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Answers to Exercises
CHAPTER 3 • CHAPTER
3
CHAPTER 3 • CHAPTER
LESSON 3.1
1.
B
A
D
C
F
E
2.
AB
9. For Exercise 7, trace the triangle. For Exercise 8,
trace the segment onto three separate pieces of
patty paper. Lay them on top of each other, and
slide them around until the segments join at the
endpoints and form a triangle.
. Copy Q and
10. One method: Draw DU
construct COY QUD. Duplicate DUA
at point O. Construct OYP UDA.
CD
U
O
A
AB ⫹ CD
3.
AB
EF
EF
Q
AB ⫹ 2EF ⫺ CD
CD
Answers to Exercises
D
C
Y
11. One construction method is to create congruent
circles that pass through each other’s center. One
side of the triangle is between the centers of the
circles; the other sides meet where the circles
intersect.
12. a 50°, b 130°, c 50°, d 130°, e 50°,
f 50°, g 130°, h 130°, k 155°, l 90°,
m 115°, n 65°
13. west
14. An isosceles triangle is a triangle that has at
least one line of reflectional symmetry.Yes, all
equilateral triangles are isosceles.
15.
4.
5. possible answer:
L
G
P
E
copy
6. m3 m1 m2; possible answer:
⬔2
16. new coordinates: A(0, 0), Y(4, 0), D(0, 2)
y
1
3
2
⬔1
6
Y
7. possible answer:
D'
Y'
C
–6
AC
D A' A
6
x
BC
–6
A
B
AB
8.
17. Methods will vary. It isn’t possible to draw a
second triangle with the same side lengths that is
not congruent to the first.
11 cm
8 cm
10 cm
32
ANSWERS TO EXERCISES
LESSON 3.2
1.
A
B
2.
Q
3.
D
Edge of the paper
Original segment
4.
D
1 CD
2
1 CD
2
AB
AB
I
1 CD
2
2AB – 1 CD
2
5.
A
AB
M
L
CD
N
MN = 1 (AB + CD)
2
8. The medians all intersect in one point.
C
N
6. Exercises 1–5 with patty paper:
Exercise 1 This is the same as Investigation 1.
Exercise 2
.
Step 1 Draw a segment on patty paper. Label it QD
Step 2 Fold your patty paper so that endpoints Q
and D coincide. Crease along the fold.
Step 3 Unfold and draw a line in the crease.
Step 4 Label the point of intersection A.
Step 5 Fold your patty paper so that endpoints Q
and A coincide. Crease along the fold.
Step 6 Unfold and draw a line in the crease.
Step 7 Label the point of intersection B.
Step 8 Fold your patty paper so that endpoints A
and D coincide. Crease along the fold.
Step 9 Unfold and draw a line in the crease.
Step 10 Label the point of intersection C.
A
M
L
B
appears to be parallel to EF
, and its length
9. GH
is half the length of EF .
F
G
D
H
E
ANSWERS TO EXERCISES
33
Answers to Exercises
C
Exercise 3 This is the same as Investigation 1.
Exercise 4
Step 1 Do Investigation 1 to get 12CD.
Step 2 On a second piece of patty paper, trace AB
two times so that the two segments form a segment
of length 2AB.
Step 3 Lay the first piece of patty paper on top of
the second so that the endpoints coincide and the
shorter segment is on top of the longer segment.
Step 4 Trace the rest of the longer segment with a
different colored writing utensil. That will be the
answer.
Exercise 5
Step 1 Trace segments AB and CD so that the two
segments form a segment of length AB CD.
Step 2 Fold your patty paper so that points A and
D coincide. Crease along the fold.
Step 3 Unfold and draw a line in the crease.
7. The perpendicular bisectors all intersect in one
point.
10. The quadrilateral appears to be a rhombus.
V
E
O
C
R
D
14. One way to balance it is along the median. The
two halves weigh the same.
sample figure:
C
S
I
11.
D
Ness Station
Umsar
Station
A
Answers to Exercises
Any point on the perpendicular bisector of the
segment connecting the two offices would be
equidistant from the two post offices. Therefore,
any point on one side of the perpendicular bisector
would be closer to the post office on that side.
12. It is a parallelogram.
F
L
Area CDB = 158 in2
Area DAB = 158 in2
B
Ruler
15. F
16. E
17. B
18. A
19. D
20. C
21. B, C, D, E, H, I, O, X (K in some fonts, though
not this one)
22. Methods will vary.
C
T
10 cm
A
40⬚
13. The triangles are not necessarily congruent,
but their areas are equal. A cardboard triangle
would balance on its median.
