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Geometry Selected Answers 1

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Geometry Selected Answers 1

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© Houghton Mifflin Harcourt Publishing Company

UNIT 1 Selected Answers UNIT 1 Selected Answers

JJJG JJJG JJJG

MODULE 1 17. DE (or DF ) and EF MODULE 1 5.
19.
Lesson 1.1 Segment Length and Lesson 1.2 Angle Measures and
Midpoints Angle Bisectors
21.
Your Turn Your Turn
6. 4. AEB, BEA 7. L, GLJ, JLG, and 2
23. 30 mi
25. approximately 104 meters 5. 9.
27. Answer B, D
29. (4, 1)
8. JK LM 13 . Therefore, JK and LM
have the same length.
10. Measure each of the segments formed by 7. 40q 11. 172q
the bisector. The two segments should 8. 105q 13.
each have a length that is half as long as
10.
the given segment.
11. M is the origin, since the x- and
y-coordinates are both 0.
12. M lies in Quadrant IV, since the
x-coordinate is positive and the
y-coordinate is negative. 15. m BXC 30q
SA1

11. 17.
Evaluate
1. line segment; defined term
3. ray; defined term
5.

Evaluate
9. AB z BC, so AB and BC do not have the 1.
same length. 19.

11. AB CD 29. Therefore, AB and CD

have the same length.
13. M lies in Quadrant I, since the x- and
y-coordinates are both positive.
15. Both midpoints have the same
coordinates, so the segments have the
same midpoint.
3. 21. y 5.5
23. m ADB 29q and m BDC 61q
Selected Answers

Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor. Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor.
1 2

Selected Answers
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25. A. No
B. Yes
UNIT 1 Selected Answers
C. Yes
MODULE 1 Evaluate
D. No
1.
E. No Lesson 1.3 Representing and
27. Yes; the construction still works. In this Describing Transformations
case, the construction produces two right
angles since each has half the measure of Your Turn
a straight angle (180q). 5. The transformation is a rotation of 180q
around the origin (x, y) o (x, y).
DE DcEc 6 mD mDc 90q
EF EcFc 45 mE mEc 27q rotation of 180q around the origin
DF DcFc 3 mF mFc 63q 3.
The transformation preserves length and
angle measure.
6. The transformation is a translation
29. Construct the bisector of the given angle. (x, y) o (x 1, y 2).
Then construct the bisector of one of the
angles that was formed. ST ScTc 5 mS mSc 76q
TU TcUc 32 mT mTc 45q
SU ScUc 17 mU mUc 59q
SA2

1
vertical compression by a factor of
The transformation preserves length and 3
angle measure.
5. (x, y) o (y, x); rotation of 90q clockwise
8. (x, y) o (1.5x, 1.5y) around the origin.
AB 2 The transformation preserves length and
AcBc 3 angle measure.
Since AB z AcBc, the transformation is not §1 1 ·
7. (x, y) o ¨ x, y ¸
a rigid motion. ©2 2 ¹
§1 · Since AB z AcBc, the transformation is not
9. (x, y) o ¨ x, 3 y ¸
©2 ¹ a rigid motion.
9.
RS 37
RcSc 18
Since RS z RcSc, the transformation is not
a rigid motion.

(x, y) o (1.5x, y)

© Houghton Mifflin Harcourt Publishing Company

11. a. Segment HK is a horizontal segment 17. The transformation is a horizontal stretch

that is 8 units long. by a factor of 3, so it preserves the length
UNIT 1 Selected Answers
b. The image of segment HK is a vertical of vertical segments but not the length of JJJG
segment that is 8 units long. horizontal or diagonal segments. In order MODULE 1 19. Because SV is an angle bisector of
to be a rigid motion, the transformation RST, mRSV mVST. Use this fact
c. The image of segment HK is a must preserve all lengths, so this
horizontal segment that is 16 units long. Lesson 1.4 Reasoning and Proof and the Angle Addition Postulate; 12 x
transformation is not a rigid motion.
d. Possible answer (x, y) o (y, x) is 19. The transformation maps all points to the Your Turn 21. Possible answer P, R, and T
rigid, because it does not change the origin, so the image of any figure under 3. If an animal is a zebra, then it belongs to 23. When two planes cross, they intersect
length of the segment. (x, y) o (2x, y) is this transformation is a single point, (0, 0). the genus Equus. each other at an infinite number of points,
not rigid because it doubles the length The transformation is not a rigid motion i.e., in a line.
of the segment. The transformation 4. If the bill gets two-thirds of the vote in the
because all line segments are mapped to Senate, then it will pass. 25. A line can be in more than one plane.
given by (x, y) o (y, x) switches the a point, so the length of the segment is
segment from horizontal to vertical, not preserved. 5. 3 4x 5 27. Four coplanar lines can intersect in up to
while (x, y) o (2x, y) does not. 6 points.
4x 8 Subtraction Property of
13. The x-coordinate does not change. The Equality
y-coordinate has a constant added to it x 2 Division Property of Equality
(for a translation up) or subtracted from it
(translation down). The coordinate 6. Multiplication Property of Equality
notation has the form (x, y) o (x, y b), Symmetric Property of Equality
where b is a real number, b z 0. Substitution Property of Equality
15. Since QR z QcRc, the transformation is not Transitive Property of Equality
a rigid motion. 8. mLMN 120q
Up to four planes can be determined by
Evaluate 4 noncollinear points.
1. The conclusion is based on observing
SA3

three numbers.
3. The conclusion is based on two
observations.
5. Counterexample 2; if x 2, then x 1 3,
which is a prime number.
7. When I draw three points that are
noncollinear, I can draw a single plane
through all three points, so they are
coplanar after all.

9. The number 14 is an even number.

11. There is exactly one plane containing
points W, X, and Y.
13. Hypothesis the ball is red.
Conclusion it will bounce higher.
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15. Hypothesis the light does not come on.

Conclusion the circuit is broken.
17. Use the Segment Addition Postulate 9 x

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UNIT 1 Selected Answers UNIT 1 Selected Answers

MODULE 2 3. MODULE 2 12.

Lesson 2.1 Translations Lesson 2.2 Reflections

Your Turn Your Turn
4. 5.

midpoints AAc (1, 1);

BBc (2, 2);
CCc (2, 2)
7. 13.
6.
5.

JJJK
SA4

9. GH , 5, 2

11. Emma Quadrant I; Tony Quadrant III

13. 2, 5 midpoints AAc (5, 5);
9.
6.
15. Possible answer He drew vectors from A BBc (1, 3);
to Ac and from B to Bc that were not
K CCc (3, 4)
parallel to or the same length as Q . The
correct vectors should each point 3 units 15. (1, 1)
right and 3 units down.
Evaluate
17. y 2x 3; (2, 1) 1.

8. 5, 4 ; the components are the

opposites of the components of the vector Sc(3, 4), Tc(3, 1), Uc(2, 1), Vc(2, 4)
in Example 3A.
10.
Evaluate
1.

19. 4, 6
21. (0, 2); (1, 3); (2, 2); (1, 1); Possible
answer I used a table to find the
coordinates of the second polygon. Then I
made a new table, using the coordinates
from the second polygon to find the Ac(2, 4), Bc(1, 1), Cc(4, 1)
coordinates of the third polygon.
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7 8

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3. 13. UNIT 1 Selected Answers

MODULE 2 Evaluate
1. SPS', TPT', and UPU'; all three
Lesson 2.3 Rotations angles measure 115q
Your Turn 3.
4.

5. 15.

17. (0, 2)
7.
19. JM 14
21. When the tab key is pressed twice, the 5.
logo is reflected into Quadrant III and then
reflected back to its original position in
SA5

Quadrant II. So after the tab key is

pressed 24 times, the logo is in its original
position. When the tab key is pressed for 7.
the 25th time, the logo is reflected across
the x-axis into Quadrant III.
9. 23. No; the translation would move rectangle
PQRS into the same position as rectangle
PcQcRcSc, but the corresponding vertices 7.
would not be in the same locations.
25. No; line A is not perpendicular to MM' ,
NN' , and PP' so it cannot be the line of
reflection. There is a translation required
in addition to a reflection to map MNP + 8.
+
to McNcPc.
11. 27. Yes; points X and Y are fixed under the
reflection, so they must lie on the line of
reflection. Since two points determine a 9. The rotation is 50q counterclockwise.
line, the line of reflection is XY .
11. The rule is (x, y) o (x, y) and the
transformation is a rotation of 180q.
13. No; according to the definition of a
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rotation, every point and its image must

11. The transformation is a 70q clockwise be the same distance from P, and that is
rotation. not the case in the given figure.

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15.
UNIT 1 Selected Answers

MODULE 2
Lesson 2.4 Investigating
Symmetry
Your Turn
M (4, 2) 6. line, rotational
4
N (1, 2) 90q, 180q, 270q
P (1, 4) 7. line
17. 18q 1
none
19. Possible answer Starting with triangle 1,
rotate clockwise 60º around the vertex at 8. rotational
the center of the hexagon. Repeat the 0
process using each successive image as 72q, 144q, 216q and 288q
a preimage. 9. none
21. a. False 0
none
b. True
c. True Evaluate
+
SA6

23. Possible answer A'B'C' should be 1. 1

rotated so that B' is at the top of the figure. 3. 1
After correctly locating the image of point
5. 45q, 90q, 135q, 180q,
A, Kevin translated the figure rather than
rotating it. 225q, 270q, 315q
25. The remainder after dividing n by 4 7. rotational symmetry
defines a pattern for the table. 9. EDCBAF
A remainder of 0 o QI
11. DCBAFE
A remainder of 1 o QII
A remainder of 2 o QIII 13. Possible answers are shown.
A remainder of 3 o QIV

15. Both have two perpendicular lines of

symmetry, and both have 180q rotational
symmetry.

© Houghton Mifflin Harcourt Publishing Company

5. 9. B
UNIT 1 Selected Answers 11. F
13. sometimes
MODULE 3 13. The reflection across the y-axis will move
15. sometimes
the rectangle from the right of the y-axis to
Lesson 3.1 Sequences of the left of it. Due to the symmetry of the 17. sometimes
rectangle, it will appear to have been 19. a. Yes
Transformations translated left 6 units. Then, translating
b. Yes
Your Turn along the vector ‫ۃ‬5, 4‫ ۄ‬will move the
rectangle right 5 units and up 4 units. c. No
7. This will bring the rectangle fully into d. Yes
Quadrant I. The final result will be a
rectangle that is the same shape and size
21. Possible answer Reflect ABC across +
the y-axis and then translate it down 7
as the preimage that has moved to sit on
units.
the x-axis in Quadrant I, closer to the
y-axis than the preimage. (x, y) o (x, y 7)
14. A horizontal stretch will pull points U and 7.
23. The order of these two reflections does
8. T away from the y-axis, making the not matter. The resulting image is the
triangle longer in the left-to-right direction. same for a reflection in the y-axis followed
The translation along the vector ‫ۃ‬2, 1‫ ۄ‬will by a reflection in the x-axis as for a
move the stretched triangle 2 units right reflection in the x-axis followed by a
and 1 unit up, which will move the triangle reflection in the y-axis.
closer to the origin with one vertex on the
x-axis and another across the y-axis. The
final image will not be the same shape or
size as the preimage.
11.
SA7

Evaluate
1.

Possible answer The translation moves

the figure down one unit and left three
units, mapping Ac to the left of the y-axis
and Cc closer to the origin. The reflection
first will map AsBsCs below the x-axis and
change the orientation. The second
12. reflection will map the figure mostly into
Quadrant III, with Acs in Quadrant IV, and
3. again change the orientation. The final
image is the same size, shape, and
orientation as the preimage.
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UNIT 1 Selected Answers UNIT 1 Selected Answers

MODULE 3 15. Yes. MODULE 3
reflection: (x, y) o (x, y)
Lesson 3.2 Proving Figures are translation: (x, y) o (x 4, y) Lesson 3.3 Corresponding Parts of Congruent Figures
Congruent Using Rigid Motions 17. No. Are Congruent
There are no transformations to map
Your Turn
2. You can map ABCD to WXYZ with a
+
KLM to WXY. + Your Turn
4. SU 43 ft.
reflection across the x-axis, so the figures 19. reflection: (x, y) o (x, y);
translation: (x, y) o (x, y 6) 5. mS 38q.
are congruent.
+
3. You can map JKL to XYZ with a + 21. Rotation: (x, y) o (x, y); 6. LM
7. mH
35 cm
98q
reflection across the y-axis, followed by a translation: (x, y) o (x 4, y)
horizontal translation, so the figures are 23. False. The figures do not have the same 8.
congruent. Statements Reasons
orientation, so the sequence of
5. Reflect JKLM across the x-axis transformations must include a reflection. 1. +SVT # +SWT 1. Given
(x, y) o (x, y). Then translate the image 25. None are congruent. No rigid motions
2. Corresponding parts of
(x, y) o (x 4, y 2). map one angle onto another. 2. VST # WST
congruent figures are congruent.
6. Reflect ABCDE across the y-axis +
27. Rotate LMN 90q counterclockwise
(x, y) o (x, y). Then translate the image about the origin: (x, y) o (y, x). 3. ST bisects VSW . 3. Definition of angle bisector.
(x, y) o (x, y 10). Reflect across the y-axis: (x, y) o (x, y).
9.
Translate: (x, y) o (x 3, y 13). Statements Reasons
SA8

7. B and C are congruent. EF and GH

are congruent. In both cases, a sequence 29. rotation: (x, y) o (y, x); 1. Quadrilateral ABCD # quadrilateral
of transformations is a reflection and a reflection: (x, y) o (x, y); 1. Given
EFGH
translation. translation: (x, y) o (x 2, y 9)
2. AD # CD 2. Given
Evaluate 31. The petals can be mapped onto each
1. (x, y) o (x, y); translation (x, y) o
other by a reflection, which is a rigid 3. Corresponding parts of congruent
transformation. 3. CD # GH
figures are congruent.
(x 8, y)
33. Both students are correct.
3. The figures are congruent. translation: 4. AD # GH 4. Transitive Property of Congruence
(x, y) o (x 2, y 7). 35. Yes; by the definition of congruence, there
is a sequence of rigid motions that maps Evaluate
5. There is no sequence of rigid + +
ABC onto DEF and another that
transformations that will map one figure
onto the other, so they are not congruent.
+
maps DEF onto GHJ. The first + 1. The same sequence of rigid motions that maps ABCD to WXYZ also maps sides and angles
of ABCD to corresponding sides and angles of WXYZ. Therefore, those sides and angles
sequence followed by the second
7. rotation: (x, y) o (x, y); sequence maps ABC onto GHJ, + + are congruent AB # WX, BC # XY, CD # YZ, AD # WZ, A # W , B # X, C # Y,
D # Z.
translation: (x, y) o (x 2, y 6) so the triangles are congruent.
3. mD 43q
9. rotation: (x, y) o (x, y);
translation: (x, y) o (x 2, y) 5. PS 2.1 cm
11. A, B and C are all congruent. The 7. mU 62q
sequence of transformations is a reflection 9. mD 92q
and a translation.
11. U
13. None of the segments are congruent.
13. mG mS
There is no rigid transformation that maps
one of them to another.

