Download Document

Answers for 5.2
For use with pages 306–309
5.2 Skill Practice
19. Sample:
1. circumcenter
2. all points on the perpendicular
bisector of }
AB
3. 15
4. 30
5. 55
6. yes
7. yes
8. no
9. B
10. You don’t know that EC 5 DC.
11. 35
12. 43
13. 50
14. 50
15. Yes; the Converse of the
Perpendicular Bisector Theorem
‹]›
guarantees L is on JP
.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
16. 9
17. 11
18. Sample answer: In the
20. Sometimes; a scalene triangle can
be acute, right, or obtuse.
21. Always; congruent sides
are created.
22. Sometimes; consider an
equilateral triangle and a
scalene triangle.
construction of the segment
bisector four congruent triangles
are created. In the process four
pairs of congruent angles are
formed which are right angles
making the bisector perpendicular
to the segment.
Geometry
Answer Transparencies for Checking Homework
ngat-0500.indd 6
A6
3/27/06 1:53:14 PM
Answers for 5.2
For use with pages 306–309
continued
26. Statements (Reasons)
23. a– c.
Statements (Reasons)
1. P is a perpendicular bisector
of }
XZ at Y; W and V lie in
plane P.
(Given)
2. AP 5 BP, m∠CPA 5 m∠CPB
5 908
(Definition of
perpendicular bisector)
2. XW 5 ZW, XV 5 ZV
(Perpendicular Bisector
Theorem)
AP > }
BP
(Definition of
3. }
segment congruence)
XW > }
ZW, }
XV > }
ZV
3. }
(Definition of segment
congruence)
4. ∠CPA > ∠CPB (Definition of
angle congruence)
WV > }
WV (Reflexive Property
4. }
of Segment Congruence)
5. nVXW > nVZW
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
‹]›
is the perpendicular
1. PC
bisector of }
AB.
(Given)
(SSS)
6. ∠VXW > ∠VZW (Corr. Parts
s are >.)
of > n
5.2 Problem Solving
CP > }
CP (Reflexive Property
5. }
of Segment Congruence)
6. nCPA > nCPB
CA > }
CB
7. }
(SAS)
(Corr. parts
s are >.)
of > n
(Definition of
8. CA 5 CB
segment congruence)
24. 59.6 m, 195.5 m; Perpendicular
Bisector Theorem
25. Theorem 5.4 shows you that
you can find a point equidistant
from three points by using the
perpendicular bisectors of the
sides of the triangle formed by
the three points.
Geometry
Answer Transparencies for Checking Homework
ngat-0500.indd 7
A7
3/27/06 1:53:15 PM
Answers for 5.2
For use with pages 306–309
continued
27. Statements (Reasons)
30. Midpoint of the hypotenuse.
(Given)
1. CA 5 CB
‹]› }
⊥ AB through
2. Draw PC
point C.
(Perpendicular
Postulate)
3. }
CA > }
CB
(Definition of
segment congruence)
CP > }
CP (Reflexive Property
4. }
of Segment Congruence)
5. ∠CPA and ∠CPB right angles.
(Definition of
perpendicular lines)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
6. nCPA and nCPB are right
triangles.
(Definition of
right triangle)
7. nCPA > nCPB
PA > }
PB
8. }
(HL)
(Corr. parts
s are >.)
of > n
9. C is on the perpendicular
bisector of }
AB. (Definition of
perpendicular bisector)
28. a. Find the intersection of the
perpendicular bisectors of
the triangle formed by the
three points.
b. approximately (7, 6.5)
29. Check students’ work;
Sample answer: Right triangle
with vertices A(2a, 0), B(0, 2b),
and C(0, 0); the midpoint of
}
AC is (a, 0) and midpoint of }
BC
is (0, b). The equation of the
perpendicular bisectors of }
AC
}
and BC are x 5 a and y 5 b.
These two lines intersect in the
point (a, b) which is the midpoint
of }
AB.