34
ANSWERS TO EXERCISES
A
70⬚
B
It is not possible to draw a second triangle with the
same angle and side measures that is not congruent
to the first.
LESSON 3.3
1.
6. The two folds are parallel.
P
B
P
Q
I
G
The answer depends on the angle drawn and where
P is placed.
C
2.
D
A
B
7. Fold the patty paper through the point so that
two perpendiculars coincide to see the side closest
to the point. Fold again using the perpendicular
of the side closest to the point and the third
perpendicular; compare those sides.
8. Draw a line. Mark two points on it, and label
them A and C. Construct a perpendicular at C.
congruent to CA
. The altitude CD
is
Mark off CB
also the angle bisector, median, and perpendicular
bisector.
A
D
C
B
9.
U
O
B
T
O
T
Answers to Exercises
3.
B
U
10.
E
Two altitudes fall outside the triangle, and one falls
inside.
4. From the point, swing arcs on the line to
construct a segment whose midpoint is that point.
Then construct the perpendicular bisector of the
segment.
L
A
B
11.
12.
Complement
of ⬔A
⬔A
5. Construct perpendiculars from point Q and
and RE
congruent to QR
.
point R. Mark off QS
Connect points S and E.
Q
R
S
E
ANSWERS TO EXERCISES
35
13. See table below.
14.
18. not congruent
6 cm
6 cm
40⬚
40⬚
8 cm
15.
F
8 cm
19. possible answer:
Y
T
I
16.
A
E
20. possible answer:
C
D
5 cm
F
9 cm
B
5 cm
Answers to Exercises
17.
9 cm
13. (Lesson 3.3)
Rectangular pattern with triangles
Rectangle
1
2
3
4
5
6
…
Number of shaded
triangles
2
9
20
35
54
77
…
n
…
35
…
2484
(2n 1)(n 1)
36
ANSWERS TO EXERCISES
LESSON 3.4
1.
2.
3.
4.
5.
6.
D
F
A
C
E
11.
12. The angle bisectors are perpendicular. The
sum of the measures of the linear pair is 180°. The
sum of half of each must be 90°.
z
z
z
7.
A
B
M
R
P
8.
P
A
P
E
M
S
M
M
E
U
M
O
S
9a, b.
18.
G
45⬚
90⬚ 45⬚
N
9c.
I
A
S
L
135⬚
O
T
45⬚
10.
Altitude
Angle
bisector
Median
ANSWERS TO EXERCISES
37
Answers to Exercises
A
R
R
13. If a point is equally distant from the sides of
an angle, then it is on the bisector of an angle. This
is true for points in the same plane as the angle.
Mosaic answers: Square pattern constructions:
perpendiculars, equal segments, and midpoints;
The triangles are not identical, as the downward
ones have longer bases.
14. y 110°
15. mR 46°
16. Angle A is the largest; mA 66°,
mB 64°, mC 50°.
17. STOP
19.
22a. A web of lines fills most of the plane, except
a U-shaped region and a V-shaped region. (The
U-shaped region is actually bounded by a section
were
of a parabola and straight lines. If AB
extended to AB , the U would be a complete
parabola.)
B
5.6 cm
130⬚
A 3.5 cm C
20.
A
A
B
6.5 cm
C
21. No, the triangles don’t look the same.
60⬚
40⬚
60⬚
40⬚
8 cm
Answers to Exercises
8 cm
38
ANSWERS TO EXERCISES
D
B C
Parabola
and half the
22b. a line segment parallel to AB
length (The segment is actually the midsegment
of ABD.)
LESSON 3.5
1.
2.
3. Construct a segment with length z. Bisect the
segment to get length 2z. Bisect again to get a
segment with length 4z. Construct a square with
each side of length 4z.
8. 1 S and 2 U by of the Alternate
Interior Angles Conjecture
9. The ratios appear to be the same.
10. 1 3 and 2 4 by the
Corresponding Angles Conjecture
11. A parallelogram
12. Using the Converse of the Parallel Lines
Conjecture, the angle bisectors are parallel:
BC
.
DAB ABC, so AD
D
B
A
1z
4
13. Construct the perpendicular bisector of each
of the three segments connecting the fire stations.
1z
2
z
x
A
x
x
A
Eliminate the rays beyond where the bisectors
intersect. A point within any region will be closest
to the fire station in that region.
O
B
M
14. Z
15.
x
x
5. sample construction:
R
P
T
D
A
16.