© Houghton Mifflin Harcourt Publishing Company

15. UNIT 2 Selected Answers

Statements Reasons
1. +ABC # +DEF 1. Given MODULE 4 Therefore, the measure of the
complement of ABC the measure of
2. BAC # DAC 2. Corr. parts of # fig. are # Lesson 4.1 Angles Formed the complements of DEF.
3. BCA # DCA 3. Corr. parts of # fig. are # by Intersecting Lines The measures of the complements of the
angles are equal, so the complements of
4. AC bisects BAD and AC bisects Your Turn the angles are congruent.
4. Definition of angle bisector
BCD. 6. 18 x 15. 2. supplement
17. mD 28q 7. 10 x 3. Definition of the supplement of an
The measure of each angle is 13q. angle.
19. 1700 ft.
9. The measure of the angle is 45q. 4. mABC mDEF
21. 92 ft.
The measure of its complement is 45q. 5. Equality
23. a. True
10. The measure of the angle is 120q. 6. Substitution Property of Equality
b. False The measure of its supplement is 60q. 7. DEF
c. False
Evaluate 17. In the diagram of intersecting lines, 2
d. False and 4 are vertical angles. Also, 2 and
1. A. Vertical
e. True 3 are a linear pair and 3 and 4 are a
B. Neither
25. The student incorrectly identified corresponding sides. Since +GHJ # +RST, GH # RS. C. Linear Pair
linear pair. By the Linear Pair Theorem,
2 and 3 are supplementary and 3
5x 2 4x 3 o x 5; GH 5(5) 2 23 m.
D. Linear Pair and 4 are supplementary. Then m2
27. Yes; m3 180q and m3 m4 180q by
E. Vertical
Statements Reasons the definition of supplementary angles. By
SA9

F. Neither the Transitive Property of Equality, m2

1. +PQR # +SQR 1. Given 3. 130q m3 m3 m4. Using the Subtraction
Property of Equality, m2 m4. So,
2. RP # RS 2. Corr parts of # figs. are # 5. m2 47.5q
2 # 4 by the definition of congruence.
m3 42.5q
3. RS # RT 3. Given 19. Yes; 90q the measure of its complement
m4 42.5q is 0q, and the measure of its supplement
4. RP # RT 4. Transitive Property 7. False. Vertical angles do not share a is 90q, so 0q 90q 90q.
common side.
5. R is the midpoint of PT 5. Definition of midpoint
9. True
11. The measure of the angle is 45q, the
measure of its complement is 45q, and the
measure of its supplement is 135q.
13. Plan for Proof If ABC # DEF, then
mABC mDEF.
The measure of the complement of
ABC 90q mABC.
The measure of the complement of
DEF 90q mDEF.
Since mABC mDEF, the measure of
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the complement of DEF 90q mABC.

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UNIT 2 Selected Answers UNIT 2 Selected Answers

MODULE 4 21. A possible diagram is shown, with two MODULE 4 Evaluate
nonparallel lines cut by a transversal. 1. lines A and m are parallel; 5.
Lesson 4.2 Transversals and I can measure the angles in my drawing Lesson 4.3 Proving Lines are
Parallel Lines with a protractor as a counterexample. Parallel If 1 # 5, then lines A and m are parallel.
4 and 5 are alternate interior angles,
Your Turn Your Turn 3. lines A and m are parallel; 4.
but m4 90q and m5 130q, so the
7. x 10, y 3; the Corresponding Angles measures are not the same when the 4. m1 120q and m2 120q. They are If 4 # 6, then lines A and m are parallel.
Theorem and the Alternate Interior Angles lines are not parallel. congruent alternate interior angles. The
5. 7 # 3; Converse of the Corresponding
Theorem. 23. B lines are parallel because of the Converse
Angles Theorem
of the Alternate Interior Angles Theorem.
Evaluate 7. m4 65q and m5 115q, so m4
5. m1 120q and m2 60q. The angles
1. A. C m5 180q. Yes, the lines are parallel by
are supplementary. The lines are parallel
the Converse of the Same-Side Interior
B. A because of the Converse of the Same-
Angles Postulate.
C. D Side Interior Angles Postulate.
7. 9. When x 25, x 25 2x 50; the
D. B alternate interior angles are congruent
3. m5 98q and the horizontal parts of the letter Z are
parallel.
5. m5 122q
11. x 28
7. m1 109q
13.
SA10

9. m10 69q
11. m7 118q
13. m14 66q
15. y 40 8. Same side interior angles; by the
Converse of the Same Side Interior
x 15
Angles Postulate.
17. y 15
9. Corresponding angles; by the Converse of
19. the Corresponding Angles Theorem.
Statements Reasons 15.
1. p q 1. Given Statements Reasons
2. m3 m5 2. Alternate Interior 1. lines A and m are cut by a transversal;
1. Given
Angles Theorem 1 # 2
3. m1 m3 3. Vertical Angles 2. m1 m2 2. Definition of congruence
Theorem
3. 2 and 3 are supplementary. 3. Linear Pair Theorem
4. m1 m5 4. Substitution
4. m2 m3 180q 4. Definition of supplementary angles
Property of Equality
5. m1 m3 180q 5. Substitution Property of Equality
6. 1 and 3 are supplementary. 6. Definition of supplementary angles
7. Converse of Same-Side Interior
7. A & m
Angles Postulate

© Houghton Mifflin Harcourt Publishing Company

UNIT 2 Selected Answers UNIT 2 Selected Answers

MODULE 4 MODULE 4 19. a. Neither
b. Perpendicular
Lesson 4.4 Perpendicular Lines Lesson 4.5 Equations of Parallel
c. Perpendicular
Your Turn and Perpendicular Lines
d. Parallel
4. DC 20 cm Your Turn e. Perpendicular
5. BC 25 cm 4. y x 7.5 21. He should have said “two nonvertical
7. AC 16 in. 3 lines” because vertical lines have
5. y x6 undefined slope. should have had a
9. m4 40q 2
negative sign on one of his expressions
2 for slope because the slopes of
Evaluate 7. y x 1
3 perpendicular lines have a product of 1.
1. Fold line A onto itself so that the crease
1
passes through point P. The crease is the 8. y x
4
required perpendicular line.
3. The midpoint of the segment is the point Evaluate
on the perpendicular bisector that is 4
closest to the endpoints of the segment. 1.
5
5. PC 5 cm
1
7. 8 cm 3.
4
9. 52q
5. y 3x 27
SA11

11. 90q
7. y 5x 3
13. 180q
9. y 3x 36
15. x 45; y 5.
3
17. The valve pistons are lines that are 11. The top line slope is . The bottom line
4
perpendicular to the same line (the lead
2
pipe), so they form right angles with the slope is . The lines do not have the
same line. By the corresponding angles 3
theorem, all the congruent right angles same slope, so they are not parallel.
mean the valve pistons are parallel to 1 1
each other. 13. y x
2 2
19. are both obtuse 15. y 2x 17
90q 17. a. y x 30
the definition of obtuse angles
b. y x 30
180q c. They are perpendicular.
m1 m2 180q
false
1 and 2 cannot both be obtuse
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UNIT 2 Selected Answers UNIT 2 Selected Answers

MODULE 5 MODULE 5
Lesson 5.1 Exploring What Lesson 5.2 ASA Triangle Congruence
Makes Triangles Congruent Your Turn
Your Turn 5. + ADB # + ADC by the ASA Triangle Congruence Theorem.
+ +
4. RST # UVW, because all six pairs of + PQR has a measure of 67q. So, + PQR is not congruent
6. None of the angles in
corresponding parts are congruent. to +STU.
5. mB z mE and mD z mG. The 8.
triangles are not congruent because some Statements Reasons
pairs of the corresponding angles are not
congruent. 1. JLM # KML 1. Given
7. x 26 2. JML # KLM 2. Given
8. y 5 3. LM # LM 3. Reflexive Property of Congruence

Evaluate +
4. JML # KLM+ 4. ASA Triangle Congruence Theorem
1. Possible answer counterclockwise 9.
rotation about point M by mQMN Statements Reasons
HJJJG
followed by a reflection across MQ. 1. S and U are right angles. 1. Given
+ +
SA12

3. ABC # DEF because all six pairs of 2. S # U 2. All right angles are congruent.
corresponding parts are congruent.
3. RV bisects SU 3. Given
5. JK z MN. Not congruent; there aren’t six
pairs of congruent corresponding parts. 4. ST # UT 4. Definition of bisector
7. z 1.6 5. RTS # VTU 5. Vertical angles are congruent.
9. w 27 +
6. RST # VUT + 6. ASA Triangle Congruence Theorem
11. LM is not congruent to QR. So, the Evaluate
triangles are not congruent.
13. The triangles have six pairs of congruent 1. a.
corresponding parts, and are congruent
by the converse of CPCTC.
15. 7 x
y 37
17. Never true
b. Yes; the triangles are congruent by ASA. Therefore, there is a sequence of rigid motions
19. Sometimes true that will map my triangle onto Natasha’s triangle.
21. a. Not congruent
+ +
3. JKL # MNP by ASA.
b. Congruent
c. Congruent
+
5. None of the angles in STU has a measure of 34q. So, +STU is not congruent to +PQR.
7. Given
d. Not congruent
CBA # DBA
e. Congruent
AB # AB
23. No; LN # LN by the Reflexive Property of
ASA Triangle Congruence Theorem
Congruence. Therefore, the triangles are
congruent.
Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor. Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor.
23 24

© Houghton Mifflin Harcourt Publishing Company

9. 21. Possible sketch:

Statements Reasons
1. H # J 1. Given
2. G is the midpoint of HJ . 2. Given
3. HG # JG 3. Definition of midpoint 23. No; the quadrilaterals in the figure meet
the conditions of the proposed ASAS
4. FG is perpendicular to HJ . 4. Given Congruence Theorem, but they are not
5. FGH and FGJ are right angles. 5. Definition of perpendicular congruent. Therefore, the quadrilaterals
serve as a counterexample to show that
6. FGH # FGJ 6. All right angles are congruent. there is no such theorem.
+
7. FGH # FGJ + 7. ASA Triangle Congruence Theorem
HJJJG
11. a. Translate LM to WX , then reflect +LMN across WX .
JJJJG
JJJG JJJJG
JJJJG
b. Since L # W and M # X, the images of LN and MN lie on WY , and XY ,
respectively. The image of N must lie on both rays, so the image is the intersection
point Y.
13.
SA13

15.
Statements Reasons
1. A # E 1. Given
2. C is the midpoint of AE. 2. Given
3. AC # EC 3. Definition of midpoint
4. ACB # ECD 4. Vertical angles are congruent.
+
5. ACB # ECD + 5. ASA Triangle Congruence Theorem
6. AB # ED 6. CPCTC
17. In each sign, the side that is 36 in. long is the included side between two 60q angles.
Therefore, by the ASA Triangle Congruence Theorem, the signs are congruent.
19. B and D are both right angles, so B # D; CB CD, so CB # CD; ACB and ECD
are vertical angles, so ACB # ECD. Therefore, ACB # ECD by the ASA Triangle + +
Congruence Theorem. Since corresponding parts of congruent triangles are congruent,
AB # ED, so AB ED. Mariela can find the distance AB across the canyon by measuring
the distance ED.
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UNIT 2 Selected Answers UNIT 2 Selected Answers

JJJK
MODULE 5 13. a. Each triangle has side lengths of 2 and MODULE 5 5. Possible answer A translation along AD,
6 and an included right angle. By SAS and then a counterclockwise rotation
Lesson 5.3 SAS Triangle they are congruent. Lesson 5.4 SSS Triangle
about D so that AB coincides with DE .
Congruence b. Possible answer Reflect the triangle Congruence JJJK
across the y-axis. Next translate it 1 7. Possible answer A translation along CF ,
Your Turn Your Turn and then a counterclockwise rotation
unit to the left. Then translate it 6 units
3. DE # GH , DF # GJ , D # G, because down. 4. AB # AD and CB # CD, so A is about F so that CA coincides with FD.
HJJG
corresponding parts have the same 15. Yes; since the opposite sides of a equidistant from B and D, and C is 9. DF is the perpendicular bisector of EG .
+ +
measure. EDF # HGJ by the SAS rectangle are congruent and the included equidistant from B and D. By the converse
of
This shows that EHJJG
maps to G by a
Triangle Congruence Theorem. angles between the sides are right angles, HJJGthe Perpendicular Bisector Theorem, reflection across DF .
4. Possible answer You are given that the two triangles are congruent by the AC is the perpendicular bisector of BD .
SAS Theorem. By the definition of a reflection, point D is 11. Not congruent
AB # AD and 1 # 2. You also know
the image of point B reflected across AC . 13. Not congruent
that AC # AC by the reflexive property.
Two sides and the included angle of The reflection also maps AC onto AC, 15. x 4, y 4; ASA
+ BAC are congruent to two sides and +
so ABC # ADC. + 17. no solution
the included angle of DAC. The + 5. The corresponding sides MN and QR 19. x 6, y 15
triangles are congruent by the SAS
are not congruent. Therefore, the triangles 21. Possible solution x 8, y 8
Triangle Congruence Theorem.
are not congruent. 23. A. not sufficient; B. sufficient; C. not
Evaluate 6. It is given that GK # GL and JK # JL, and sufficient; D. sufficient; E. sufficient;
SA14

JJJG JJJG F. sufficient

1. No; she can map AC to DF by a GJ # GJ by the Reflexive Property.
HJJG
reflection across DE , but C will map 25. No; At least one pair of sides must be
7. ASA. x 3; y 8 congruent. For example, two triangles that
to F only if AC DF.
have three 60 degree angles may not
3. AB # DB, BC # BC, and ABC # DBC, Evaluate
have at least one pair of congruent
+ +
ABC # DBC by SAS.
1. corresponding sides, such as a triangle
5. AB # AB,CB # DB, ABC # ABD + with side lengths of 4 inches and another
triangle with side lengths of 8 inches.
+
ABC # ABD by SAS. 27.
+ +
7. ABC # DEF by SAS when x is 7. Statements Reasons
9. You are given that BD # FD and CD #
1. BFC # ECF , 1. Given
ED. You also know that D # D by the
reflexive property. Two sides and the BCF # EFC
+
included angle of BDE are congruent 3.
2. FC # FC 2. Reflexive
to two sides and the included angle of Property of
+ FDC. The triangles are congruent by Congruence
the SAS Triangle Congruence Theorem.
So, by CPCTC, the braces BE and FC
3. +BFC # +ECF 3. ASA Triangle
Congruence
and are also congruent. Theorem
11. a. Measure AB and BD; so he can 4. FB # CE 4. CPCTC
confirm that a pair of sides and their
included angles are congruent. 5. AB # DE , AF # DC 5. Given
b. Measure ACB and DCB; so he can
confirm that a pair of angles and their
6. +ABF # +DEC 6. SSS Triangle
Congruence
included sides are congruent. Theorem

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UNIT 2 Selected Answers UNIT 2 Selected Answers

MODULE 6 17. Fold the circle so that one half coincides MODULE 6 Sides AC and DF are equal. Therefore,
with the other. The crease is a diameter. because two angles and a non-included
Lesson 6.1 Justifying Then fold the diameter onto itself to make Lesson 6.2 AAS Triangle +
side are congruent, ABC # DEF by +
another crease that is the perpendicular the AAS Theorem.
Constructions bisector of the diameter. The two creases
Congruence
determine the four vertices of the square Evaluate
Your Turn Your Turn
on the circle. 1. Congruent, by AAS Congruence.
8. You need to add an extra step to say that 5. Because BC is parallel to EF, this means
19. a. B; AB; P; T 3. Congruent, by ASA Congruence.
because PQ is the perpedicular bisector that ACB is congruent to DFE, using
b. AP # BT ; AB # PT 5. Cannot be determined.
of MN, the point of intersection will be the the Corresponding Angles Theorem.
c. Draw PB PB # PB by the Reflexive Since ABC is congruent to DEF and 7. AB # DE , or BC # EF
midpoint of MN by the definition of
Property of Congruence.
midpoint.
+ PAB # +
BTP by SSS, and therefore
AC is congruent to DF, then two angles
9. AB # DE , AB # DF , AC # DE , or AC # DF
9. First construct a perpendicular bisector of
+ ABP # +
TPB by CPCTC. So by the
and two non-included sides are
congruent. This means that ABC is + 11. AB # DE , or BC # DC
a segment. This creates 90q angles.
Choose one of the angles to bisect. This
Alternate Interior Angles Theorem,
PT & AB.
+
congruent to DEF using the AAS
13. 1. Reflexive Property of Congruence;
will construct two 45q angles. Triangle Congruence Theorem.
3. ACB # ACD; 4. B # D ;
23. 7. It is given that A # D and B # E.
Evaluate 5. AAS Triangle Congruence Theorem
5 (1)
2
1. She should adjust the angle of the BC (5 2)2 15. Congruent; AB DE 4; BC DF 2;
reflective device until the image of line A AC EF 4.47
coincides with line A. Then she should 16 9
17. Not congruent. Segments AB and DF are
draw a line along the edge of the reflective
5 corresponding, but they have different
device. This is the required line.
SA15

lengths (AB 4, DF 5).

