Download Answers for the lesson “Use Median and Altitude”

Answers for the lesson “Use Median and
Altitude”
LESSON
5.4
11. Z
Skill Practice
1. circumcenter: when it is an
acute triangle, when it is a right
triangle, when it is an obtuse
triangle; incenter: always, never,
never; centroid: always, never,
never; orthocenter: when it is
an acute triangle, when it is a
right triangle, when it is an
obtuse triangle.
C
Y
X
12.
O
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2. Both are perpendicular to a
side of the triangle, although the
altitude contains the vertex
opposite the side while a
perpendicular bisector bisects
the side but does not necessarily
contain the opposite vertex; both
bisect one side of a triangle,
although the perpendicular
bisector does not necessarily
contain the opposite vertex while
the median is not necessarily
perpendicular to the side but does
contain the opposite vertex.
3. 12
4. 9
6. 5
7. D
5. 10
8. a. (8, 1); (5, 1)
b. (5, 21); SQ 5 4 and SR 5 6,
2
therefore SQ 5 }3SR.
9. (3, 2)
10. (3, 4)
B
A
C
13. no; no; yes
14. yes; yes; yes
15. no; yes; no
16. T is the orthocenter, but the
centroid is needed to reach
the conclusion.
17. altitude
18. angle bisector
19. median
20. perpendicular bisector, angle
bisector, median, altitude
21. perpendicular bisector, angle
bisector, median, altitude
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152
22. perpendicular bisector, angle
31.
bisector, median, altitude
O
23. 6, 228; nABD > nCBD by HL,
use corr. parts of > ns are >.
24. 908, 228; nABD > nCBD by
SSS, use definition of a linear pair
and corr. parts of > ns are >.
25. 3
3
27. }
2
26. 2
28. If the base angles of the isosceles
triangle are placed at (2a, 0) and
(0, a), the vertex angle will be on
the y-axis.
C
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29.
O
C
B
A
32. Sample answer: The midpoint of
} is L(3, 7), so the equation of
FG
the median from H(6, 1) to L(3, 7)
is y 5 22x 1 13. P(4, 5) lies on
}
this median. The midpoint of GH
is J(5, 5), so the equation of the
median from F(2, 5) to J(5, 5) is
y 5 5. P(4, 5) lies on this median,
so all three medians intersect at
the centroid.
5
33. }
2
34. 9
}
B
A
O
30.
35. 4
}
9Ï 19 9Ï 19
36. }, }; yes; the height and
2
2
base of both triangles will always
be the same.
B
Problem Solving
37. B; it is the centroid of the triangle.
A
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153
4. ŽADB and ŽCDB are right
angles. (Definition of altitude)
38. Sample:
y
y
(n, h)
(0, h)
x
(2n, 0)
5. nABD and nCDB are right
triangles.
(Definition of
right triangles)
x
(0, 0) (2n, 0)
(n, 0)
39. 6.75 in.2; altitude
40. Right; the orthocenter is on the
41. (0, 2);
y
8
3
y2 4 x 5
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(0, 5)
2
x
3
y3 x
4
2
42. a. Statements (Reasons)
} is an
1. nABC is equilateral, BD
altitude of nABC.
(Given)
2. }
AB > }
BC
8. AD 5 CD
(Definition of
congruent segments)
b. Statements (Reasons)
2. BA 5 BC
(2, 2)
(0, 4)
y1 3x 4
(Corr. parts
s are >.)
of > n
1. nABC is equilateral. (Given)
(4, 2)
(2, 1)
7. }
AD > }
CD
(HL)
9. }
BD is a perpendicular bisector
of }
AC.
(Definition of
perpendicular bisector)
right angle.
(4, 8)
6. nABD > nCBD
(Definition of
equilateral triangle)
3. }
BD > }
BD (Reflexive Property
of Segment Congruence)
3. }
BD > }
AC
(Definition of
equilateral triangle)
(Definition
of altitude)
4. B is on the perpendicular
bisector of }
AC.
(Converse of the Perpendicular
Bisector Theorem)
5. Only one line exists through
B perpendicular to }
AC, so the
perpendicular bisector and the
altitude are contained in the
same line.
(Perpendicular Postulate)
Geometry
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154
43. a. Check students’ work.
b. Statements (Reasons)
b. Their areas are the same.
c. They weigh the same; it means
the weight of nABC is evenly
distributed around its centroid.
2
44. a. 2}
3
1
2
28
3
2
3
b. y 5 }x, x 5 8, y 5 2}x 1 },
(8, 4)
c. Find the equation of each
perpendicular bisector of each
side and solve the system.
45. a. Statements (Reasons)
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1.
}
LP and }
MQ are medians of
]›
scalene nLMN, R is on LP
such that }
LP > }
PR, S is
›
]
on MQ such that }
MQ > }
QS.
(Given)
2. }
MP > }
NP, }
QL > }
QN
(Definition of median)
1. nMPL > nNPR,
nMQL > nSQN
(Exercise 45a)
2. ŽMLP > ŽNRP,
ŽMLQ > ŽSQN (Corr. parts
s are >.)
of > n
} (Converse
} i RN
}, LM
} i SN
3. LM
of Alternate Interior
Angles Theorem)
c. Statements (Reasons)
1. }
LM i }
RN, }
LM i }
SN
(Exercise 45b)
2. There is exactly one line
through N parallel to }
LM, so
‹]›
‹]›
RN and SN are the same line.
(Parallel Postulate)
3. S, N, and R are collinear.
(Definition of collinear points)
3. ŽMPL > ŽNPR,
ŽMQL > ŽSQN
(Vertical
Angles Congruence Theorem)
4. nMPL > nNPR,
nMQL > nSQN
(SAS)
5. }
ML > }
NR, }
ML > }
NS
s are >.)
(Corr. parts of > n
} (Transitive Property
} > NS
6. NR
of Segment Congruence)
Geometry
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155