I
R
R
60⬚
O
E
W
T
K
R
perpendicular to
6. Draw a line and construct ML
it. Swing an arc from point M to point G so that
MG RA. From point G, swing an arc to construct
. Finish the parallelogram by swinging an arc of
RG
length RA from R and swinging an arc of length GR
from M. There is only one possible parallelogram.
M
G
L
A
R
RC = KE = 8 cm
C
17. a 72°, b 108°, c 108°, d 108°, e 72°,
f 108°, g 108°, h 72°, j 90°, k 18°, l 90°,
m 54°, n 62°, p 62°, q 59°, r 118°;
Explanations will vary. Sample explanation:
c is 108° because of the Corresponding Angles
Conjecture. Using the Vertical Angles Conjecture,
2m 108°, so m 54°. p n because of the
Corresponding Angles Conjecture. Using the
Linear Pair Conjecture, n 62°, so p 62°.
Using the Linear Pair Conjecture, r p 180°.
Because p 62°, r 118°.
7. 1 S, 2 U
ANSWERS TO EXERCISES
39
Answers to Exercises
4.
C
USING YOUR ALGEBRA SKILLS 3
Answers to Exercises
1. perpendicular
2. neither
3. perpendicular
4. parallel
5. possible answer: (2, 5) and (7, 11)
6. possible answer: (1, 5) and (2, 12)
7. Ordinary; no two slopes are the same, so no
EM
because their
sides are parallel, although TE
slopes are opposite reciprocals.
8. Ordinary, for the same reason as in Exercise 7—
none of the sides are quite parallel.
RO
9. trapezoid: KC
61;
10a. Slope HA slope ND
slope NA
6. Quadrilateral HAND
slope HD
is a rectangle because opposite sides are parallel
and adjacent sides are perpendicular.
40
ANSWERS TO EXERCISES
midpoint AD
21, 3. The
10b. Midpoint HN
diagonals of a rectangle bisect each other.
11a. Yes, the diagonals are perpendicular.
1; slope VR
1.
Slope OE
midpoint OE
(2, 4).
11b. Midpoint VR
The diagonals of OVER bisect each other.
11c. OVER appears to be a rhombus. Slope
slope RE
51 and slope OR
slope
OV
VE 5, so opposite sides are parallel. Also, all of
the sides appear to have the same length.
12a. Both slopes equal 21.
12b. The segments are not parallel because they
are coincident.
12c. distinct
13. (3, 6)
LESSON 3.6
1. Sample description: Construct one of the
segments, and mark arcs of the correct length from
the endpoints. Draw sides to where those arcs meet.
M
angle at each end of that segment congruent to one
of the angles in the book. Where they meet is the
third vertex of the triangle.
5.
C
A
A
B
C
A
M
A
S
S
A
M
S
2.
O
D
B
Sample description: Construct A and mark off
the distance AB. From B swing an arc of length BC
to intersect the other side of A at two points.
Each gives a different triangle.
C
6.
A
y ⫺x
____
2
O
O
O
T
x
T
y
T
Sample description: Construct O. Mark off
distances OD and OT on the sides of the angle.
Connect D and T.
I
3.
Sample description: Mark the distance y, mark
back the distance x, and bisect the remaining
length of y x. Using an arc of that length, mark
arcs on the ends of segment x. The point where
they intersect is the vertex angle of the triangle.
7.
I
Y
G
Y
I
Y
. Construct I at
Sample description: Construct IY
I and Y at Y. Label the intersection of the rays
point G.
4. Yes, students’ constructions must be either
larger or smaller than the triangle in the book.
Sample description: Draw an angle. Mark off equal
segments on the sides of the angle. Use a different
compass setting to draw intersecting arcs from the
ends of those segments.
8. Sample description: Draw an angle and mark
off unequal distances on the sides. At the endpoint
of the longer segment (not the angle vertex), swing
an arc with the length of the shorter segment. From
the endpoint of the shorter segment, swing an arc
the length of the longer segment. Connect the
endpoints of the segments to the intersection
points of the arcs to form a quadrilateral.
Sample description: Draw one side with a different
length than the lengths in the book. Duplicate an
ANSWERS TO EXERCISES
41
Answers to Exercises
D
y ⫺x
____
2
x
12. new coordinates: E(4, 6), A(7, 0), T(1, 2)
9.
y
–5
Sample description: Draw a segment and draw an
angle at one end of the segment. Mark off a
distance equal to that segment on the other side of
the angle. Draw an angle at that point and mark off
the same distance. Connect that point to the other
end of the original segment.
10.