3. 1. BC, BD 1 5 2 ( 1)
2 2
EF
2. Converse of the Perpendicular Bisector 19. Not congruent. Segments AC and EF
Theorem 16 9 are corresponding, but they have different
3. Converse of the Perpendicular Bisector lengths (AC 9, EF 10).
Theorem 5
HJJG 21. A; B; D
4. CD Sides BC and EF are equal. Therefore,
HJJG HJJG 23. They are not necessarily congruent.
5. CD bisects AB ; perpendicular bisector because two angles and a non-included
5. They are congruent. +
side are congruent, ABC # DEF by + 25. The two triangles cannot be congruent.
27. Yes, the sum of the measures in each
the AAS Theorem.
7. They are congruent. triangle must be 180q, which makes it
8. It is given that A # D and B # E.
9. They are congruent. possible to solve for x and y. The value of
x is 15, and the value of y is 12. Each
4 (1)
2
11. Line m is a transversal of lines A and n. AC (0 2)2
Angles 1 and 2 are congruent triangle has angles measuring 82q, 68q,
corresponding angles, because angle 2 is 94 and 30q. VU # VU by the Reflexive
a constructed copy of angle 1. By the Property of Congruence. So
Converse of the Corresponding Angles 13 + VSU # +
VTU by ASA or AAS.
Postulate, A || n.
2 5 3 (1)
2 2
DF
13. 30q
15. Fold the paper so that the segment’s 94
endpoints coincide. The fold line is the
segment’s perpendicular bisector, so any 13
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three points on the fold line will be

equidistant from the endpoints.

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4 0
2
UNIT 2 Selected Answers BC (1 3)2 20,

and EF 4 0
2
( 2 0) 2
20.
MODULE 6
Therefore, AB # DE and BC # EF . By the Slope Formula,
Lesson 6.3 HL Triangle Congruence
34 1 1 4
slope of AB , slope of AC 3,
Your Turn 0 ( 3) 3 4 ( 3)
+ +
3. Yes. VWX and YXW are right triangles that share hypotenuse WX < WX # WX by the 0 ( 3) 2 ( 3) 1
Reflexive Property of Congruence. It is given that WV # XY , therefore VWX # YXW + + slope of DE
0 1
3, slope of DF
4 1 3
.
by the HL Triangle Congruence Theorem.
5. It is given that and CAB and DBA are right angles and AD # BC. AB # AB by the

Since slope of AB < slope of AC 1, AB A AC and A is a right angle.

Reflexive Property of Congruence. Then ABC # + +

BADby the HL Triangle Congruence Since slope of DE < slope of DF 1, DE A DF and D is a right angle.
Property.
Evaluate
+ +
So, ABC # DEF by the HL Triangle Congruence Theorem.
17. a. No; there is not enough information to use any of the triangle congruence theorems.
+ +
1. No.
3. No. b. Yes; KJL # MJL by the HL Triangle Congruence Theorem since KJ # MJ and
5. Yes. JL # JL .
7. +
19. The perimeter of PQR is 16 ft.
Statements Reasons 21. Yes. Let the remaining leg of the first triangle have a length of x inches. Then by the
1. FGH and JHK are right angles. Pythagorean Theorem, x 2 102 262. So, x 2 576, and x 24. Therefore, the hypotenuse
SA16

1. Given
and a leg of the first right triangle are congruent to the hypotenuse and a leg of the second
2. H is the midpoint of GK. 2. Given right triangle, so the triangles are congruent by the HL Triangle Congruence Theorem.
3. GH # HK 3. Definition of midpoint 23. There is an LL Triangle Congruence Theorem.

4. FH # JK 4. Given Statements Reasons

+
5. FGH # JHK + 5. HL Triangle Congruence Theorem 1. A and D are right angles. 1. Given

9. 2. A # D 2. All right angles are congruent.

Statements Reasons 3. AB # DE 3. Given
1. ADC and BDC are right angles. 1. Given 4. AC # DF 4. Given
2. AC # BC 2. Given 5. +ABC # +DEF 5. SAS Triangle Congruence Theorem

3. DC # DC 3. Reflexive Property of Congruence

+
4. ADC # BDC + 4. HL Triangle Congruence Theorem
5. Corresponding parts of congruent
5. AD # BD
triangles are congruent.
11. x 11.
13. x 10.
15. By the Distance Formula,

0 ( 3)
2
AB (3 4)2 10,

1 0
2
DE ( 3 0)2 10,

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UNIT 2 Selected Answers UNIT 2 Selected Answers

MODULE 7 27. By the Triangle Sum Theorem, mL MODULE 7 19.
mM mN 180q and mR mS 1. M is the midpoint 1. Given
Lesson 7.1 Interior and Exterior mT 180q. Since each set of angle Lesson 7.2 Isosceles and of BC.
Angles measures total 180q, they are equal using Equilateral Triangles 2. BM # CM 2. Definition of
the substitution property of equality. So, midpoint
Your Turn mL mM mN mR mS Your Turn
3. AB # AC 3. Given
6. Each unknown angle measure is 135q. mT. Since L # R and M # S, then 5. 51q
mL mR and mM mS by the 4. AM # AM 4. Reflexive Property
7. 87q 6. 15 cm of Congruence
definition of congruence. Subtracting
8. 68q and 136q equals from both sides gives mN Evaluate 5. +AMB # +AMC 5. SSS Triangle
11. 100q mT. Then N # T by the definition of Congruence
1.
congruence. Theorem
12. 37q
6. B # C 6. CPCTC
29.
Evaluate Number of 21. Triangles ABD and CBD are congruent by
1. They must be complementary. 3 4 5 6 7 8 ASA.
Sides, n
3. 1980q Number of 23.
0 2 5 9 14 20 CAB # CBA, so opposite sides CA a
5. 103q Diagonals, d
nd CB are congruent. Therefore, it is an
The number of diagonals increases by 2,
7. 146q isosceles triangle.
then 3, 4, 5, etc. A formula relating n and
9. 120q n( n 3) 3.
d is d .
11. w 136 2
The three line segments drawn are radii,
SA17

13. 59q which have the same length in both

15. A. A; B. B; C. D; D. E; E. C circles, since the circles are the same
size. Therefore, all of the line segments
17. 135q are congruent and form the three sides of
19. 1440q an equilateral triangle.
144q; 36q 5. 76q
21. x 20q; y 45q; z 115q 7. 57q
23. A regular hexagon; if the construction 9. KL 33
continues and the sides are kept
congruent, the polygon will include six 2
11. BC
120q angles and six congruent sides, so it 13
is a regular hexagon. 13. xq will equal 112q
25. 360q; 360q; 72q, 72q(5) 360q; 720q, 60q, 15. By the Angle Addition Postulate,
60q(6) 360q mATB 80q 40q 40q. mBAT 40q
It appears from the table that the sum of by Alt. Int. Thm. ATB # BAT by the
the measures of the exterior angles of any definition of congruence and BA # BT by
polygon is always 360q. the Converse of the Isosceles Triangle
Theorem. Then BA BT 2.4 mi.
17. 48q
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UNIT 2 Selected Answers UNIT 2 Selected Answers

MODULE 7 21. The safest route is to avoid sailing MODULE 8 3.
between the islands at X and Y.
Lesson 7.3 Triangle Inequalities 23. The towers at Q and R are closest
Lesson 8.1 Perpendicular
Your Turn together. Bisectors of Triangles
6. No; 12 4 ! 17 25. AB BC ! x Your Turn
7. Yes; 24 8 ! 30, 8 30 ! 24, AB x ! BC 5. ZJ 65
and 24 30 ! 8 x ! BC AB 6. GM 60 5. ZC 85
9. 7 x 35 BC x ! AB ZG 65 AC 154
10. 9 x 27 x ! AB BC 7. BC 136
8.
11. m A, m B, m C Since AB ! BC, BC AB 0, so the 9. 1. Given; 2. Given; 3. PB; 4. PB;
second inequality is not relevant. 5. PB PC; Transitive Property of Equality
12. m B, m C, m A
Combining the first and last inequalities
11. (2.5, 3.5)
13. CB, AC, AB gives AB BC x AB BC.
13.
14. AC, BC, AB The constructions show that AC
approaches but is always greater than
Evaluate AB BC, and that AC approaches but is
1. No always less than AB BC.
3. No 27. B, C, D
2, 3
SA18

5. No 29. Write two equations, AD2 CD2 AC 2 Let the three towns be vertices of a
7. Yes triangle. By the Circumcenter Theorem,
and BD CD 2 2 2
BC . Equating 9.
the circumcenter of the triangle is
9. 5 x 11 expressions for CD 2 , AC 2 AD2 equidistant from the vertices. Trace the
11. 5.3 x 49.9 BC 2 BD2 and therefore
outline of the lake. Draw the triangle
formed by the towns. To find the
13. m A, m B, m C AC 2 BC 2 AD2 BD2 . Since the right circumcenter, find the perpendicular
15. m P, m R, m Q side is positive, so is the left side, which bisectors of each side. The position
leads to BC AC. of the boat is the circumcenter, F.
17. DE, EF, DF
15. a. To find AB, note that DB is a leg of
19. PR, PQ, QR
+
right triangle ZBD and ZB is the
3.5, 4 hypotenuse. Use the Pythagorean
Theorem to find DB and multiply by 2
Evaluate
because D is the midpoint of AB . To
1. find AC, use the same method, noting
first that ZC ZB because C is the
circumcenter of ABC. Also, ZF is a
+
leg of right triangle ZCF and ZC is
the hypotenuse.
b. AB 210; AC 288
c. No; the only information given about
+
isosceles ZBC is the length of two
sides, which is insufficient for finding BC

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UNIT 2 Selected Answers UNIT 2 Selected Answers

MODULE 8 19. MODULE 8 17. 5, 3
Lesson 8.2 Angle Bisectors Lesson 8.3 Medians and 19.
of Triangles Altitudes of Triangles
Your Turn Your Turn
4. QS 14.7 6. 12 units; SP 12; PW 6
5. mKJM 58q 8. The centroid is located 7 feet above
8. 19.2 the ground along AD, or 14 feet below
Draw the bisectors of two angles of the point A. 21.
9. mPQX 52q triangular park. The monument should be
9. S(4, 5)
at the intersection of the bisectors. This
Evaluate point is the incenter of the triangle. By the 10. S(0, 4)
1. Use the compass to measure both Incenter Theorem, it is equidistant from
perpendicular segments from P to the the sides of the triangle. 13. 2, 3
sides of A 23. The coordinates of the orthocenter
21. a. False b. True c. False d. True 15. 4, 1
3. mLKM 31.5q e. False are (2, 2).
5. mHFJ 90q 23. The circle will not necessarily pass Evaluate 25. The coordinates of the orthocenter
are (8, 0).
7. through the points where the angle 1.
bisectors intersect the sides of the 27. Yes. The orthocenter is B.
triangle. Instead, the student should have 29. As the triangle is stretched to the right,
used S as the center of the circle and the centroid moves to the right along a
SA19

made a circle that just touches the three 3. No; no matter how you change the shape horizontal line. The orthocenter remains
sides of the triangle. +
of ABC, the point at which the medians at C, the origin.
intersect is always in the interior of the
triangle.
5. 24 FH;

9. mFED 46q 7. GX 13.5

9. KY 3.6 in
11. mRTJ 55q
13. VY 17 11. a. x 11; b. y 5; c. BP 10; d. BD 15;
e. CP 18; f. PE 9
15. mGDF 39q
13. 5, 0.5
17. 1. Given; 2. Definition of perpendicular;
3. Reflexive Property of Congruence; 15. x 5
4. HL Triangle Congruence Theorem;
5. Corresponding parts of congruent
triangles are congruent.;
JJJG
6. YV bisects XYZ; 8. Definition
of angle bisector
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UNIT 2 Selected Answers UNIT 3 Selected Answers

MODULE 8 5. The slope of JK and BA is 0, so the lines MODULE 9 7. 80
5 9. 23
Lesson 8.4 Midsegments of are parallel. The length of JK is and Lesson 9.1 Properties of
2 11. The diag. of PRQS bisect each other,
Triangles 1
Parallelograms so QT ST 18.
the length of BA is 5, so JK BA.
Your Turn 2 Your Turn 13. Opp. angles of PRQS are congruent,
4. 7. 7.9 8. QR 20 so mSPQ # mQRS 110q.
9. 68q 9. PR 28 15. 1. PSTV is a parallelogram
11. n 4 2. STV # P
Evaluate
13. n 6
1. Possible answer: He can use the ruler to 3. PQ # RQ
15. a. True b. False c. True d. False e. False
f. True
draw JL and KM , label their intersection
as point N, and use the ruler to find that
4. +PQR is isosceles
17. An equilateral triangle with sides 4 units JN LN and KN MN. His conjecture 5. P # R
long would be that the diagonals of a 6. STV # R
19. No. If AC is 36 feet long, DE is 18 feet parallelogram bisect each other.
1. Given
long. Because he needs to insert two 3. ABCD is a parallelogram.
2. Opp. angles of a . are congruent
crossbars, he needs 2 u 18, or 36, feet of AB || DC, AD || BC; Draw DB.
slope of UW 5; slope of YZ 5; Since the timber. He needs 4 more feet. 3. Given
ADB # CBD, 4. Definition of isosceles triangle
SA20

slopes are the same, UW & YZ. 21. Possible answer: First construct a triangle DB # DB
with a straightedge. Next use the ABD # CDB; 5. Isosceles Triangle Theorem
1
UW 26; YZ 2 26; So UW
2
YZ. compass to find the midpoint of two sides
of the triangle. Finally, connect the two
+ABD # +CDB 6. Transitive Property of Congruence
6. JL 78; PM 47.5; mMLK 105q midpoints to create a midsegment parallel AB # CB and AD # CB 17. Sometimes; opposite sides of a
to the third side. parallelogram are congruent, but
Evaluate
1. Drawings will vary. Students should
23. +QXY # +XPZ # +YZR # +ZYX; 5. 1. ABCD is a parallelogram consecutive sides, such as RS and ST ,
may or may not be congruent.
2. AB || DC
area of +XYZ # area of +PQR
1
conclude that the midsegment is half the 19. Always; opposite sides of a parallelogram
length of the third side. 4 3. ABE # CDE, BAE # DCE
are congruent.
3. The slope of XY and GH is undefined, 4. AB # DC 21. Never; diagonals of a parallelogram bisect
so the lines are parallel. The length of 5. +ABE # +CDE each other.
XY is 3 and the length of GH is 6, so 23. a. 3, 6, 8;
1 6. AE # CE and BE # DE
XY GH. b. 1 is supplementary to 2, 4,
2 1. Given 5, and 7.
2. Definition of parallelogram 25. The side lengths of the tile are 6 inches
3. Alt. Int. Angles Thm. and 8.5 inches.
4. Opposite sides of a parallelogram are 27. y 2x
congruent
5. ASA Triangle Cong. Thm.
6. CPCTC

