Download Answers for the lesson “Find Segment Lengths in Circles”

Answers for the lesson “Find Segment
Lengths in Circles”
LESSON
10.6
Skill Practice
21. Statements (Reasons)
1. external segment
2. A tangent segment intersects
the circle in only one point while
the secant segment intersects the
circle in two points.
3. 5
4. 23
5. 4
6. 4
7. 6
8. 5
9. 12
10. 36
11. 4
12. The wrong segment lengths are
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being multiplied together;
FD + CF 5 AF + BF,
4 + CF 5 3 + 8, CF 5 6,
CD 1 DF 5 CF,
CD 1 4 5 6, CD 5 2.
13. 5
16. D
14. 30
15. 1
17. 18
18. 17.7
}
19. 2Ï 10
Problem Solving
20. about 108 ft
1. Two intersecting chords in the
same circle.
(Given)
2. Draw }
AC and }
BD.
(Two points determine
a line.)
3. ŽACD > ŽABD,
ŽCAB > ŽCDB
(If two
inscribed angles of
a circle intercept the
same arc, then the
angles are congruent.)
4. nAEC , nDEB
(AA Similarity Postulate)
EA
EC
5. }
5}
EB
ED
(If two triangles
are similar, the ratios of
corresponding sides
are equal.)
6. EA + EB 5 EC + ED
(Cross Products Property)
AD + (AD 1 DE)
22. BC 5 }} 2 AB
AB
Geometry
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316
23.
AC , which implies
mŽD 5 }2mC
1
A
x
E
y
r
C
r
P
r
D
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} is a tangent segment to
Given: EA
} is a secant segment of
circle P, ED
} is a diameter of
circle P, and CD
circle P. Prove: EA2 5 EC ? ED.
} > AP
},
By Theorem 10.1, EA
so nEAP is a right triangle.
By the Pythagorean Theorem,
(y 1 r)2 = x2 1 r2. So,
y2 1 2yr 1 r2 5 x2 1 r2.
By the Subtraction Property
of Equality, y2 1 2yr 5 x2.
Factoring this, y(y 1 2r) 5 x2,
or EC + ED 5 EA2.
24. 4 cm/sec; by the Segments of
Chords Theorem, 6 + 8 5 4 + CN,
so CN 5 12 cm. It takes 3 sec for
sparkles to move the 6 cm from C
to D, so the sparkles need to travel
12 cm in 3 sec, or 4 cm/sec.
25. Given: }
EB and }
ED are secant
segments. Draw }
AD and }
BC.
Using the Measure of an
Inscribed Angle Theorem,
mŽB 5 }
mC
AC and
2
1
ŽB > ŽD. Using the
Reflexive Property of Angle
Congruence, ŽE > ŽE.
Using the AA Similarity
Postulate, nBCE , nDAE.
Using corresponding sides of
similar triangles are proportional,
EA
EC
ED
EB
} 5 }. Cross multiplying
gives EA + EB 5 EC + ED.
26. Given: A secant segment and a
tangent segment sharing an
endpoint outside of a circle.
Draw }
AC and }
AD. ŽADC is
inscribed, therefore
mŽADC 5 }2 mC
AC .
1
ŽCAE is formed by a secant
and a tangent, therefore
mŽCAE 5 }2 mC
AC . This
1
implies ŽADC > ŽCAE.
ŽE > ŽE by the Reflexive
Property of Angle Congruence,
therefore n AEC , n DEA using
the AA Similarity Postulate.
Using corresponding sides of
similar triangles are proportional,
EA
EC
ED
EA
} 5 }. Cross multiplying
gives EA2 5 EC + ED.
Geometry
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317
27. a. 608
b. Using the Vertical Angles
Theorem, ŽACB > ŽFCE.
Since mŽCAB 5 608 and
mŽEFD 5 608,
then ŽCAB > ŽEFD.
Using the AA Similarity
Postulate, n ABC , n FEC.
y
x 1 10
x 1 10
c. } 5 }; y 5 }
6
2
3
d. y 2 5 x(x 1 16)
e. 2, 6
2
CE
f. Since } 5 }, let CE 5 2x
1
CB
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and CB 5 x. Using Theorem
10.14, 2x2 5 60 which
}
implies x 5 Ï30 which
}
implies CE 5 2Ï 30 .
}
25Ï 29
28. }
29
Geometry
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318