E
T'
A'
–5
A
–5 T
E'
13.
Reflectional
symmetries
Rotational
symmetries
Trapezoid
0
0
Kite
1
0
Parallelogram
0
2
Rhombus
2
2
Rectangle
2
2
Answers to Exercises
Figure
Sample description: Draw an angle and mark off
equal lengths on the two sides. Use that length to
determine another point that distance from the
points on the sides. Connect that point with the
two points on the side of the angle.
11. Answers will vary. The angle bisector lies
between the median and the altitude. The order of
the points is either M, R, S or S, R, M. One possible
conjecture: In a scalene obtuse triangle the angle
bisector is always between the median and the
altitude.
m⬔ABC = 111⬚
14. half a cylinder
15. 503
16.
110⬚
E
R
B
3.2 cm
R
A
S
Altitude
42
Median
M
Angle
bisector
ANSWERS TO EXERCISES
110⬚
C
x
A
5.5 cm
110⬚
C
LESSON 3.7
1. incenter
Because the station needs to be equidistant from
the paths, it will need to be on each of the angle
bisectors.
2. circumcenter
3. incenter
8. Yes, any circle with a larger radius would not
fit within the triangle. To get a circle with a larger
radius tangent to two of the sides would force the
circle to pass through the third side twice.
9. No, on an obtuse triangle the circle with the
largest side of the triangle as the diameter of the
circle creates the smallest circular region that
contains the triangle. The circumscribed circle of
an acute triangle does create the smallest circular
region that contains the triangle.
Stove
Fridge
Sink
To find the point equidistant from three points,
find the circumcenter of the triangle with those
points as vertices.
5. Circumcenter. Find the perpendicular bisectors
of two of the sides of the triangle formed by the
classes. Locate the pie table where these two lines
intersect.
6.
10. For an acute triangle, the circumcenter is
inside the triangle; for an obtuse triangle, the
circumcenter is outside the triangle. The
circumcenter of a right triangle lies on the
midpoint of the hypotenuse.
11. For an acute triangle, the orthocenter is inside
the triangle; for an obtuse triangle, the orthocenter
is outside the triangle. The orthocenter of a right
triangle lies on the vertex of the right angle.
12. The midsegment appears parallel to side MA
and half the length.
A
S
M
7.
H
T
13. The base angles of the isosceles trapezoid
appear congruent.
A
T
O
M
ANSWERS TO EXERCISES
43
Answers to Exercises
The center of the circular sink must be equidistant
from the three counter edges, that is, the incenter of
the triangle.
4. circumcenter
14. The measure of A is 90°. The angle inscribed
in a semicircle appears to be a right angle.
21.
Y
40⬚ 40⬚
A
4.8 cm
6.4 cm
M
T
K
15. The two diagonals appear to be perpendicular
bisectors of each other.
A
T
E
22. construction of an angle bisector
T
16.
y
9
23. construction of a perpendicular line through a
point on a line
Answers to Exercises
x+y=9
9
x
17.
Construct the incenter by bisecting the two angles
shown. Any other point on the angle bisector of the
third angle must be equidistant from the two
unfinished sides. From the incenter, make congruent
arcs that intersect the unfinished sides. The
intersection points are equidistant from the incenter.
Use two congruent arcs to find another point that
is equidistant from the two points you just
constructed. The line that connects this point and
the incenter is the angle bisector of the third angle.
18. Answers should describe the process of
discovering that the midpoints of the altitudes are
collinear for an isosceles right triangle.
19. a triangle
M
20.
6.0 cm
R
6.0 cm
60⬚ 6.0 cm 60⬚
O
60⬚
60⬚
6.0 cm
6.0 cm
H
44
ANSWERS TO EXERCISES
24. construction of a line parallel to a given line
through a point not on the line
25. construction of an equilateral triangle
26. construction of a perpendicular bisector
LESSON 3.8
1. The center of gravity is the centroid. She needs
to locate the incenter to create the largest circle
within the triangle.
2. AM 20; SM 7; TM 14; UM 8
3. BG 24; IG 12 4. RH 42; TE 45
5. The points of concurrency are the same point
for equilateral triangles because the segments are
the same.
9. circumcenter
10. The shortest chord through P is a segment
perpendicular to the diameter through P, which is
the longest chord through P.
O
P
11.
A
B'
C
6.
B
12. rule: 2n 2, possible answer:
H
Centroid
Incenter
7. ortho-/in-/centroid/circum-; the order changes
when the angle becomes larger than 60°. The
points become one when the triangle is equilateral.