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UNIT 3 Selected Answers 17. 1. Draw DB 21. Possible answer: When you draw AC,

2. AB # CD +
you form DAC and BAC. LM is a +
MODULE 9 3. Possible answer: It is given that 3. AB || CD
+
midsegment of DAC, so LM || AC and

AE # CE and DE # BE. Since vertical 1

4. ABD # CDB LM AC. JK is a midsegment of
Lesson 9.2 Conditions for angles are congruent, AEB # CED and 2
Parallelograms AED # CEB. Therefore, AEB # + 5. DB # DB
Your Turn
+ +
CED and AED # CEB by the SAS + 6. +ABD # +CDB +BAC, so JK || AC and JK 1
2
AC.
Triangle Congruence Theorem. Since
6. PQ 16.8 This shows that LM || JK and LM JK, so
corresponding parts of congruent triangles 7. AD # CB
RS 16.8 are congruent, AB # CD and AD # CB. LM || JK . A pair of opposite sides of
8. ABCD is a parallelogram.
m Q 74q
So, ABCD is a parallelogram because if JKLM are both parallel and congruent, so
both pairs of opposite sides of a 1. Through any two points, there is exactly JKLM is a parallelogram.
mR 106q quadrilateral are congruent, then the one line.
quadrilateral is a parallelogram. 2. Given
So, PQ RS. Also, PQ || RS since
same-side interior angles are 5. JK 27 3. Given
supplementary. PQRS is a parallelogram LM 27 4. Alternate Interior Angles Theorem
since a pair of opposite sides are parallel 5. Reflexive Property of Congruence
KL 13
and congruent.
MJ 13 6. SAS Triangle Congruence Theorem
7. JN 14
So, JK LM and KL MJ. JKLM is a 7. CPCTC
LN 14 8. If both pairs of opposite sides of a
parallelogram.
KN 22 7. No. One pair of opposite sides are quadrilateral are congruent, then the
MN 22 parallel. A different pair of opposite sides quadrilateral is a parallelogram.
19. Possible answer: The student used an
SA21

So JN LN and KN MN. JKLM is a are congruent.

parallelogram since the diagonals bisect 9. Yes. A pair of alternate interior angles are invalid reason in Step 7. The student
each other. congruent, so a pair of opposite sides are should also show that AD # CB. This is
parallel. The same pair of opposite sides true because 1 # 2 and these are
Evaluate are congruent by SAS and CPCTC. alternate interior angles. Then the student
1. Draw DB 11. Yes. The 73q angle is supplementary to can conclude that ABCD is a
both of the 107q angles. This shows that parallelogram because a pair of opposite
AB # CD AD # CB DB # DB sides are both parallel and congruent.
both pairs of opposite sides are parallel by
+ABD # +CDB the Converse of the Same-Side Interior
Angles Postulate.
ADB # CBD, ABD # CDB
13. He should draw KL . KL is a midsegment
AB || DC, AD || BC of 'PQR because it connects the
ABCD is a parallelogram. midpoints of two sides of the triangle.
1
KL is parallel to PQ and KL PQ by
2
the Triangle Midsegment Theorem.
15. No. The figure shows a counterexample.
The diagonals are congruent, but the
quadrilateral does not have two pairs of
parallel opposite sides, so it is not a
parallelogram.
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UNIT 3 Selected Answers UNIT 3 Selected Answers

MODULE 9 3. BD AC 0 MODULE 9 5. Rhombus
5. 48 in. 7. You need to know that JKLM is a
Lesson 9.3 Properties of Lesson 9.4 Conditions for parallelogram.
7. m1 29q, m2 61q, m3 90q, m4
Rectangles, Rhombuses, and 29q, and m5 90q. Rectangles, Rhombuses, and 9. The conclusion is valid.
Squares 1 Squares 11. 6.5
9. 32
Your Turn 3 Your Turn 13. parallelogram, rhombus, rectangle, square
5. 12.5 cm 11. m1 126q, m2 27q, m3 27q, 4. Yes; the figure is a parallelogram because 15. parallelogram, rhombus
6. 21.25 cm m4 126q, and m5 27q. of congruent opposite sides, and it is a 17. Since both pairs of opposite sides are
13. A. always; B. sometimes; C. always; rectangle because it is a parallelogram congruent, PQRS is a parallelogram.
9. Since JKLM is a rhombus, with congruent diagonals.
D. sometimes; E. sometimes Since PZ, QZ, RZ, and SZ are all equal
JM # LM and JK # LK . MK # MK by the
15. You cannot use a theorem that assumes 5. No; by the Angle Addition Postulate, lengths, PZ RZ QZ SZ. So QS # PR
Reflexive Property of Congruence. So, mFEH 45q 50q 95q, so FEH is not
+ +
MJK # MLK by the SSS Triangle
the quadrilateral is a parallelogram to
justify the final statement because you do a right angle and EFGH is not a rectangle.
Since the diagonals are congruent, PQRS
is a rectangle.
Congruence Theorem. Therefore, JMK # not know that JKLM is a parallelogram.
LMK and JKM # LKM by CPCTC, so 6. It is given that JK # KL. Because opposite 19. parallelogram
That is what you are trying to prove. sides of a parallelogram are congruent,
MK bisects JML and JKL. A similar Instead, use the converse, which states rhombus
argument shows that JL bisects MJK that if both pairs of opposite sides of a KL # MJ and JK # LM. By substituting the
21. ABC is a right angle. And since
and MLK. quadrilateral are congruent, then the sides JK for KL and visa versa, DE A EF, DEF is a right angle. By the
quadrilateral is a parallelogram. So JK # MJ and KL # LM. So,
10. mVYX 73q Hypotenuse-Leg (HL) Triangle
SA22

11. mXYZ 36.5q

therefore, JKLM is a parallelogram.
17. 25.3 feet
JK # KL # LM # MJ, making JKLM a Congruence Theorem, ABC # DEF . + +
12. By definition, a rhombus is a quadrilateral rhombus. By CPCTC, BC # EF . Since the opposite
with four congruent sides. Since a square D%RWKSDLUVRIRSSRVLWHVLGHVDUH
8. The conclusion is not valid. You must also sides of EBCF are parallel and congruent,
is also a quadrilateral that has four parallel.
first be given that ABCD is a it is a parallelogram. Since BE A EF, then
congruent sides, then a square is a %RWKSDLUVRIRSSRVLWHVLGHVDUH parallelogram.
congruent. BEF is a right angle, which makes EBCF
rhombus.
Evaluate a rectangle.
13. Possible answer: All four angles of a %RWKSDLUVRIRSSRVLWHDQJOHVDUH
square are right angles. All right angles congruent. 1. Square; because the diagonals are
are congruent. Any quadrilateral with both 2QHDQJOHLVVXSSOHPHQWDU\WRERWKRI congruent, it is a rectangle and because
pairs of opposite angles congruent is a its consecutive angles. the diagonals are perpendicular, it is a
parallelogram, so a square is a rhombus. A figure that is both a rectangle
7KHGLDJRQDOVELVHFWHDFKRWKHU
parallelogram. and a rhombus must be a square.
E7KHGLDJRQDOVDUHFRQJUXHQW
3. No information is known about its sides or
Evaluate F7KHGLDJRQDOVDUHSHUSHQGLFXODU angles, so it may not be a parallelogram.
1. parallelogram (DFKGLDJRQDOELVHFWVDSDLURI So, it cannot be determined if it is a
all right angles are congruent opposite angles. rectangle
it is a quadrilateral that has congruent 21. 180° rotation around its center; reflectional
opposite angles symmetry across a line that contains the
opposite sides of a parallelogram are midpoints of opposite sides (two lines)
congruent 23. A square has the reflectional properties of
DC # DC both a rhombus that is not a square and a
rectangle that is not a square. Because
D # C squares have all angles congruent and all
congruent sides congruent, as opposed to only all
SAS Triangle Congruence Theorem sides congruent (rhombuses) or only all
AC # BD angles congruent (rectangles), a square
has 90° rotational symmetry.
Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor. Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor.
43 44

© Houghton Mifflin Harcourt Publishing Company

UNIT 3 Selected Answers UNIT 3 Selected Answers

MODULE 9 17. 1. JKLN is a parallelogram. (Given); MODULE 10 Quadrilateral KLMN is a parallelogram
2. KL # NJ (Opposite sides of a because both pairs of opposite sides are
Lesson 9.5 Properties and parallelogram are congruent.); Lesson 10.1 Slope and Parallel parallel.
Conditions for Kites and 3. JKMN is an isosceles trapezoid. Lines 5. U is at (2, 2)
Trapezoids (Given); 4. NJ # MK (Definition of Your Turn 1
Check slope of TU
Your Turn isosceles trapezoid); 3
1
4. slope of JK
5. KL # KM (Transitive Property of 3 1
4. 86q
7. BC # AD ; Parallel Postulate; BA; CE
+
Congruence); 6. KLM is an isosceles
1
slope of SV
3
triangle. (Definition of isosceles triangle) slope of LM
3 7. slope of FG is 0
CED; D 19. 14
slopes are the same, so JK is 3
CED # D; CD 21. mA 90q; mC 45q slope of GH
parallel to LM 2
ABCE; opposite sides mB mD 112.5°
23. Terrence mistakenly reasoned that ABC Quadrilateral JKLM is a trapezoid slope of HJ is 0
CE; CD because it is a quadrilateral with exactly
# ADC; only one pair of opposite angles slope of JF 3
ABCD; isosceles trapezoid in a kite are congruent, and they are one pair of parallel sides.
adjacent to the bisected diagonal, not the 3 Slope of FG HJ , so FG is parallel to
8. mR 103q bisecting diagonal. To find mABC: Since 5. slope of AB
2 HJ .
9. y 4 BAE and ABE are complementary,
mABE can be found by 90q 66q 24q. slope of BC 2 Slope of GH z JF, so GH is not parallel
10. 12.6
+
Then, since AEB # CEB so that + 3 to JF.
Evaluate ABE # ‫ס‬CBE, mABC is twice mABE, slope of CD
SA23

or 2(24°) 48°. 2 The quadrilateral has only one pair of

1. mABE 62q parallel sides, so it is a trapezoid but not a
25. Rhombuses and kites both have pairs of slope of DA 2 parallelogram.
3. mABC 95q congruent consecutive sides, but in a
Quadrilateral ABCD is a parallelogram
5. CDA rhombus, this is because all four sides are
because both pairs of opposite sides are
9. Sometimes
congruent. In a kite, two distinct pairs of 11. Always
AD # AD consecutive sides are congruent. The parallel.
CPCTC diagonals of both types of quadrilaterals 7. (4, 0) 13. Neither
are perpendicular. In a kite, exactly one 15. Trapezoid
7. MN 11.2 diagonal is bisected by the other, while in 8. (3, 1)
9. mE 43° a rhombus, each diagonal bisects the 17. p 5
Evaluate
other. Finally, in a kite, exactly one pair of q 9
11. XY 15.6 opposite angles are congruent, while both 1. She should use the fact that
13. XY 11.5 pairs of opposite angles of a rhombus are corresponding parts of congruent triangles 19. w 1
congruent. are congruent to show that KL ST, z 3
15. A. kite; B. kite; C. trapezoid; D. kite;
which will allow her to conclude that
E. trapezoid 21. No; since the slopes of RU and ST are
KL ST
. undefined, these two segments are
JL RT vertical. Since all vertical lines are parallel,
3. slope of KL 1 this means a pair of sides of the
quadrilateral are parallel, and therefore
1
slope of LM the quadrilateral is a trapezoid.
2
1
slope of MN 1 23. y x3
2
Selected Answers

1
slope of NK
2

Selected Answers
Selected Answers
Selected Answers

3 ( 3)
25. The slope of QP should be 2 . UNIT 3 Selected Answers
03
1
The slope of QR and the slope of
3 MODULE 10 42 2
7. slope of KL ;
3 1 4 3
RS . No sides of PQRS are parallel, Lesson 10.2 Slope and
2 1 2
so PQRS is not a trapezoid. Perpendicular Lines slope of LM ; slope of MN ;
4 3
27. Possible answer Plot the points A(4, 2), Your Turn slope of NK 3; A pair of opposite sides
B(2, 4), C(4, 0), D(0, 2). do not have the same slope, so these
2
midpoint of AB M(1, 3); midpoint 5. slope of DE sides are not parallel. Therefore, KLMN is
3
of BC N(3, 2); midpoint of CD not a parallelogram.
3
P(2, 1); midpoint of DA Q(2, 0); slope of EF 9. Always
JJJG 2
1 11. Never
slope of MN ; slope of NP 3; 2
4 slope of FG 13. Never
1 3
slope of PQ ; slope of QM 3. 15. PQ RS 10.5 and QR SP 31.5
4 3
slope of GD 17. PQ RS 6 and QR SP 36
Since opposite sides are parallel, MNPQ 2
1
is a parallelogram. This is true when Consecutive sides are perpendicular since 19. y x2
starting with any set of four points. the product of the slopes is 1. 2
Conjecture The quadrilateral formed by Quadrilateral DEFG is a rectangle 21. The coordinates of C are (0, 4).
connecting the midpoints of consecutive because it is a quadrilateral with four right The coordinates of D are (4, 2).
SA24

sides of a quadrilateral is a parallelogram. angles.

7. The quadrilateral is a parallelogram since
both pairs of opposite sides are parallel.
Evaluate
a
1. a.
b
b
b.
a
c. The product of the slopes of the lines
a b
is < 1.
b a
2
3. slope of KL slope of MN ;
3
3
slope of LM ; slope of NK 5;
2
KL A LM and LM A MN .
KLMN is a trapezoid with two right angles
because a pair of opposite sides are
parallel and two pairs of consecutive sides
are perpendicular.
5. The quadrilateral is a parallelogram since
both pairs of opposite sides are parallel.

