè 12è 2
èèè 2. BCèè8è 1èèèèèèè2. Given
èèèèè╓╓ = ╓╓ = ╓
èèèèèHPè 16è 2
èèè 3. ACèè9è 1èèèèèèè3. Given
èèèèè╓╓ = ╓╓ = ╓
èèèèèEPè 18è 2
èèè 4. ABè BCè ACèèèèèè 4. Transitive axiom
èèèèè╓╓ = ╓╓ = ╓╓
èèèèèEHè HPè EP
èèè 5. ΦABC ~ ΦEHPèèèèèèè5. Similar by SSS
Ç A
èèè