answers to exercises - cerritos collegeweb.cerritos.edu/imccance/sitepages/worksheets and...

32 ANSWERS TO EXERCISES

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LESSON 3.1

1.

2.

3.

4.

5. possible answer:

6. m�3 � m�1 � m�2; possible answer:

7. possible answer:

8.

A B

C

AC BC

AB

1 2 3�1

�2

copy

EG

L

AB � 2EF � CD

EFEFAB

CD

AB

AB � CD

CD

EF

CD

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.

10. One method: Draw DU��. Copy �Q and

construct �COY � �QUD. Duplicate �DUA

at point O. Construct �OYP � �UDA.

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.

16. new coordinates: A�(0, 0), Y�(4, 0), D�(0, 2)

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

y

xD A

Y

–6

–6 6

6

A'

D'

Y'

UA

DQ

OP

YC

Answers to Exercises

CHAPTER 3 • CHAPTER CHAPTER 3 • CHAPTER3

ANSWERS TO EXERCISES 33

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LESSON 3.2

1.

2.

3.

4.

5.

6. Exercises 1–5 with patty paper:

Exercise 1 This is the same as Investigation 1.

Exercise 2Step 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 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 10 Label the point of intersection C.

AB CD

M N

12

MN = (AB + CD)

C D

CD12

12

CD12

CD12

AB AB

2AB – CD

Edge of the paper

Original segment

Q D

A B

Exercise 3 This is the same as Investigation 1.

Exercise 4Step 1 Do Investigation 1 to get �

12

�CD.

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 5Step 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.


7. The perpendicular bisectors all intersect in one

point.

8. The medians all intersect in one point.

9. GH�� appears to be parallel to EF�, and its length

is half the length of EF�.

D EH

G

F

C

B

N

M

L

A

LA

I

10. The quadrilateral appears to be a rhombus.

11.

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.

13. The triangles are not necessarily congruent,

but their areas are equal. A cardboard triangle

would balance on its median.

F L

AT

Ness Station

Umsar Station

V

R

I

C

E

D

O

S

14. One way to balance it is along the median. The

two halves weigh the same.

sample figure:

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.

It is not possible to draw a second triangle with the

same angle and side measures that is not congruent

to the first.

10 cm

40� 70�A B

C

Area CDB = 158 in2

Area DAB = 158 in2

RulerA

C

D

B


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LESSON 3.3

1.

The answer depends on the angle drawn and where

P is placed.

2.

3.

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.

5. Construct perpendiculars from point Q and

point R. Mark off QS� and RE� congruent to QR��.

Connect points S and E.

ES

RQ

B

O TU

C

AD

B

G

PB

I

6. The two folds are parallel.

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.

Mark off CB� congruent to CA�. The altitude CD�� is

also the angle bisector, median, and perpendicular

bisector.

9.

10.

11.

12.

Complement of �A

�A

A B

E L

T

O B

U

C

A

D

B

P

Q

13. See table below.

14.

15.

16.

17.

C D

E

F

A

B

Y

F

I T

18. not congruent

19. possible answer:


5 cm

9 cm

5 cm

9 cm

40�

8 cm

6 cm

40�

8 cm

6 cm


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Rectangle 1 2 3 4 5 6 … n … 35

Number of shaded 2 9 20 35 54 77 … … 2484triangles

Rectangular pattern with triangles

13. (Lesson 3.3)

(2n � 1)(n � 1)


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LESSON 3.4

1. D

2. F

3. A

4. C

5. E

6.

7.

8.

9a, b.

9c.

10.AnglebisectorAltitude

Median

45�135�

45�45�

90�

M S

U

O

E

M

M

E

S

M

A

A

B

P

PR

M

R AR P

z

z

z

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°.

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. m�R � 46°

16. Angle A is the largest; m�A � 66°,

m�B � 64°, m�C � 50°.

17. STOP

18.G

T

IN

SA

OL

19.

20.

21. No, the triangles don’t look the same.

8 cm 8 cm40� 40�60�

60�

6.5 cmB C

A

3.5 cm

5.6 cm

130�A

B

C

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

of a parabola and straight lines. If AB� were

extended to AB��, the U would be a complete

parabola.)

22b. a line segment parallel to AB� and half the

length (The segment is actually the midsegment

of �ABD.)

A

B

D

C Parabola


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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

�.

4.

5. sample construction:

6. Draw a line and construct ML�� perpendicular to

it. Swing an arc from point M to point G so that

MG � RA. From point G, swing an arc to construct

RG�. Finish the parallelogram by swinging an arc of

length RA from R and swinging an arc of length GR

from M. There is only one possible parallelogram.

7. �1 � �S, �2 � �U

RL

M A

G

R

AP

T

x

x

x

x

x

A

A

z12

z14

z

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:

�DAB � �ABC, so AD�� BC��.

13. Construct the perpendicular bisector of each

of the three segments connecting the fire stations.

Eliminate the rays beyond where the bisectors

intersect. A point within any region will be closest

to the fire station in that region.

14. 15.