© Houghton Mifflin Harcourt Publishing Company

0 3a 0 b
2 2
UNIT 3 Selected Answers and NB 9a 2 b 2 , 15. The distance between the ships is

NB and MC # NB . AB | 128 nautical miles.

so MC
AP | 64 nautical miles and
MODULE 10 So, PQ # RS , QN # SM , and
BP | 64 nautical miles.
Lesson 10.3 Coordinate Proof PN # RM . Therefore, PQN # RSM + + 1
by the SSS Triangle Congruence Since AP BP AB,
Using Distance with Segments 2
Theorem and PQN # RSM since
and Triangles corresponding parts of congruent triangles
P is the midpoint of AB .
are congruent. +
17. The perimeter of is JKL twice the
Your Turn
5. Possible answer Let the coordinates of Evaluate
+
perimeter of MNP.
7. No; these coordinates result in a triangle
the vertices be A(0, 4), B(0, 0), and 19. The proof is incorrect because the
1. The coordinates of point R are R (x1, y2). that is a right triangle, so the proof would
C(2, 0). assigned coordinates do not result in a
not hold for triangles in general.
y2 y1 and RQ x 2 x1
+
area of ABC 4 units2
PR
+PQR is a right triangle with hypotenuse 9. DE 37 , EF 8 , DF 17 ,
general right triangle. For a general right
triangle, the coordinates of R should be
The coordinates of D, the midpoint of AC , ST 37 , TU 8 , SU 17 . (a, b).
are (1, 2). PQ , PQ 2
RQ PR 2 2
21. Possible answer (0, 0), (0, 2s) and
+
area of DBC 2 units2 By substitution, PQ2 (x2 x1)2 (y2 y1)2
So, DE # ST , EF # TU , and DF # SU .
+
Therefore, DEF # STU by the SSS + (2s, 0). This results in an easier
+
The area of DBC is one half the area of ( x2 x1 )2 ( y 2 y1 )2 .
coordinate proof since there are no
+ ABC.
3. By the Midpoint Formula, the coordinates
Triangle Congruence Theorem and FDE
# UST since corresponding parts of
negative coordinates and more of the
coordinates are 0. If the proof does not
6. Possible answer Let the coordinates of congruent triangles are congruent. involve midpoints, and even easier choice
the vertices be A(0, 8), B(0, 0), and of X, Y, and Z are as follows.
+
C(8, 0) area of ABC 32 units2. § 2a 0 2b 0 ·
X¨ , X (a, b ) ,
11. The coordinates of U are U is (0, 0), (0, s), and (s, 0).

The coordinates of D, the midpoint of AC , © 2 2 ¸¹ § 1 3 4 0 ·

¨ 2 , 2 ¸ U(1, 2) and the
are (4, 4). © ¹
§ 2a 4a 2b 0 ·
SA25

+
area of DBC 16 units2 Y¨
© 2
,
2 ¸¹
Y (3a, b ) , coordinates of V are
§ 3 1 6 2 ·
+
The area of DBC is one half the area of V ¨¨ ¸¸ V(1, 4).
+
,
ABC. § 0 4a 0 0 · © 2 2 ¹
Z¨ ,
2 ¸¹
Z (2a, 0)
10. By the Distance Formula, AB 17 ,
© 2
JK 32 , KV 20 , JV 68 ,
BC 5 , AC 34 , DE 17 , By the Distance Formula,
PQ 32 , QU 20 , PU 68 .
EF 5 , DF 34 . So, AB # DE , XZ 2a a 0 b
2 2
a b
2 2
and
So, JK # PQ , KV # QU , and JV # PU .
BC # EF , and AC # DF . Therefore,
+ABC # +DEF by the SSS Triangle
YZ 2a 3a 0 b
2 2
a2 b2 . +
Therefore, JKV # PQU by the SSS +
Congruence Theorem and BCA # EFD Since XZ YZ, XZ # YZ and + XYZ is Triangle Congruence Theorem and
KVJ # QUP since corresponding parts
since corresponding parts of congruent isosceles. of congruent triangles are congruent.
triangles are congruent.
5. Assign coordinates as shown in the figure.
11. By the Midpoint Formula, the coordinates 13. JK 17, KL 34, and JL 17.
By the Midpoint Formula, the coordinates
§ 3 ( 1) 5 ( 1) · of M and N are as follows. Since JK and JL have the same length,
of M are M ¨ ,
© 2 2 ¸¹ § 0 2a 0 2b · JK # JL , and therefore the triangle is
M ¨ , M a, b ;
M (2, 2) and the coordinates of N are © 2 2 ¸¹ isosceles.
§ 4 2 5 ( 1) ·
N ¨ , § 2a 4a 2b 0 ·
2 ¸¹
N (3, 2).
© 2 N ¨ , N 3a, b
© 2 2 ¸¹
By the Distance Formula, PQ 40 ,
By the Distance Formula,
Selected Answers

QN 5, PN 45 , RS 40 , SM 5,
4a a 0 b
2 2
MC 9a 2 b 2
RM 45.

Selected Answers
Selected Answers
Selected Answers

11. The diagonals are perpendicular. ABCD is

UNIT 3 Selected Answers a rhombus.
The lengths of the diagonals are equal.
MODULE 10 9. The diagonals are congruent; KLMN is a ABCD is a rectangle. Since ABCD is both
rectangle. a rhombus and a rectangle, it is also a
Lesson 10.4 Coordinate Proof The diagonals are perpendicular; KLMN is square.
Using Distance with a rhombus. 13. C must be (5, 4)
Quadrilaterals Since KLMN is both a rectangle and a 15. D is at (4, 2)
rhombus, it is also a square.
Your Turn 17. BD 38 in.
10. The diagonals are not congruent. So
4. Possible answer CE 24 in.
PQRS is not a rectangle, and thus not a
BE | 36.88 in.
8 3 2 2 square.
2 2
AB 5;
The diagonals are perpendicular, so 19. Possible answer To know that the
5 0 0 0
2 2
DC 5 PQRS is a rhombus. reflecting pool is a parallelogram, the
congruent sides must be opposite each
Evaluate
3 0 2 0
other. If this is true, then knowing that one
2 2
AD 13;
1. A angle in the pool is a right angle or that
the diagonals are congruent proves that
8 5 2 0 5 2
2 2
BC 13 3. Slope of AB ; Slope of BC ; the pool is a rectangle.
2 5
Since the opposite sides have the same 21. Midpoint of AB (10, 50)
lengths, ABCD is a parallelogram. 5 2
Slope of DC ; Slope of DA
2 5 Since P is the midpoint of AB it lies on
§0 8 0 2·
5. Midpoint of DB : ¨ , 4, 1 The slopes of opposite sides are equal. AB . Therefore, points A, B, and P are
2 ¸¹
SA26

© 2 collinear and the port and the two cruise

This means that opposite sides are
ships are in a line.
§35 20· parallel, so ABCD is a parallelogram.
Midpoint of AC : ¨ , 4, 1
© 2 2 ¸¹ 5. In Evaluate 3 we proved that ABCD is a
23. Point C is at (3.1).

Since the diagonals share a midpoint, parallelogram. Point D is at (6, 3.5).

they bisect each other.
Slope of AC
7
; Slope of DB
3
25. Slope of OQ
a b 0 ab
1;
7. WX 4 0
2
3 0
2
25 5
3 7 a b 0 ab
§ 7 · § 3 · ab ab
Since ¨ ¸ ¨ ¸ 1, the diagonals are Slope of PR 1
XY 9 4
2
3 3
2
25 5 © 3 ¹© 7 ¹ ba 1 a b
perpendicular. So, ABCD is a rhombus.
Because the product of the slopes is 1,
5 9 0 3
2 2
YZ 25 5
2 5 5 0 the diagonals are perpendicular.
2 2
7. WY 74
0 5 0 0
2 2
ZW 25 5
2 5 0 5
2 2
XZ 74
Since all four sides have the same length,
WXYZ is a rhombus. So the diagonals are congruent.
30 1 9. midpoint of
8. Slope of WY § 2 5 5 0 · §3 5·
90 3 WY ¨ 2 , 2 ¸ ¨ 2,2¸
03 3 © ¹ © ¹
Slope of XZ
54 1 midpoint of XZ
(Slope of WY ) (Slope of XZ ) § 5 2 5 0 · §3 5·
¨¨ , ¸ ¨ 2,2¸
1 3 © 2 2 ¸¹ © ¹
< 1
3 1
Since they share a midpoint, the
So, WY is perpendicular to XZ . diagonals bisect each other.

© Houghton Mifflin Harcourt Publishing Company

UNIT 3 Selected Answers UNIT 4 Selected Answers

MODULE 10 MODULE 11 11. Rigid motions preserve angle measure,

Lesson 10.5 Perimeter and Area

area of +ABD A 1
(BD)(AE); betweenness, and collinearity. Dilations
2 Lesson 11.1 Dilations preserve all of these except distance.
on the Coordinate Plane
area of +CBD A 1
2
(BD)(CE) Your Turn 13. It is a dilation
1
2
.
Your Turn 3
5. It is a dilation. The scale factor is . 15.
4. Area of JKLM 15 1 1 2
area of ABCD A (BD)(AE) (BD)
6. 34 units2 2 2 6. This is not a dilation.
1 1
7. 32 units2 (CE) (BD)(AE CE) (BD)(AC) 8.
2 2
8. area of wastage 0.625 in2
Evaluate
1. 14.8 units
3. 19.6 units
5. 34 units2
7. 32 units2
9. 27.5 units2 1 to 2.
11. cost of fencing $1493.90 17. 3 to 1.
13. B, C, and D 19.
15.
SA27

1 to 3.
Evaluate
1. The ratios of the lengths of the
corresponding sides are equal.
3. No, this is not a dilation.
5. (c)
P | 10.2 units
2
7.
A 4.5 units
17. JK 18 3 2 ; KL 45 3 5;
LM 45 3 5 ; JM 18 3 2
So JK # JM # KL # LM , and therefore
JKLM is a kite. a. Perimeter is 12 units, Area is 6 square
A 27 units2 units

19. Since ABCD is a kite, AD # AB and b. Perimeter is 36 units, Area is 54 square

9.
units
CD # CB. The kite can then be divided
into two triangles with the same base. perimeter 3
c. scale factor
Therefore, AE is an altitude for ABD + perimeter 1
Selected Answers

with base BD , and CE is an altitude for area 9

+ CBD with base BD . area 1
scale factor squared

53 54

Selected Answers
Selected Answers
Selected Answers

15. Since the base and the height of the

UNIT 4 Selected Answers triangles are equal, no dilation has
occurred. The triangles are 180q rotations
MODULE 11 of each other around the center of the
11. Step A Dilate ABCD with center of rectangle (where the diagonals intersect).
y The triangles are similar to each other.
Lesson 11.2 Proving Figures are dilation A and scale factor , producing
x
Similar Using Transformations square AcBcCcDc. Square ABCD has four 17. 180 in., or 15 ft
Your Turn sides of length x. Square AcBcCcDc will
19. D
y
3. Yes have four sides of length ( x ) y . These
x AX DX
4. No, the angles are different. are the same side lengths as EFGH. The 21. Step A Rearrange into
BX CX
5. Yes angles are all 90q in each square, so AX BX AX
8. Reflection (x, y) o (x, y) AcBcCcDc is congruent to EFGH. . Let K .
DX CX DX

+
§1 1 · Step B Translate AcBcCcDc with
JJJJG a
Dilation (x, y) o ¨ x , y ¸ Step B Rotate DXC 180q around point
©3 3 ¹ translation along the vector AcE, producing
AsBsCsDs. Through this translation, As is X so that DXC coincides with AXB.
9. Rotation (x, y) o (x, y)
mapped to E. It may be true that Bs is
Step C Dilate DXC by a factor of k about
§3 3 · mapped to F, Cs is mapped to G, and Ds
Dilation (x, y) o ¨ x, y ¸ the center X. This dilation moves the point
©2 2 ¹ is mapped to H. If not, rotate AsBsCsDs
D to A, since k(DX) AX, and moves C to
about E so that Bccc is mapped to F.
Translation (x, y) (x 3, y 1.5) B, since k(CX) BX. Since the dilation is
Then, C ccc lands on G and Dccc lands on H. through point X and dilations take line
10. Translate JKLMN right 7 units so that 13. Step A Dilate +ABC with center of segments to line segments, DXC is +
SA28

J maps to V.
HJJG x +
mapped to AXB. So DXC is similar to +
Reflect JKLMN across JN.
dilation B and scale factor K
a
, producing
+ AXB.
Dilate JKLMN with center J and scale + +
AcB cC c. After the dilation AcB cC c will
x 23. Step A Draw the bisector of AXC.
1 have sides of length B cC c ka (a ) x,
factor .
2
x bx
a
+
Step B Reflect CXD across the angle
A cC c kb (b ) , and bisector. This maps XC to XA. Since
Evaluate a a
x cx reflections preserve angles, it also maps
1. EFGH can be mapped onto ABCD with
a dilation of 4 with center at the origin.
A cB c kc
a
(c )
a
. These are the XD onto XB. Since XCD # XAB, + +
same side lengths of XYZ. By SSS + the image of CD is parallel to AB.
3. JKLMN is similar to JPQRS.
Triangle Congruence, AcB cC c is +
5. Reflection (x, y) o (x, y)
congruent to XYZ. + Step C Dilate +
XCD about point XHJJG
. This
Translation (x, y) o (x, y 6)
+
Step B Translate AcB cC c along the
JJJJG
moves the HJJG
new point C to A. Since AB is
HJJG HJJG
7. Reflection (x, y) o (x, y) +
vector BcY, producing AccB ccC cc. Through
parallel to CD, the new CD, moves to AB .
Therefore, the new point D is mapped to B
§1 1 ·
Dilation (x, y) o ¨ x, y ¸
this translation, Bs is mapped to Y. It may and +
XCD is mapped to XAB. So +
©2 2 ¹ be true that As and Cs are mapped to X
+
and Z. If not, rotate AccB ccC cc about Y so
+XCD is similar to XAB. +
9. Reflection (x, y) o (x, y)
that Bccc is mapped to Y. Then, Cccc is
Dilation (x, y) o (3x, 3y) mapped to Z.
Translation (x, y) o (x 1, y 2)

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UNIT 4 Selected Answers UNIT 4 Selected Answers

MODULE 11 MODULE 11 15. The student did not compare
corresponding sides of the two triangles.
Lesson 11.3 Corresponding Lesson 11.4 AA Similarity
Parts of Similar Figures of Triangles AB is the shortest side of +ABC, so its
corresponding side is DE the shortest
Your Turn
5. The similarity transformation is
Your Turn
6. AB 10 feet
+
side of DEF. The ratios
AB BC
,
DE EF
and
(x, y) o (1.5x, 1.5y). AC
7. PQ 11 are equal, so the triangles are similar
6. No DF
10. The two triangles cannot be proven
8. 12 x by SSS Similarity.
similar. Although the two given sides are
9. y 1.4 in proportion, there is not a pair of 17. + +
ABD a ACB by the AA Triangle
Evaluate included congruent angles. Similarity Theorem. AB 8
1. Yes; yes; yes 11. By the Pythagorean Theorem, NO 6 +
19. Apply a dilation to ABC with scale
3. No; no; no and GH 4, so
HJ
NO MN
GH GJ
MO
1
2
. factor k
XY
AB
and let the image of ABC +
5.
BC
NK
CD
KL + +
MNO GHI by the SSS Triangle +
be AcBcCc. Then Ac # A. It is given
Similarity Theorem that A # X, so by transitivity Ac # X.
7. X # X , Y # V , and Z # W
XY
1 Evaluate Also AcBc k < AB <AB XY and
9. AB
5 4.5 3
SA29

1. or 1.5 XY XZ
3 2 AcCc k < AC < AC < AC XZ.
11. XZ 42 AB AC
13. YZ n2 4n 4 2.1 3
or 1.5 +
Therefore, AcBcC # XYZ by SAS +
1.4 2 Congruence. So a sequence of rigid
15. x
17. a
7
2; b 3 3.9 3
or 1.5
+
motions maps AcBcC to XYZ. The +
2.6 2 dilation followed by this sequence of rigid
19. Yes, corresponding angles are congruent. motions shows that there is a sequence
The ratio of corresponding sides is 3. By the AA Triangle Similarity
constant. + +
Theorem, ABC a DEF.
of similarity transformations that maps
+ +
ABC to XYZ. So ABC a XYZ. + +
21. x o 48x 5. +ABC a +EDC by the AA Triangle 21. Possible Answer For XB to be as small
23. Possible answer (x, y) o (2x, 2y) o Similarity Theorem.
(x 14, y 8) or (x, y) o (x 7, y 4) o as possible, it should correspond to the
(2x, 2y)
7. The triangles are similar by the AA
Triangle Similarity Theorem. It is not
+
shortest side of ABC, which is AB.
25. The statement is false. A rectangle Thus, X corresponds to A.
possible to find the indicated length.
measuring 5 units by 2 units is not similar
to a rectangle measuring 4 units by 9. The triangles are similar by AA Similarity.
3 units. QR 0.96
27. The statement is false.
11. The ratios are not equal, so the two
triangles are not similar.
13. +ABC a +BDC by SSS Similarity.
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UNIT 4 Selected Answers UNIT 4 Selected Answers

MODULE 12 MODULE 12 11.