Orthocenter
Centroid
Circumcenter
8. Start by constructing a quadrilateral, then make
a copy of it. Draw a diagonal in one, and draw a
different diagonal in the second. Find the centroid
of each of the four triangles. Construct a segment
connecting the two centroids in each quadrilateral.
Place the two quadrilaterals on top of each other
matching the congruent segments and angles.Where
the two segments connecting centroids intersect is
the centroid of the quadrilateral.
C
D
H
C
C
H
H
H
C
C
H
H
H
C
C
C
H
H
H
H
14. a 128°, b 52°, c 128°, d 128°,
e 52°, f 128°, g 52°, h 38°,
k 52°, m 38°, n 71°, p 38°
15. Construct altitudes from the two accessible
vertices to construct the orthocenter. Through the
orthocenter, construct a line perpendicular to the
southern boundary of the property. This method
will divide the property equally only if the southern
boundary is the base of an isosceles triangle.
Altitude to
missing vertex
16. 1580 greetings
C⬘
D⬘
M4
M2
M3
M1
A
H
C
13.
Orthocenter
Incenter
H
B A⬘
B⬘
ANSWERS TO EXERCISES
45
Answers to Exercises
Circumcenter
CHAPTER 3 REVIEW
1. False; a geometric construction uses a straightedge and a compass.
2. False; a diagonal connects two non-consecutive
vertices.
3. true
4. true
5. false
B
A
28.
_1 z
2
y
y
Segment
29.
5x
P
C
D
R
3x
6. False; the lines can’t be a given distance from a
segment because the segment has finite length and
the lines are infinite.
7. false
C
x
_1 z
2
4x
Q
30. mA mD.You must first find B.
mB 180° 2(mA).
2y
B
Answers to Exercises
A
D
B
8. true
9. true
10. False; the orthocenter does not always lie
inside the triangle.
11. A
12. B or K 13. I
14. H
15. G
16. D
17. J
18. C
19.
20.
2y
⬔D
⬔B
⬔A
A
31.
21.
4x
A
Copy
B
y
22.
y
D
F
32.
23. Construct a 90° angle and bisect it twice.
24.
I
y
25. incenter
26. Dakota Davis should locate the circumcenter
of the triangular region formed by the three stones,
which is the location equidistant from the stones.
B
27.
A
46
z
ANSWERS TO EXERCISES
C
R
x
T
33. rotational symmetry
34. neither
35. both
36. reflectional symmetry
37. D
38. A
39. C
40. B
41. False; an isosceles triangle has two congruent
sides.
42. true
43. False; any non-acute triangle is a
counterexample.
44. False; possible explanation: The orthocenter is
the point of intersection of the three altitudes.
45. true
46. False; any linear pair of angles is a
counterexample.
47. False; each side is adjacent to one congruent
side and one noncongruent side, so two consecutive
sides may not be congruent.
48. false;
58c. no
59.
Q
Q'
49. False; the measure of an arc is equal to the
measure of its central angle.
50. false; TD 2DR
51. False; a radius is not a chord.
52. true
53. False; inductive reasoning is the process of
observing data, recognizing patterns, and making
generalizations about those patterns.
54. paradox
55a. 2 and 6 or 3 and 5
55b. 1 and 5
55c. 138°
56. 55
57. possible answer:
58a. yes
58b. If the month has 31 days, then the month is
October.
60. (Chapter 3 Review)
n
f(n)
1
2
3
4
5
6
…
1
2
5
8
11
14
…
n
…
20
…
56
f(n) 3n 4
61. (Chapter 3 Review)
n
1
2
3
4
5
6
…
f(n)
0
3
8
15
24
35
…
n
f(n) n2
…
20
…
399
1
ANSWERS TO EXERCISES
47
Answers to Exercises
60. See table below.
61. See table below.
62. a 38°, b 38°, c 142°, d 38°, e 50°,
f 65°, g 106°, h 74°.
Possible explanation: The angle with measure c is
congruent to an angle with measure 142° because
of the Corresponding Angles Conjecture, so
c 142°. The angle with measure 130° is
congruent to the bisected angle by the
Corresponding Angles Conjecture. The angle with
measure f has half the measure of the bisected
angle, so f 65°.
63. Triangles will vary. Check that the triangle is
scalene and that at least two angle bisectors have
been constructed.
64. mFAD 30° so mADC 30°, but
its vertical angle has measure 26°. This is a
contradiction.
65. minimum: 101 regions by 100 parallel lines;
maximum: 5051 regions by 100 intersecting,
noncurrent lines