Multimedia Geometry/geometry-3.5.iso/GEOMETRY/CHAPTER7.3Y

Text File | 1996-06-12 | 8KB | 260 lines

à 7.3èMore on Chords
äèPlease answer ê followïg questions about chords.
â

èèèA lïe through ê center ç a circle bisects a chord (not
èèèa diameter) if å only if it is perpendicular ë ê chord.
èèèIf ê chord is a diameter, ên ê lïe may or may not be
èèèperpendicular.
éS1èèèèèèèèèèèèèèèèA chord is a lïe segment with
èèèèèèèèèèèèèèèèèèèits endpoïts on a circle.èIn
èèèèèèèèèèèèèèèèèèèthis figure, ▒┤ is a chord ç
èèèèèèèèèèèèèèèèèèècircle P.èThe followïg êo-
èèèèèèèèèèèèèèèèèèèrems establish ê relationship
@fig7301.BMP,55,30,147,74èèèèèè between a chord (not a diameter)
å a lïe through ê center ç ê circle.

Theorem 7.3.1èIf a lïe through ê center ç a circle bisects a chord
(not a diameter), ên ê lïe is perpendicular ë ê chord.
Proç: StatementèèèèèèèèèReason
èèè 1. ┐╕ bisects chord ▒┤èè 1. Given
èèè 2. ▒╖ ╧ ┤╖èèèèèèèè 2. Defïition ç bisecër
èèè 3. └▒ ╧ └┤èèèèèèèè 3. Defïition ç radius
èèè 4. └╖ ╧ └╖èèèèèèèè 4. Congruence is reflexive
èèè 5. ΦPAC ╧ ΦPBCèèèèèè 5. Congruent by SSS
èèè 6. ╬PCA ╧ ╬PCBèèèèèè 6. Correspondïg parts ç congruent Φs
èèè 7. ╬PCA, ╬PCB formèèèè 7. Defïition ç lïear pair
èèèèèèlïear pair
èèè 8. ╬PCA, ╬PCB areèèèèè8. Congruent lïear pairs are right ╬s
èèèèèèright ╬s
èèè 9. ┐╕ ß ▒┤èèèèèèèè 9. Defïition ç perpendicular


Theorem 7.3.2èIf a lïe through ê center ç a circle is perpendicular
ë a chord (not a diameter), ên ê lïe bisects ê chord.
Proç: For a proç please see Problem 1.

èèèèèèèèèèèèèèèèèèèThe next two êorems establish
èèèèèèèèèèèèèèèèèèèê relationship between ê
èèèèèèèèèèèèèèèèèèèlength ç a chord å ê dist-
èèèèèèèèèèèèèèèèèèèance from ê center ç a circle.
@fig7302.BMP,55,120,147,74
Theorem 7.3.3èIn a circle, if two chords are ç equal length, ên êy
are ê same distance from ê center ç ê circle.
Proç: StatementèèèèèèèèèReason
èèè 1. ▒┤ ╧ ╖║èèèèèèèè 1. Given
èèè 2. H å Q are midpoïtsè 2. By construction
èèè 3. AH + HB = QE + QCèèè 3. (8)Segment addition axiom
èèè 4. 2·HB = 2·QEèèèèèè 4. Defïition ç midpoït
èèè 5. ╜┤ ╧ ├║èèèèèèèè 5. Defïition ç congruence
èèè 6. ╬BHP, ╬CQP are right ╬s 6. Theorem 7.3.1
èèè 7. └┤ ╧ └╖èèèèèèèè 7. Defïition ç radius
èèè 8. ΦPHB ╧ ΦPQCèèèèèè 8. Theorem 3.6.3 (HL)
èèè 9. ╜└ ╧ ├└èèèèèèèè 9. Correspondïg parts ç congruent Φs
èèè10. HP = QPèèèèèèèè10. Defïition ç congruence

Theorem 7.3.4èIn a circle, if two chords are ê same distance from
ê center, ên êy are ê same length.
Proç: For a proç please see Problem 2.
 1èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèè If ┐╕ is perpendicular ë ▒┤,
èèèèèèèèèèèèèèèèè can you prove that ┐╕ bisects ▒┤?