31. Case 1:
Given: D, E, and B are collinear.
AB > }
BC
Prove: }
Statements (Reasons)
1. D, E, and B are collinear;
}
AD > }
CD, }
AE > }
CE. (Given)
‹]›
containing point E.
2. Draw DB
(Two points determine a line.)
DE > }
DE (Reflexive Property
3. }
of Segment Congruence)
4. nDAE > nDCE
(SSS)
5. ∠AED > ∠CED (Corr. parts
s are >.)
of > n
6. ∠AED and ∠AEB, ∠CED
and ∠CEB are linear pairs.
(Definition of linear pair)
Perpendicular Bisector Theorem.
Geometry
Answer Transparencies for Checking Homework
ngat-0500.indd 8
A8
3/27/06 1:53:16 PM
Answers for 5.2
For use with pages 306–309
continued
31. (cont.)
7. ∠AED and ∠AEB are
supplementary, ∠CED and
∠CEB are supplementary.
(Linear Pair Postulate)
8. ∠AEB > ∠CEB (Congruent
Supplements Theorem)
EB > }
EB (Reflexive Property
9. }
of Segment Congruence)
10. nAEB > nCEB
AB > }
BC
11. }
(SAS)
(Corr. parts
s are >.)
of > n
AB > }
BC
Case 2: Given: }
Prove: D, E, and B are collinear.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Statements (Reasons)
DE, }
EB, and }
AC.
1. Draw }
(Two point determine a line.)
AB > }
BC, }
AD > }
CD,
2. }
}
}
AE > CE
(Given)
3. AB 5 CB, AD 5 CD,
AE 5 CE
(Definition of
congruent segments)
4. B is on the perpendicular
bisector of }
AC, D is on the
perpendicular bisector of }
AC,
E is on the perpendicular
bisector of }
AC. (Converse of
the Perpendicular
Bisector Theorem)
5. There exists only one line that
is the perpendicular bisector
of }
AC, so B, D, and C are all
on the same line. (Through a
point not on a line, there exists
only one line through the
point perpendicular to the
given line.)
6. D, E, and B are collinear.
(Definition of collinear)
32. Statements (Reasons)
1. PQRST is a regular polygon,
}
SV > }
RV.
(Given)
TP > }
QP, }
TS > }
QR (Definition
2. }
of regular polygon)
PW > }
PW, }
PV > }
PV
3. }
(Reflexive Property of
Segment Congruence)
4. VPTS > VPQR (Definition of
congruent polygons)
5. ∠TPW > ∠QPW (Corr. parts
of > polygons are >.)
6. nPWT > nPWQ
TW > }
QW
7. }
(SAS)
(Corr. parts
s are >.)
of > n
8. ∠PWT > ∠PWQ are a linear
pair. (Definition of linear pair)
Geometry
Answer Transparencies for Checking Homework
ngat-0500.indd 9
A9
3/27/06 1:53:17 PM
Answers for 5.2
For use with pages 000–000
continued
32. (cont.)
41. Starting with 3, zero is added to
PV ⊥ }
TQ (If two lines intersect
9. }
to form a linear pair of
congruent angles, the
lines are perpendicular.)
get 3, then one is added to get 4,
then 2 is added to get 6, next
add 3; 9.
PV is the perpendicular
10. }
bisector of }
TQ. (Definition of
perpendicular bisector)
33. No; unless the four points
determine a rectangle there is no
single point to locate the school
so that it is equidistant from the
four towns.
Cedar Falls
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Shady Hills
Lake City
Willow Valley
5.2 Mixed Review
34. 612
35. 68
37. 18; 1588
}
36. 62Ï 7
38. 9; 908
39. The numbers are decreasing
by 5; 1.
40. The next number in the sequence
is the previous one multiplied by
3; 162.
Geometry
Answer Transparencies for Checking Homework
ngat-0500.indd 10
A10
3/27/06 1:53:19 PM