Lesson 12.1 Triangle Lesson 12.2 Subdividing a

Proportionality Theorem Segment in a Given Ratio
Your Turn Your Turn
§ 24 · 960 5. (1, 0)
5. 40 ¨ ¸ DG; DG 30
© 32 ¹ 32 6. (2, 7)
1 9.
6. RN 6 13. 12.5
4
VT 90 5 VU 67.5 135 5 15. 9
9. , 17. Points P and Q are the same point.
TR 72 4 US 54 108 4
VT VU 2
, so RS & TU. Sample explanation: Point P is of the
TR US 3
1
distance from A to B. Point Q is of the
Evaluate 3
10.
1 distance from B to A. This means the
3. 5 points lie at the same location along the
3
line segment.
SA30

5. 20
§ 1· § 1·
7. MN & QR 19. ¨ 1, ¸ and ¨ 1, 1 ¸
© 3¹ © 3¹
9. LN & AB 21. B
11. 0.24 kilometer E
13. Determine a line 23. Point P must be closer to point B than to
Evaluate point A, so the coordinate of point P
Triangle Proportionality Theorem
1. 20 should be positive.
Triangle Proportionality Theorem
3. 17 Sample answer: The student found the
Transitive 2
5. (6, 1) of the
coordinate of the point that is
15. C and E 3
7. (3, 3)
17. a. midsegment distance from B to A.
9.
median §8 1·
25. If point P is on RS, ¨ , 2 ¸ . There is
substitution © 9 9¹
Triangle Proportionality also a point P, not on RS, that lies
b. Can repeat the same process twice to beyond point S (32, 11).
2
show that Z is also located of the
3
distance from vertices A and B of
+ ABC to the midpoints of their
opposite sides.

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UNIT 4 Selected Answers UNIT 4 Selected Answers

MODULE 12 MODULE 12
Lesson 12.3 Using Proportional Lesson 12.4 Similarity in Right
Relationships Triangles
Your Turn Your Turn
3. The tree is 21 feet tall. 6. 12
5. 24 meters 7. 2 15
Evaluate 10. 35
1. Indirect measurement
Evaluate
3. 52 ft
5. 52 1. +PQR a +SPR a +SQP
7. 312 meters 3. +XYZ a +XWY a +YWZ
9. 35.2 meters 5. 6

1 7. 70
11. 10 feet, or 10 feet 4 inches
3 3 5
9.
13. 304 inches or 25 feet 4 inches 10
15. 69 inches or 5 feet 9 inches 11. x 20 3
17. +XYZ a +VUZ by the SAS Similarity
SA31

XY XZ y 10 21
Criterion, so . Then
VU VZ z 20 7
XY 800 13. e
, so XY 1,000 ft.
500 400
15. d
19. A 17. a
B 19. 2.4
E
21. | 4.62
21. AB is the shortest side of right ABE, so + 23. 4 inches and 16 inches
corresponding side DC of DCE must +
be shorter than DE, that is, DE 200.
The triangles are similar, but Andy must
have used the wrong proportion. The
d 200
correct proportion is , so
147 300
§ 200 ·
d 147 ¨ 98. The distance across
© 300 ¹¸
the river is 98 ft.
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UNIT 5 Selected Answers UNIT 5 Selected Answers

MODULE 13 MODULE 13 25. 0 BC AB
0

BC AB
0
AB AB AB
Lesson 13.1 Tangent Ratio Lesson 13.2 Sine and Cosine sin A 1
Your Turn Ratios The same argument shows that
0 cos A 1.
5 Your Turn
5. | 0.42 27. Two segments A to the same line are & to
12 8. sin 17q | 0.292
12 9. cos 45q | 0.707 each other, so BC & DE. By the Triangle
6. 2.4
5 11. | 16.5 Proportionality Theorem, BC divides
7. 3.6 ft 13. mY | 53q sides AD and AE of ADE +
proportionally
8. 26q 14. mZ | 37q
BD CE BD CE AB BD
Evaluate 1 1
Evaluate AB AC AB AC AB
1. 0.40 1. 0.866 | cos 30q AC CE AD AE AD AB

0.40; The ratios are the same. AC AB AC AE AC
12
3. tanF 1.8 3. | 0.923
13 These ratios determine cos A, and since
tanD 0.56 12
they are equal, cos A is the same when
5. | 0.923 calculated in either triangle.
5. tanR 0.16 13
SA32

tanP 6.33 7
7. | 0.28
7. tanA 2.0 25
9. QR 4.0 9. 9.0 | x
11. PQ 0.79 11. | 54.1
13. AB 8.4 13. | 64q
15. mA 66q 15. | 47q
17. mB 43q 17. A # A and ABC # ADE, since both
38 are right angles.
19. PQ
tan75q
10.2
+
By AA~, ABC~ ADE, so +
corresponding sides are proportional
21. 60q
AC BC (BC )( AE ) AC AE
23. The student’s calculations are correct only AC
if the triangle is a right triangle. AE DE DE BC DE
25. 10.2q with the first road and 79.8q with the These ratios determine sin A, and since
second road. they are equal, sin A is the same when
calculated in either right triangle.
19. 9.5 ft
21. 66q and 24q
Check 66q 24q 90q
23. So mBAC 2mBAD
§ 100 ·
2sin1 ¨ ¸ | 9.7q
© 1185 ¹

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UNIT 5 Selected Answers UNIT 5 Selected Answers

MODULE 13 19. Given BC AC 3. Since tan A
BC
, MODULE 13 17. Suppose AB # DE and AC # DF . Solving
AC either of these right triangles determines
Lesson 13.3 Special Right use an inverse tangent ratio Lesson 13.4 Problem Solving the length of the hypotenuse in the same
Triangles with Trigonometry way, e.g., using the Pythagorean
BC AC 3 Theorem, so BC EF and therefore, by
tan A 3
Your Turn AC AC Your Turn +
SSS #, ABC # DEF. +
5. 2 3 JK m A tan 1
3 60q 3. | 50.3 mm2 Suppose B # E. The given
JL 6 A and B are complementary, so 4. | 8.6 cm2 corresponding side lengths allow the
unknown sides to be calculated in the
mB 90q mA 90q 60q 30q. 6. | 65 m
6. PR 2 6 same way using trigonometric ratios, so
21. 10 x 7. | 73 m that all corresponding side lengths are
PQ 4 3
23. | 4,160 cm2 8. 63q equal and therefore all corresponding
sides are congruent. Again, by SSS #,
9. AB
AC
5 2
5 2
25. No; if the two shorter side lengths are odd,
then their squares are odd, because the
10. JK | 6.71; JL | 4.47; KL | 8.06 + ABC # DEF. +
square of an odd number is always odd. mJ 90q. 19. sin 2T 2sinT cosT
5 2 2 But the sum of their squares is even, mK | 34q.
sin 45q cos 45q because the sum of two odd numbers is
10 2 mL | 66q
always even. Therefore the sum of the
tan 45q 1 squares of the two shorter side lengths Evaluate
11. PR 30 cannot itself be the square of an odd
1. | 5.8 cm2
12. YZ 60 number.
3. AC | 1.5 cm
SA33

Evaluate mA | 61q

1. No; 3 3 : 6 : 6 3 z 1: 3 : 2 mB | 29q

3. Yes; 4 3 : 12 : 8 3 1: 3 : 2 5. mR 38q

37.0 mm | PR 28.9 mm | PQ
5. 9 2 AB BC
7. mJ 90q; JK | 3.6; JL | 10.8; KL | 11.4;
7. UV 6 3 mL | 18qmK | 72q
6 VW 9. | 12,700 yd2
9. 7 2 AC 7 2 BC 11. 1.8 ft | AB 29.9 ft | BC
13. Z(5, 1)
11. JL 4 6 KL 4 3
mZ 90q
13. AB 65
XY 5 2, XZ 5, YZ 5
15. Not possible 9 21 2 2
522; 522 is not a
perfect square. mX mY 45q.

17. x 6 3 15. mCAE | 37qmAEC | 27q

mACE | 27q
Area | 13.5 mm2
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UNIT 5 Selected Answers UNIT 5 Selected Answers

13. mC 10qa | 51.8; b | 35.5 5. p | 25.0; mR | 42.1q; mQ | 14.9q
15. No triangles are possible. 7. mP | 25.2q; mR | 58.4q; mQ | 96.4q
17. There are two triangles. 9. mR | 93.8q; mQ | 29.9q; mP 56.3q
mB | 39.6qmC 105.4qc | 15.1 m 11. b | 6.5; mA | 64.9q; mC 37.1q
Second triangle 13. mA |94.8q; mB | 39.8q;
mB |140.4qmC 4.6qc | 1.3 m mC 45.4q
19. a. | 123.7 mi 15. mA | 56.6q; mB | 46.4q; mC 77.0q
b. Possible answer Calculate the area 17. 3.8 miles
using the sine formula. Use the value of
the area to solve for h using the other 19. D. 45q
form of the area formula. 21. mA | 45q; mB | 60q; mC | 74q
| 120 mi 23. 1.2 km
21. | 5.2 mi 25. Abby did not make an error. The three
23. Transfer corresponding angle measures lengths cannot form a triangle because
from each triangle to the other. a b c.
25. f | 2.1; e | 3.7

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UNIT 6 Selected Answers UNIT 6 Selected Answers

MODULE 15 MODULE 15 15. Cannot be inscribed in a circle.
17. 105q
Lesson 15.1 Central Angles and Lesson 15.2 Angles in Inscribed
Inscribed Angles Quadrilaterals 19. mM 99q; mJ 108q; mK 81q;
mL 72q.
Your Turn Your Turn 21. a. 100 square feet
p 11q
4. mBD 7. b. (50S 100) square feet
7. mABE 37q
Evaluate
1. DE, EF ; DEF; DCF

3. DF , DG, EF , EG; DGE, DFE, FDG,

FEG; none
5. 42q
Evaluate
7. 141q
1. No. A and C are still supplementary.
9. 321q Opposite angles of an inscribed
11. 48q quadrilateral are always supplementary.
13. 85q 3. Yes; mHGJ 115q;
15. 236q H and J measure 90°
SA35

17. 20q 5. 95q

19. 324q 7. Yes. Since the opposite angles of
quadrilateral GHIJ are supplementary, the
21. C, D, E, G, H quadrilateral can be inscribed in a circle.
23. 135q 9.
25. 144q

Construct the perpendicular bisector of

the diameter. Connect the points where
the perpendicular bisector and the
diameter intersect the circle.
11. mA 70q; mB 115q; mC 110q;
mD 65q
13. mV 101q; mT 79q; mU 86q;
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mW 94q

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UNIT 6 Selected Answers UNIT 6 Selected Answers

MODULE 15 MODULE 15
Lesson 15.3 Tangents and Lesson 15.4 Segment
Circumscribed Angles Relationships in Circles
Evaluate Your Turn
1. 90q 3. 1.93 | x
3. The tangents are perpendicular. CB | 15.93
5. approximately 1400 mi AD 12 (given)
7. 3 units 6. x 9
(3, 1) PT 13; PR 10.4
x 3 7. x 3.5
9. AC 32 HL 10; NL 12
AB 32 9. 400 yards
11. P and Q are supplementary angles. Evaluate
13. mP 45q 1. The product of the lengths of the
15. 28q segments on one chord will equal the
product of the lengths of the segments on
SA36

17. It is not possible.

the other chord RV < VS TV < VU.
p
19. m MJ 82
3. DF < CF BF < EF
21. ABCD is a kite.
23. From point B, construct two tangents to 5. y 6. DE 7; FG 8.
circle O. From point C, construct two 7. a. PM 6 in.
tangents to circle O. Label the point of
b. PQ 12 in.
intersection of the tangent from B and the
tangent from C as point D. Triangle DEB 9. I agree.
is circumscribed about circle O. 11. y 14.3; HL 24.3; NL 27
(Also, triangle FEC is circumscribed about 13. z 2 21
circle O.)
15. y 4 10
17. B
19. 1770 feet
21. x 18
23. about 3.2 in
25. Method 1 By the Secant-Tangent
Product Theorem, BC2 12 < 4 and so
BC 48 4 3.
Method 2 Because a line tangent to a
circle is a line A to the radius, ABC is a
right angle. By the Pythagorean Theorem,
BC2 42 82. Thus BC2 64 16 48
and BC 48 4 3.
Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor. Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor.
71 72

UNIT 6 Selected Answers UNIT 6 Selected Answers

MODULE 15 mACD
1 p 1 p
m AD mBD. Thus by the MODULE 16 25. about 12 cm
2 2 27. The range of areas is about 177 square
Lesson 15.5 Angle Distributive Property, mACD Lesson 16.1 Justifying miles to about 314 square miles.
Relationships in Circles 1 p
2

m AD mBDp .
Circumference and Area of
a Circle
29. Disagree; the total distance is 5280 <
Your Turn 12 63,360 inches and the number of
23. Case 1 Assume AB is a diameter of the revolutions equals the total distance
4. 86q Your Turn
p 180q, and ABC is a
circle. Then mAB divided by the circumference, which is
5. 34q 4. about 6.4 feet
right angle, because a diameter is 63,360
6. 122q perpendicular to a tangent at the point of 5. about 10.2 feet | 775.7 revolutions.
26S
7. 119q 1 p 9. 254 ft2
tangency. Thus mABC m AB. 1
8. 47 2 31. The circumference of circle E is the
Evaluate 4
9. 45q Case 2 Assume AB is not a diameter of 1. A regular polygon with 40 sides. circumference of circle C because the
the circle. Let X be the center of the circle 1
10. 116q 3. | 150.7 cm radius of circle E is the radius of
and draw radii XA and XB. Since they
+
4
Evaluate are radii, XA # XB so AXB is 5. | 56.5 in. circle C.
1. The measure of the angle will be half the isosceles. Thus XAB # XBA, and 7. | 301.4 ft
measure of its intercepted arc; the 2mXBA mAXB 180q. This means 9. 19.625 m2
Tangent-Secant Interior Angle Theorem. 1
that mXBA 90q mAXB. Because 11. 78.5 in2.
3. 64.5q 2
a line tangent to a circle is perpendicular 13. 5 r; A S (5)2| 78.5 ft2
5. 135q
to the radius at the point of tangency, 15. 11 r; A S (11)2| 379.9 ft2
SA37

7. 56q XBC is a right angle, so mXBA 17. C | 15.7 ft.