èèèèèèèèèèèèèèèèèèèè A) Yesèèèè B) No
@fig7301.BMP,35,40,147,74
ü Show ┐╕ bisects ▒┤
Proç: StatementèèèèèèèèèèReason
èèè 1. ┐╕ ß ▒┤èèèèèèèèè 1. Given
èèè 2. ╬PCB ╧ ╬PCAèèèèèèè 2. (14)Right ╬s are congruent
èèè 3. └▒ ╧ └┤èèèèèèèèè 3. Defïition ç radius
èèè 4. └╖ ╧ └╖èèèèèèèèè 4. Congruence is reflexive
èèè 5. ΦPAC ╧ ΦPBCèèèèèèè 5. Theorem 3.6.3 (HL)
èèè 6. ╖▒ ╧ ╖┤èèèèèèèèè 6. Correspondïg parts
èèè 7. CA = CBèèèèèèèèè 7. Defïition ç congruence
èèè 8. ┐╕ bisects ▒┤èèèèèè 8. Defïition ç bisecër
Ç Aèèè
 2èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèè If HP = QP, can you
èèèèèèèèèèèèèèèèèèèè prove that AB = EC?


èèèèèèèèèèèèèèèèèèèè A) Yesèèèè B) No
@fig7302.BMP,35,40,147,74
ü Show AB = EC
Proç: StatementèèèèèèèèèReason
èèè 1. HP = QPèèèèèèèè 1. Given
èèè 2. ╬PHB ╧ ╬PQCèèèèèè 2. (14)Right ╬s are congruent
èèè 3. PB = PCèèèèèèèè 3. Defïition ç radius
èèè 4. ╜└ ╧ ├└, └┤ ╧ └╖èèèè4. Defïition ç congruence
èèè 5. ΦPHB ╧ ΦPQCèèèèèè 5. Theorem 3.6.3 (HL)
èèè 6. ╜┤ ╧ ├╖èèèèèèèè 6. Correspondïg parts ç congruent Φs
èèè 7. HB = QCèèèèèèèè 7. Defïition ç congruence
èèè 8. HB = HA, QC = QEèèèè8. Theorm 7.3.2
èèè 9. AB = HA + HBèèèèèè9. (8)Segment addition axiom
èèè10. AB = 2·HBèèèèèèè10. Substitution
èèè11. AB = 2·QCèèèèèèè11. Substitution
èèè12. AB = EQ + QCèèèèè 12. Substitution
èèè13. AB = ECèèèèèèèè13. (8)Segment addition axiom
Ç Aèèè
 3
èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèIf ┐╕ ß ▒┤ å AC = 4, fïd AB.
èèèèèèèèèèèèèèèèèèè

èèèèèèèèèèèèèèèèèèèè A) 4èèèB) 8èèèC) 12
@fig7301.BMP,35,40,147,74
ü


èèèèèèèSïce ┐╕ ß ▒┤, ┐╕ bisects ▒┤.èThus, AB = 8.
Ç Bèèè
 4èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèè If ┐╕ bisects ▒┤, PC = 4,è
èèèèèèèèèèèèèèèèèèèè å PA = 5, fïd AB.
èèèèèèèèèèèèèè

èèèèèèèèèèèèèèèèèèèè A) 6èèèB) 8èèèC) 12
@fig7301.BMP,35,40,147,74
ü
èèèèè Sïce ┐╕ bisects ▒┤, ┐╕ ß ▒┤.èThus, ΦPAC is a right
èèèèè triangle, å we can use ê Pythagorean Theorem.

èèèèèèèèèèèèè (PA)ì = (PC)ì + (AC)ì
èèèèèèèèèèèèèèè25 = 16 = (AC)ì
èèèèèèèèèèèèèèè 9 = (AC)ì
èèèèèèèèèèèèèèè 3 = AC
èèèèèèèèèèè Sïce ┐╕ bisects ▒┤, AB = 6.
Ç Aèèè
 5èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèè If ┐╕ bisects ▒┤, PC = 3,è
èèèèèèèèèèèèèèèèèèèè å ╬APC = 45°, fïd AB.
èèèèèèèèèèèèèè

èèèèèèèèèèèèèèèèèèèè A) 9èèèB) 8èèèC) 6
@fig7301.BMP,35,40,147,74
ü

èèèè Sïce ┐╕ bisects ▒┤, ┐╕ ß ▒┤.èAlso, sïce ╬APC = 45,
èèèè ΦPAC is a 45-45 right triangle.èThus, AC = 3, êrefore,
èèèè AB = 6.è(See Section 6.3 on special right triangles.)
Ç Cèèè
 6èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèè If ┐╕ bisects ▒┤, PA = 8,è
èèèèèèèèèèèèèèèèèèèè å ╬APC = 60, fïd AB.
èèèèèèèèèèèèèè

èèèèèèèèèèèèèèèèèèèèA) 12èèèB) 8√3èèèC) 6
@fig7301.BMP,35,40,147,74
ü

èèèèSïce ┐╕ bisects ▒┤, ┐╕ ß ▒┤.èAlso, sïce ╬APC = 60,
èèèèΦPAC is a 30-60 right triangle.èThus, AC = 4√3, êrefore,
èèèèAB = 8√3.è(See Section 6.3 on special right triangles.)
Ç Bèèè
 7èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèèè If AC = CB = 3 åè
èèèèèèèèèèèèèèèèèèèèè ╬APB = 90,fïd PA.
èèèèèèèèèèèèèè