16.

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°.

R E

W

RC = KE = 8 cmK C

MB

R O60�

Z

D

O

TR

I

A C

BD

USING YOUR ALGEBRA SKILLS 3

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

sides are parallel, although TE�� EM�� because their

slopes are opposite reciprocals.

8. Ordinary, for the same reason as in Exercise 7—

none of the sides are quite parallel.

9. trapezoid: KC� � RO�

10a. Slope HA�� slope ND�� 16

�;

slope HD�� slope NA�� 6. Quadrilateral HAND

is a rectangle because opposite sides are parallel

and adjacent sides are perpendicular.

10b. Midpoint HN�� midpoint AD�� 12

�, 3�. The

diagonals of a rectangle bisect each other.

11a. Yes, the diagonals are perpendicular.

Slope OE�� 1; slope VR�� 1.

11b. Midpoint VR�� midpoint OE�� (�2, 4).

The diagonals of OVER bisect each other.

11c. OVER appears to be a rhombus. Slope

OV�� slope RE�� 15

� and slope OR�� slope

VE�� 5, so opposite sides are parallel. Also, all of

the sides appear to have the same length.

12a. Both slopes equal �12

�.

12b. The segments are not parallel because they

are coincident.

12c. distinct

13. (3, �6)


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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.

2.

Sample description: Construct �O. Mark off

distances OD and OT on the sides of the angle.

Connect D and T.

3.

Sample description: Construct IY�. Construct �I at

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 one side with a different

length than the lengths in the book. Duplicate an

I Y

Y

GY

I

I

O

D

T

TO

O D

O

M

A

M

S

S

A

M A

S

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.

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.

6.

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.

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.

y

C

A

T

x

xy �x____

2

y �x____2

A

A

B

B

C

C

9.

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.

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.

CA

Altitude

Median

Anglebisector

S MR

B

m�ABC = 111�

12. new coordinates: E�(4, �6), A�(7, 0), T�(1, 2)

13.

14. half a cylinder

15. 503

16.

110� 110�

110�

A 5.5 cm

3.2 cm

C

ER

y

x

E

A

T

–5

–5

–5

A'

E'

T'


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Reflectional RotationalFigure symmetries symmetries

Trapezoid 0 0

Kite 1 0

Parallelogram 0 2

Rhombus 2 2

Rectangle 2 2


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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

The center of the circular sink must be equidistant

from the three counter edges, that is, the incenter of

the triangle.

4. circumcenter

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.

7.

Stove

Fridge Sink

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.

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.

13. The base angles of the isosceles trapezoid

appear congruent.

A O

MT

M H T

S

A

14. The measure of �A is 90°. The angle inscribed

in a semicircle appears to be a right angle.

15. The two diagonals appear to be perpendicular

bisectors of each other.

16.

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

20.

6.0 cm

6.0 cm 6.0 cm

6.0 cm6.0 cm

60� 60�

60�60�

H

M

OR

x

x + y = 9

9

9

y

A

T

A

TM

21.

22. construction of an angle bisector

23. construction of a perpendicular line through a

point on a line

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

40� 40�

4.8 cm

6.4 cm

Y

E

TK


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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.

6.

7. ortho-/in-/centroid/circum-; the order changes

when the angle becomes larger than 60°. The

points become one when the triangle is equilateral.

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.

BA

D

C

M1

M2

B �A�

D �

C �

M4

M3

Circumcenter

CentroidIncenter

Orthocenter

Orthocenter

Incenter

Centroid

Circumcenter

9. circumcenter

10. The shortest chord through P is a segment

perpendicular to the diameter through P, which is

the longest chord through P.

11.

12. rule: 2n � 2, possible answer:

13.

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.

16. 1580 greetings

Altitude to missing vertex

H CCCC

H H H

H H H

HCCCC

H H H

H H H

A

C

B

B'

O

P

CHAPTER 3 REVIEW

1. False; a geometric construction uses a straight-

edge and a compass.

2. False; a diagonal connects two non-consecutive

vertices.

3. true 4. true

5. false

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

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.

21. 22.

23. Construct a 90° angle and bisect it twice.

24.

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.

27.

A

B

Cz

Copy

A B

CD

A C

B

28.

29.

30. m�A � m�D.You must first find �B.

m�B � 180° � 2(m�A).

31.

32.

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.

y

I

RTx

y

A B

D F

y

4x

2y

2y

�B �A

�D

A

B D

5x

3x 4x

P R

Q

1_2

z

1_2

z

y y x

Segment


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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;

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


58a. yes

58b. If the month has 31 days, then the month is

October.

58c. no

59.



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. m�FAD � 30° so m�ADC � 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

Q

Q'

n 1 2 3 4 5 6 … n … 20

f(n) �1 2 5 8 11 14 … … 56

n 1 2 3 4 5 6 … n … 20

f(n) 0 3 8 15 24 35 … … 399

60. (Chapter 3 Review)

61. (Chapter 3 Review)

f(n) � 3n � 4

f(n) � n2 � 1

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