9. 90q mABC 90q or mABC 90q
A | 19.6 ft2
mXBA. By substitution,
11. 57.5q
§ 1 · 19. Possible answer The circular table would
13. 18q m ABC 90q ¨ 90q m AXB ¸ . fit at least as many people as the
© 2 ¹
15. a. 110q rectangular table. At the rectangular table,
1
Simplifying gives mABC mAXB. 2 people would fit at each of the 4 ft sides
b. 116q 2 and 3 people would fit at each of the 6 ft
17. 90q mAXB m AB p because AXB is a sides, for a total of 10 people. Each
1 q person would have 2 ft of space. The
19. 142q central angle. Thus mABC m AB. circumference of the circular table is
2
21. Since though any two points, there exists C Sd S(6) | 18.8 ft. If 11 people sat at
exactly one line, then BD can be drawn. the circular table, each person would have
By the Exterior Angle Theorem, 18.8
mABD mACD mBDC, so | 1.7 ft, or about 1 ft 8 in. of space.
11
mACD mABD mBDC.
21. a. 267 ft
1 p
mABD m AD by the Inscribed Angle b. | 6 min
2
1 p 23. The calculation shown in A is incorrect
Theorem, and mBCD mBD because because the diameter, instead of the
2
the measure of an angle formed by a radius, is used to find the area.
tangent and a secant intersecting on a
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circle at the point of tangency is half the

measure of the intercepted arc. By subst.,

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UNIT 6 Selected Answers UNIT 6 Selected Answers

MODULE 16 19. 160q MODULE 16 27. Disagree; the original area is
m
< Sr 2
360
21. 15 in.
Lesson 16.2 Arc Length and Lesson 16.3 Sector Area m
23. The student used the diameter instead of and the new area is < S (2r )2 or
Radian Measure the radius in the circumference formula. Your Turn 360
2S m. m
Your Turn 4. | 0.32S 4< < S r 2, so the area becomes
360
3. | 37.7 in. 25. | 21.5 ft | 1.01 cm 2
4 times greater.
S 5. 36S 29. Arc length requires using the
7. radians
2 | 113.10 mm2 circumference of the circle, whereas area
requires the circle's area. In each, the
S 6. | 12, 723 ft2
8. radians central angle is used to find the fraction of
4 the circumference or area of the circle.
Evaluate
Evaluate 1. sector
1. The circumference of the circle. 3. A S r 2, where A represents the area and
3. The diameter of the circle. r represents the length of the radius.
5. | 31.4 m 5. A 63S | 197.7 cm2
7. | 5.8 cm S
7. A | 0.3 mm2
9. 34.8 m 9
SA38

11. 10 m 9. A 100S | 314.2 in2

2 11. 18 pieces
13. S
9 13. | 628.3 in2
5S 15. 24S
15.
9 | 75.40 cm2
17. For 0q: 0 47
17. S
S 90
For 30q:
6 | 1.64 ft 2
S x 9
For 45q: 19. Let mAOB < S (6)2
xq, then S.
4 360 2
S Solving for x shows that mAOB 45q.
For 60q:
3 21. Buying a party wheel is the better value.
S 23. B, F
For 90q: 25. 2.57 in2
2
2S
For 120q:
3
3S
For 135q:
4
5S
For 150q:
6
For 180q: S

UNIT 6 Selected Answers UNIT 6 Selected Answers

MODULE 17 7. center (3, 4); radius 3 MODULE 17 Evaluate
1 2 1 2
Lesson 17.1 Equation of a Circle Lesson 17.2 Equation of a 1.
4(2)
x
8
x

Your Turn Parabola

5. (x 4)2 (y 3)2 16 Your Turn
6. (x 1)2 (y 1)2 3 1 2
6. y x
8. Center (1, 4), radius 2 4

9. center (3, 3); radius 2

1 1 2
3. x2 x
4( 5) 20

9. Center (0, 2), radius 3 8. (x 3)2 12(y (1)) or (x 3)2

12(y 1)

11. The point 2, 3 does not lie on the circle
SA39

because its x- and y-coordinates do not

satisfy the equation
3
2
22 43 7 z 9.

13. The graph of the circle intersects the 5. Focus (0, 6); directrix is y 6.
x-axis at both (3, 0) and (3, 0).
§ 1· 1
12. The point
18, 4 does not lie on the 15. (x 1)2 (y 2)2 4 7. Focus ¨ 0, ¸ ; directrix is y
© 2¹ 2
.

circle because the point’s x- and 17. The point does lie on the equation
because its x- and y coordinates satisfy 9. (x 2)2 16(y 5) 9. (x 2)2 8 (y 5)
y-coordinates do not satisfy the equation
of the circle. the equation 02 4(0) 42 4(4) 0
13. The point (4, 4) lies on the circle 19. False; r 7
because the point’s x- and y-coordinates 21. False; the center is (6, 4), which is in the
satisfy the equation of the circle. fourth quadrant.
Evaluate 23. False; the equation is x2 y2 32 9
1. (h, k) 25. a. (x 3)2 (y 3)2 10
3. (x 1)2 (y 3)2 64 b. It does lie on the circle because x 0
5. (x 9)2 y2 3 and y 4 satisfies the equation
(x 3)2 (y 3)2 10
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11. equation of parabola (x 4)2 20(y 3)

UNIT 7 Selected Answers

MODULE 18
Lesson 18.1 Volume of Prisms
and Cylinders
Your Turn
4. 88 cubic units
5. 72k 3 cubic units
13. Focus is (2, 3) and directrix is y 5. 6. 20,357.5 in.3
15. Focus is (5, 3.5) and directrix is y 3. 7. 20 x2 (x 2) cm3
1 2 1 2 1 2 1 2 r
17. y x ; y x ; y x ; y x 9.
4 8 12 16 3S h r
As the focus and directrix move apart, the
parabola is vertically compressed. Evaluate
1 2 1. A. rectangular prism; C
19. y x B. cylinder; A
24
21. 3 C. triangular prism; B
23. a. x t 0, y t 0 3. 810 yd3
SA40

b. Since 11.25 ! 6, Amber’s ball went 5. | 2580.72 m3

higher. 7. | 1130.97 ft3
Amber's ball traveled the farthest 9. | 1209.1 ft3
horizontal distance. Amber’s x-intercepts 11. 792 cm3
are 14.25 and 15.75. Her ball traveled
1.5 feet horizontally. James’s x-intercepts 13. | 2 yd3; $50
are 11.5 and 12.5. His ball traveled 1 15. | 14.9
foot horizontally. Because 0.9 of a candle would not make
an entire candle, 14 candles.
17. 2 in. radius h | 1.1 in.
1.5 in. radius h | 2.0 in.
19. V 840 cm3
21. The cylinder’s volume is greater than the
rectangular prism’s volume by 5.6 in3.

UNIT 7 Selected Answers UNIT 7 Selected Answers

MODULE 18 MODULE 18
Lesson 18.2 Volume of Lesson 18.3 Volume of Cones
Pyramids
Your Turn
Your Turn 2. r 1.2 in.
4. D 10.5 h | 3.710 in.
3
5. 192 in 3. | 5.596 in3
6. 150 ft2
4. | 3.1 fl oz
Evaluate 6. 43.769 cm3
1 7. | 29.2 cm3
1. The volume of the square pyramid is
3
Evaluate
the volume of the square prism.
1. It will take three cones to fill the cylinder
1
3. The volume of PABCD is the volume of with sand. Because the volume formula
3 for a cylinder is V Sr2h, and the volume
the cube. 1 2
formula for a cone is V Sr h, the
5. 136 in3 3
7. 96 cm3 1
volume of a cone is the volume of the
9. 2343.8 cm3 3
cylinder.
SA41

11. The volume doubles.

3. | 51.0 ft3
13. h 14 m
5. 1440S in3
g
15. | 10.5 7. | 703.7 in3
cm3
17. The volume would be 216 times larger; 9. | 42.9 m3
dividing the volume of the enlarged 11. A. B
pyramid by the volume of the original B. D
pyramid gives 216.
3 C. C
19. 10, 125 ft
D. A
375 ft3
13. B; E
21. Let s be the length of a side of the
pyramid’s base. Then the area of the base 15. 904.8 in3
1 17. a. | 33.5 in3
is s 2, and s 2 (5) 60. Solving shows
3 b. | 134.0 in3
that s 6 ft. c. $5; the large size holds 4 times as
much.
19. 10S 3 cm
21. h 9 in.
23. The calculation show in A is incorrect
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because it uses the slant height of the

cone instead of the height.

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UNIT 7 Selected Answers UNIT 7 Selected Answers

MODULE 18 MODULE 19 Evaluate
1. Figure B
Lesson 18.4 Volume of Spheres Lesson 19.1 Cross Sections and 3. triangular pyramid
Your Turn Solids of Rotation 5. square
3. 10,305.9947... m3 Your Turn 7. (isosceles) triangle
4. | 10,305,995 L 3. The cross section is a rectangle. Its base 9. rectangle
is smaller than the diameter of the
6. | 44.9 in3 11. circle
cylinder, and its height is the same as the
cylinder’s. 13. cone
Evaluate
1. The volume of the cylinder is 1.5 times the 5. A composite of two cones with a cylinder
volume of the sphere. in between.
3. 696.9 ft3
5. 1333.3S cm3
7. 972S in3
15.
560S 3
9. in
3
11. | 1441 mm3
SA42

13. 243 mm3

15. | 358,000,000 BTU
17. A 6.
19. Volume of Venus Volume of Mars disc (circle and its interior)
Volume of Earth 17. Yes
3
21. 14,294 in 19. For a hexagon, cut parallel to the ground;
23. approximately 1 in. for a rectangle, cut perpendicular to the
ground.
21. Equilateral triangle; each side of the
triangle has length V 2 .

a sphere

UNIT 7 Selected Answers UNIT 7 Selected Answers

MODULE 19 MODULE 19 23. a. 500S cm2 ; b. 100S cm2 ; c. 25S cm2 ;
d. 425S cm2
Lesson 19.2 Surface Area of Lesson 19.3 Surface Area of
25. A triangle is formed with 2 vertices at the
Prisms and Cylinders Pyramids and Cones midpoints of opposite sides of the square
Your Turn Your Turn base and the third vertex at the vertex of
the pyramid. The side lengths of the
4. 410 in2 3. lateral area | 102.5 ft 2
triangle are A, A, and s, the edge length of
2
5. 139.68 in surface area | 138.5 ft 2 the base. By the Triangle Inequality
7. paper needed | 565.5 cm2 4. lateral area 36 cm2 Theorem, A A ! s, so 2A ! s. Therefore
aluminum needed | 791.7 cm2 surface area | 42.9 cm2 1
A! s.
8. paper needed | 18,095.6 mm 2 2
7. paper | 47.1 in 2
aluminum needed | 26,238.6 mm 2 paper and plastic | 75.4 in 2
10. 344 in2 8. paper | 11.8 in 2
11. |628.3 mm 2 paper and plastic | 18.8 in 2
Evaluate 10. | 192.9 m 2
1. L 72 ft 2 11. |234.5 m 2
2
S 142 ft Evaluate
3. L 200 cm2 1. B; D
SA43

3. In an oblique cone, the distance from a

S 250 cm2
point on the edge of the base to the vertex
5. L 24S ft 2
is not the same point for each point on the
base.
S 42S ft 2
5. L 544 ft 2 ; S 800 ft 2
7. S | 953.1 ft 2
7. L 900 cm2 ; S | 1592.8 cm2
9. S | 352.0 cm2
11. The 23 inch bulb will produce more light. 9. L 2564.5S cm2 ; S 2696.75S cm2

13. L 80S m 2 11. L 444S in 2 ; S 588S in 2

S 208S m 2 13. S 1056S m 2

15. h 11 ft 15. L 324 cm2 , S 936 cm2
17. 4 gallons; $100 17. S 456S m 2
19. 198 cm2 19. d 6 in.
21. The triangular-prism-shaped frame will 21. Possible Answer 528, 000 ft 2
take more plastic.
Selected Answers

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UNIT 7 Selected Answers UNIT 7 Selected Answers

MODULE 19 19. About twice as great. MODULE 20 19. The volume is multiplied by 27.
21. S 108S cm2 21. Multiply the base or height by 5; Multiply
Lesson 19.4 Surface Area of Lesson 20.1 Scale Factor
23. The surface area of the sphere is equal to the base and height by 5 .
Spheres the lateral area of the cylinder. Your Turn 23. The area of the new circle is
Your Turn S 4. Possible answer The original area is 36. A (9S x 2 54S x 81S ) in.2
25. s 2r < , or s | 1.4r After the transformation the area is 144.
3. | 277.0 in2 6
When the base length changes by a factor
4. | 25.8 in2 of 4, the area changes by a factor of 4.
5. ratio of amounts of material | 10.736 5. Original area is 20. After the transformation
ratio of circumferences 3.277 the area is 100. When the width changes
square of ratio of circumferences 10.743 by a factor of 5, the area changes by a
| ratio of amounts of material factor of 5.
7. 40S in2 7. Possible answer The original
circumference is 24S, the original area is
8. 32S in2
144S. After the transformation, the
9. |326.7 in2 circumference is 6S, and the area is 9S.
The circumference changes by a factor of
Evaluate
0.25, and the area changes by a factor of
1. 3S r 2 0.25
2
.
SA44

3. 1764S in2
9. For the model, the volume is 64S cu. in.,
5. 196S cm2 and the surface area is 60Ssq. in. For the
7. 180S in2 silo, the volume is 2985984S cu. in. and
the surface area is 77760S sq. in. The
9. | 1332.0 mm 2
volume changes by a factor of 363 , and
11. 2S 15 in. the surface area changes by a factor
13. 9S in2 of 362.
4 Evaluate
15. Susana used , which is part of the
3 1
volume formula, not the surface area 1. The area is multiplied by .
2
formula. The surface area formula is
4S r 2 . 3. The area is multiplied by
1
.
4
17. 144S in2
5. The area is doubled.
2
7. The area is multiplied by .
3
9. The perimeter is tripled. The area is
multiplied by 9.
11. Volume is multiplied by 125.
13. A. non-proportional; B. proportional;
C. non-proportional; D. non-proportional;
E. proportional
15. | 800,000 acres
17. 36 in.
Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor. Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor.
87 88

UNIT 7 Selected Answers UNIT 7 Selected Answers

MODULE 20 MODULE 20 15.