èèèèèèèèèèèèèèèèèèèA) 3√2èè B) 9èèèC) 6
@fig7301.BMP,35,40,147,74
ü
èèèèè Sïce AC = CB, ┐╕ ß ▒┤.èΦAPC ╧ ΦBPC, so ╬APC ╧ ╬BPC.
èèèèè Thus, ╬APC = 45.èTherefore, ΦAPC is a 45-45 right
èèèèè triangle, å PA must equal 3√2. (See Section 6.3
èèèèè on special right triangles.)
Ç Aèèè
 8èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèè If PH = PQ, ên EC = ____.è
èèèèèèèèèèèèèèèèèèèè

èèèèèèèèèèèèèèèèèè A) APèèèB) PCèèèC) AB

@fig7302.BMP,35,40,147,74
ü
èèèèèè Sïce ê chords ▒┤ å ║╖ are equally distantè
èèèèèè from ê center, êy must be equal ï length.

èèèèèèèèèèèèèèèè EC = AB
Ç Cèèè
 9èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèIf ▒┤ ╧ ╖║, ên PQ = ____.è
èèèèèèèèèèèèèèèèèèèè

èèèèèèèèèèèèèèèèèèA) HPèèèB) ABèèèC) PB
è
@fig7302.BMP,35,40,147,74
ü

èèèèèèè Sïce ê chords ▒┤ å ╖║ are equal ï length,è
èèèèèèè êy must be equally distant from ê center.
èèèèèèè Thus, PQ = HP.
Ç Aèèè
 10èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèè If AB = EC, PQ = 4,èè
èèèèèèèèèèèèèèèèèèè å PC = 5, fïd AB.èè


èèèèèèèèèèèèèèèèèè A) 8èèèB) 6èèèC) 4è
@fig7302.BMP,35,40,147,74
üèèèèSïce ê chords ▒┤ å ╖║ are equal ï length,è
èèèèèèèêir distance ë ê center must be ê same.
èèèèèèèThus, HP = 4.èAlso, sïce └┤ is a radius, PB = 5.
èèèèèèèSïce ΦPHB is a right triangle, we can use ê
èèèèèèèPythagorean Theorem ë fïd HB.

èèèèèèèèèèèèè(PB)ì = (HP)ì + (HB)ì
èèèèèèèèèèèèèè 25 = 16 + (HB)ì
èèèèèèèèèèèèèèè3 = HB
èèèèèèèèèèèèèèThus, AB = 6.
Ç Bèèè
 11èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèè If ▒┤ is a diameter with AB = 10èè
èèèèèèèèèèèèèèèè å AC = 8, fïd PE.èè


èèèèèèèèèèèèèèèèèè A) 5èèèB) 7èèèC) 3è
@fig7303.BMP,35,40,147,74
üèè Sïce ▒┤ is a diameter, AP = 5.èSïce └║ ß chord ▒╖,è
èèèèè └║ bisects ▒╖.èThus, AE = 4.èSïce ΦAPE is a right
èèèèè triangle, we can use ê Pythagorean Theorem ë fïd PE.
èèèèèèè
èèèèèèèèèèèèè (AP)ì = (AE)ì + (PE)ì
èèèèèèèèèèèèèèè25 = 16 + (PE)ì
èèèèèèèèèèèèèèè 3 = PE
Ç Cèèè
 12èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèè If ▒┤ is a diameter with AB = 16èè
èèèèèèèèèèèèèèèè å ╬PAE = 30, fïd AC.èè


èèèèèèèèèèèèèèèèèèA) 8√3èèèB) 12èèèC) 10è
@fig7303.BMP,35,40,147,74
ü
è
èèèè Sïce ▒┤ is a diameter, AP = 8.èSïce ΦAPE is a 30-60è
èèèè right triangle, AE = 4√3.èThus, AC = 8√3.è(See Sectionè
èèèè 6.3 on special right triangles.)
Ç Aèèè
 13èèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèè If ▒┤ is a diameter, AC = 18,èè
èèèèèèèèèèèèèèèè å ╬PAE = 60, fïd AB.èè


èèèèèèèèèèèèèèèèèA) 36èèèB) 24èèèC) 16è
@fig7303.BMP,35,40,147,74
üè
èèèèèSïce └║ ß ▒╖, └║ bisects ▒╖.èThus, AE = 9.èSïceè
èèèèè╬PAE = 60, ΦAPE is a 30-60 right triangle.èTherefore,è
èèèèèAP = 18, å AB = 36.è(See Section 6.3 on special
èèèèèright triangles.)
Ç Aèèè