Lesson 20.2 Modeling and Lesson 20.3 Problem Solving

Density with Constraints
Your Turn Your Turn
5. | 13,600 persons/ mi2 4. Radius | 1.5 in.
8. | 1,129,245 Height 6.0 in.
5. almost 7 in. tall
Evaluate x 0.5 ft
1. Juniper is denser than Aspen. 7. | 4.3 m (4 ft) (1 ft) (0.5 ft)
3. Liquid oxygen is denser than liquid Evaluate
hydrogen.
1. 37,125 cm3
5. 150 persons/ mi2
3. 67,500 cm3
7. | 684,000 BTUs
5. r | 1.1 in. and h | 4.4 in.
9. 125,000 BTUs
7. r | 3.6 cm and h | 10.8 cm.
11. | 0.70 cm 3
9. | 46.8 ft
13. | 993.5 kg/ m 3
11. | 8.8 m
(2 ft) (1 ft) (1.5 ft)
15. | 14.9 mi2 13.
17. | 11.8 m
SA45

17. 406, 4; 156, 2; 52, 1; 300, 3

19. 76% of the cork’s volume, will remain 19. | 73 ft
above the surface. 21. Weight multiplier 125,000
Cross-section multiplier 2,500
No, because the ratio of leg cross-section
to weight has been reduced by a factor of
1
x 30 cm .
50
y 40 cm
20.76
(60 cm) (40 cm) (30 cm) 23. h
Sr 2
r | 1.5 in. and h | 3.0 in.
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UNIT 8 Selected Answers UNIT 8 Selected Answers

MODULE 21 21. a.
5 MODULE 21 25. Yes
36 27. n(A) should be 1 since the tiles must
Lesson 21.1 Probability and 31
Lesson 21.2 Permutations and appear in the order B-E-A-D. The correct
Set Theory b. Probability
36 1
probability is .
Your Turn 12 Your Turn 360
23.
1 13 1
3. 3.
3 25. 0 56
14 27. No; choosing a black tile is not the 5
4. complement of choosing a white tile since 4.
30 28
the bag also contains gray tiles.
1 5. 34,650
5. 29. Assume A is a subset of S. Then 0 d n(A)
30 6. 210
d n(S). For example, if S has 10 elements,
2 the number of elements of A is greater 1
6. 7.
3 than or equal to 0 and less than or equal 15
to 10. No subset of S can have fewer than
3 2
8. 0 elements or more than 10 elements. So 8.
4 n( A) 5
3 0d d 1. When all the outcomes are
9. n(S ) Evaluate
5
SA46

n( A ) 1. 479,001,600 different orders

equally likely, P ( A) . Therefore 0 d
Evaluate n(S ) 3. 3024 ways
P(A) d 1.
1. Yes, because every element of D is also 1
an element of A. 5.
165
3. ^2, 4, 6, 12` 1
7.
5. ^1, 2, 3, 4, 6, 8, 10, 12` 56
9. 1680
7. ^5, 7, 9, 10, 11`
11. 15,120
3 1
9. 13.
5 20,160
4
11. 1
5 15.
35
2
13. 1
5 17.
21
5
15. 1
6 19.
720
11
17. 1
12 21.
5525
2
19. 23. A. C
3
B. A
C. B

UNIT 8 Selected Answers UNIT 8 Selected Answers

MODULE 21 23. In permutations, order matters. In MODULE 21
combinations, order does not matter.
Lesson 21.3 Combinations and In a permutation of A, B, and C, ABC is Lesson 21.4 Mutually Exclusive
Probability different from CBA, so they would be and Overlapping Events
counted as two different permutations. In
Your Turn a combination, ABC is the same as CBA, Your Turn
1 and would not be counted again. 13
4. a. 210; 210 5.
4 25
11 b. 10 C6 10C4 210
5. Evaluate
230 c. In general, n Cr c Cn r . 3
11 1.
6. 25. n Cn 1 4
32 5
3.
1 13
7.
2 97
5.
Evaluate 125
1. 15 ways 28
7.
3. 27,405 groups 125
10 9
5. 9.
143 125
SA47

7. | 11.8% 11.
1
193 15
9.
512 58
13.
93 75
11. 61
256 15.
75
1
13. 5
220 17.
18
4
15. 7
21 19.
18
1
17. 7
435 21.
12
14
19. 23. 85%
969
25. When finding n( A B ), n( A) should be
21. a. 84
added to n(B), not multiplied.
5
b. 4
21
13
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Selected Answers
Selected Answers
Selected Answers

UNIT 8 Selected Answers UNIT 8 Selected Answers

MODULE 22 1 MODULE 22 3. The events are not independent.
21. P ( A~Ǻ ) ! ! P (B~A )
2 5. | 5.6%
Lesson 22.1 Conditional 23. n( A B ); n(not A B ) ; n(B ) Lesson 22.2 Independent Events
7. 9%
Probability Your Turn
The conditional probability 9. | 8.3%
Your Turn n( A B ) 3. The events are not independent. 11. | 24.3%
P ( A~Ǻ ) restricts the
1 n(B ) 4. The events are not independent.
5. 1
3 discussion to event B because that event 6. 6%
13.
8
is assumed to have occurred. The
1 numbers used to calculate P ( A~Ǻ ) both 7. | 11% 15. The events are not independent.
6.
4 8. The events are not independent. 17. The events are not independent.
come from the highlighted row in the
Evaluate table: n( A B ) is the number of outcomes 9. The events are independent. 19. 40; 60
1. | 56% in event B that are also in event A, while 20; 30
Evaluate
n(B ) is the number of all outcomes in
3. | 8% 3 2 Sample answer: Let W be the event that a
event B. The rest of the table is irrelevant. 1. a. P (R ) and P (B ) person is a woman. Let Pe be the event
5. | 88% 5 5
that a person prefers writing with a pen.
7. P(Pa°L); | 55%, and P(F°M) 2 60 2 100 2
b. P (W ) , P (Pe ) , and
| 8%, so the probability that a student 5 150 5 150 5
who got less than 6 hours of sleep passed 8 2 2 4
c. P (W ) < P (Pe ) < . Since
the exam is greater.
SA48

19 5 3 15
1 40 4
9. d. Events R and B are independent when P (W Pe ) and
2 the first chip is returned to the bag 150 15
because P (B~R ) P (B ) in that case P (W Pe ) P (W ) < P (Pe ), the events
1
11. are independent.
4
1 21. P ( A ~B ) 1 P ( A~B )
c
Definition of complementary events
13.
2 1 P ( A) Definition of independent events
15. A. B P ( Ac ) Definition of complementary events
B. D So, events Ac and B are also independent.
C. A
D. C
17. 45%
5
19.
6
Given that not rolling a 1 on the red
number cube has occurred, there are 30
possible outcomes:
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1),
(4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1),
(5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1),
(6, 2), (6, 3), (6, 4), (6, 5), and (6, 6).
all but 5 outcomes—(2, 1), (3, 1), (4, 1),
(5, 1), and (6, 1)—are successful

UNIT 8 Selected Answers UNIT 8 Selected Answers

MODULE 22 3.
3 MODULE 23 11. Possible answer Write the names on slips
19 of paper, but for each hour that a student
Lesson 22.3 Dependent Events 2 Lesson 23.1 Using Probability worked, write his or her name on an extra
5. to Make Fair Decisions slip. Then draw a slip at random. The
Your Turn 57
probability of Paige winning is 4 out of
4 7. Events A and B are independent. Your Turn 45, or about 0.089.
3.
15 9. a. D 4. Possible answer Choose an employee’s 1
name at random from a list of all 13.
2 b. A, C 500
4. employees. Each month, remove the
5 c. B 15. | 1.89%
previous winners from the list. Once every
1 34 employee has won once, begin again with 17. | 86.79%
7. 11. a.
5 285 the list of all employees.
19. Possible answer Team A should receive
1 34 8. Possible answer If the game is 3.7 points. Team B should receive
8. b.
5 57 interrupted when the players are tied, 6.3 points.
they each have a probability of winning
Evaluate 1
equal to .
1. a. The events C and S are dependent. 2
5 5 Evaluate
b. P (C~S ) and P (S~C ) .
6 7
1. Possible answer You could split the cost
Multiplying P(C) and P (S~C ) of a sixth pass to the climbing gym so that
gives P (C S ). you each get a third visit for half of the
SA49

price of an additional pass. Or you could

Multiplying P(S) and P (C~S )
toss a coin and the winner gets the fifth
gives P (C S ). pass. Tossing a coin gives each of you
a 50% (or equal) chance of winning the
13. P ( A B C ) P (( A B ) C ) Group events A and B as one event. last pass.
P ( A B ) < P (C~A B ) Apply the Multiplication Rule to A B and C. 1
3. Not fair; 3(Meri wins) ;
P ( A) < P (B~A) < P (C~A B) Apply the Multiplication Rule to A and B. 3
1
3(Riley wins)
2
1 1
5. Fair; 3(Meri wins) ; 3(Riley wins)
2 2
1 1
7. Fair; 3(Meri wins) ; 3(Riley wins)
2 2
9. 0.30; 0.12
0.23; 0.12
0.18; 0.05
Find the total number of games won
30 23 18 12 12 5 100. Assign
numbers from 1100 to the members so
each has as many numbers assigned as
Selected Answers

the number of games won. Then use a

random number generator to choose an
integer from 1 to 100.

Selected Answers
Selected Answers
Selected Answers

UNIT 8 Selected Answers

MODULE 23 5. Allergy 196; 4; 200
No allergy 490; 9,310; 9,800
Lesson 23.2 Analyzing
686; 9,314
Decisions
No; Only 196 people out of the 686 who
Your Turn tested positive actually have the allergy.
7. This is about 29% of those who test
positive.
1
7. ; 0
3
1 2
;
3 3
9.

8. 69%
The probability of the test being correct in
this case increases about 2%.
9. 31%
The probability of the test being correct in
SA50

this case increases about 30%.

It would not be reasonable to assume that
Evaluate this is true.
1. No; Of the 1,015 products that Helen 11.
completed, 50 were defective, which is
about 4.9%. Of the 370 products that Kyle
completed, 20 were defective, which is
about 5.5%.
3. 9,000(0.38) 3,420
1,000(0.98) 980
4,400
9,000(0.62) 5,580 About 67%
1,000(0.02) 20
5,600
10,000(0.90) 9,000
10,000(0.10) 1,000
10,000
Only 22% of the coupons will go to the
intended target of students living off
campus. Therefore, this is not a good
decision.

Original content Copyright © by Houghton Mifflin Harcourt. Additions and changes to the original content are the responsibility of the instructor.
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Geometry Selected Answers 1

Uploaded by

Geometry Selected Answers 1

Uploaded by

© Houghton Mifflin Harcourt Publishing Company

UNIT 1 Selected Answers UNIT 1 Selected Answers

11. AB CD 29. Therefore, AB and CD

© Houghton Mifflin Harcourt Publishing Company

11. a. Segment HK is a horizontal segment 17. The transformation is a horizontal stretch

9. The number 14 is an even number.

15. Hypothesis the light does not come on.

UNIT 1 Selected Answers UNIT 1 Selected Answers

Lesson 2.1 Translations Lesson 2.2 Reflections

midpoints AAc  (1, 1);

11. Emma Quadrant I; Tony Quadrant III

8. 5, 4 ; the components are the

© Houghton Mifflin Harcourt Publishing Company

3. 13. UNIT 1 Selected Answers

Quadrant II. So after the tab key is

rotation, every point and its image must

23. Possible answer A'B'C' should be 1. 1

15. Both have two perpendicular lines of

© Houghton Mifflin Harcourt Publishing Company

Possible answer The translation moves

UNIT 1 Selected Answers UNIT 1 Selected Answers

7. B and C are congruent. EF and GH

© Houghton Mifflin Harcourt Publishing Company

15. UNIT 2 Selected Answers

F. Neither the Transitive Property of Equality, m2 

the complement of DEF 90q  mABC.

UNIT 2 Selected Answers UNIT 2 Selected Answers

© Houghton Mifflin Harcourt Publishing Company

UNIT 2 Selected Answers UNIT 2 Selected Answers

UNIT 2 Selected Answers UNIT 2 Selected Answers

© Houghton Mifflin Harcourt Publishing Company

9. 21. Possible sketch:

UNIT 2 Selected Answers UNIT 2 Selected Answers

JJJG JJJG F. sufficient

© Houghton Mifflin Harcourt Publishing Company

UNIT 2 Selected Answers UNIT 2 Selected Answers

lengths (AB 4, DF 5).

three points on the fold line will be

Reflexive Property of Congruence. Then ABC # + +

4. FH # JK 4. Given Statements Reasons

9. 2. A # D 2. All right angles are congruent.

3. DC # DC 3. Reflexive Property of Congruence

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UNIT 2 Selected Answers UNIT 2 Selected Answers

13. 59q which have the same length in both

UNIT 2 Selected Answers UNIT 2 Selected Answers

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UNIT 2 Selected Answers UNIT 2 Selected Answers

9. mFED 46q 7. GX 13.5

UNIT 2 Selected Answers UNIT 3 Selected Answers

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AE # CE and DE # BE. Since vertical 1

So JN LN and KN MN. JKLM is a are congruent.

UNIT 3 Selected Answers UNIT 3 Selected Answers

11. mXYZ 36.5q

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UNIT 3 Selected Answers UNIT 3 Selected Answers

or 2(24°) 48°. 2 The quadrilateral has only one pair of

sides of a quadrilateral is a parallelogram. angles.

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NB and MC # NB . AB | 128 nautical miles.

The coordinates of D, the midpoint of AC , © 2 2 ¸¹ § 1  3 4  0 ·

11. The diagonals are perpendicular. ABCD is

© 2 collinear and the port and the two cruise

Since the diagonals share a midpoint, parallelogram. Point D is at (6, 3.5).

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UNIT 3 Selected Answers UNIT 4 Selected Answers

Lesson 10.5 Perimeter and Area

19. Since ABCD is a kite, AD # AB and b. Perimeter is 36 units, Area is 54 square

with base BD , and CE is an altitude for area 9

15. Since the base and the height of the

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UNIT 4 Selected Answers UNIT 4 Selected Answers

UNIT 4 Selected Answers UNIT 4 Selected Answers

Lesson 12.1 Triangle Lesson 12.2 Subdividing a

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15. Hypothesis the light does not come on.

midpoints AAc (1, 1);

11. Emma Quadrant I; Tony Quadrant III

8. 5, 4 ; the components are the

23. Possible answer A'B'C' should be 1. 1

Possible answer The translation moves

7. B and C are congruent. EF and GH

F. Neither the Transitive Property of Equality, m2

the complement of DEF 90q mABC.

9. 2. A # D 2. All right angles are congruent.

9. mFED 46q 7. GX 13.5

11. mXYZ 36.5q

The coordinates of D, the midpoint of AC , © 2 2 ¸¹ § 1 3 4 0 ·

Since the diagonals share a midpoint, parallelogram. Point D is at (6, 3.5).

Evaluate mA | 61q

3. Yes; 4 3 : 12 : 8 3 1: 3 : 2 5. mR 38q

17. x 6 3 15. mCAE | 37qmAEC | 27q

3. DF , DG, EF , EG; DGE, DFE, FDG,

7. 56q XBC is a right angle, so mXBA 17. C | 15.7 ft.

9. center (3, 3); radius 2

9. Center (0, 2), radius 3 8. (x 3)2 12(y (1)) or (x 3)2

11. equation of parabola (x 4)2 20(y 3)