Answers
Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers
A45
31. yes 32. no 33. yes 34. no
35. a. 1.5 ft
b. $1.60 per ft
c. $2.0
36. a. 12 in. by 18 in.
b. 2216 in.
c. $6.48
d. $13.52
37. 80° 38. 100° 39. 30° 40. 150°
41. 20° 42. 50° 43. 70° 44. 180°
45. 100° 46. 80° 47. 271.5 ft 48. 2363 yd
49. 2717.5 mm 50. 240 cm 51. 254 in.
52. 2735 mi 53. 3101.25 in. 54. 32132230 in.
55. a. $3.75 b. $0.08
56. 3 24y x= − − 57. 4 7y x= −
58. 12 17y x= + 59. 5 11y x= − −
60. 14 11y x= − 61. 2
3 9y x= − −
62. 12 2y x= + 63. 3
8 2y x= −
64. 57 7y x= + 65. 1
4 13y x= − −
66. 2y x= − + 67. 13 3y x= −
68. 3 5y x= − 69. 12 7y x= − +
70. 12 4y x= − + 71. 4 8y x= −
72. 13y x= − 73. 3 20y x= − −
74. 14 3y x= + 75. 2 12y x= − −
76. 4 19y x= − 77. 15 2y x= +
78. 5y = − 79. 12 5y x= − −
80. (5, 0), (2, 3), (4, 7)A B C′ ′ ′ −
81. (3, 6), (0, 9), (2, 1)A B C′ ′ ′ −
82. (1, 4), ( 2, 7), (0, 3)A B C′ ′ ′− −
83. (7, 2), (4, 5), (6, 5)A B C′ ′ ′ −
84. (2, 3), ( 1, 6), (1, 4)A B C′ ′ ′− −
85. (4, 2), (1, 1), (3, 9)A B C′ ′ ′− −
86. (5, 8), (2, 11), (4, 1)A B C′ ′ ′
87. ( 2, 7), ( 5, 10), ( 3, 0)A B C′ ′ ′− − −
Chapter 5 5.1 Start Thinking
If 120 ,m A∠ = ° then 60m B∠ = ° because together
they make a straight angle. If 40 ,m D∠ = ° then
140m E∠ = ° under the same reasoning. If the sum of
, ,m B m C∠ ∠ and m D∠ is 180°, then 80 .m C∠ = °
5.1 Warm Up
1. 1 31m∠ = ° 2. 2 59m∠ = °
3. 3 59m∠ = ° 4. 4 90m∠ = °
5.1 Cumulative Review Warm Up
1. 69x = 2. 6x =
3. 60x = 4. 30x =
5.1 Practice A
1. scalene; right 2. isosceles; acute
3. scalene; not a right triangle
4. isosceles; right triangle
5. 32x = 6. 29x = 7. 30x = 8. 17x =
9. 19.5 , 70.5 , 90° ° °
10. no; An exterior angle will be acute when it is adjacent to the obtuse angle of an obtuse triangle.
11. 145°
5.1 Practice B
1. scalene; obtuse 2. isosceles; acute
3. 14x = 4. 9x =
5. 1169
x = 6. 7; 19.5x y= =
Answers
Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A46
7. The measure of each exterior angle is equal to the sum of the measures of the nonadjacent interior angles. So, 1 , 2 ,m A C m B C∠ = + ∠ = +
and 3 .m A B∠ = +
( )( )
Sum of exterior angles 1 2 3
2
2 180
360
A C B C A B
A B C
= ∠ + ∠ + ∠= + + + + += + += °= °
8. 67; 124x y= = 9. yes; 15 , 60 , 105 .° ° °
10. no; For instance, in a 10 -40 -130° ° ° triangle, at the 130° angle you have a 50° exterior angle which is complementary to the 40° angle.
11. ;C T∠ ≅ ∠ The sum of the angle measures of each
triangle is 180 ,° so if two pairs of corresponding angles are congruent, then the third pair of corresponding angles must also be congruent.
5.1 Enrichment and Extension
1. 63 , 36 , 81 ; acute° ° ° 2. 10; 71x y= =
3. 50; 33x y= = 4. 29; 64x y= =
5. 12.9; 51.4x y= = 6. 1m A m∠ = ∠
5.1 Puzzle Time
THE LETTUCE WAS A “HEAD” AND THE TOMATO WAS TRYING TO “KETCHUP”
5.2 Start Thinking
AB corresponds to ;DE BC corresponds to ;EF AC
corresponds to .DF
, ,A D B E C F∠ ≅ ∠ ∠ ≅ ∠ ∠ ≅ ∠
To get ,DEF multiply the side lengths of ABC by a scale factor of 2 and rotate 180 .ABC °
5.2 Warm Up
1. 9x = 2. 69x =
5.2 Cumulative Review Warm Up
1.
5.2 Practice A 1. ,A E∠ ≅ ∠ ,B F∠ ≅ ∠ ,C G∠ ≅ ∠ ,D H∠ ≅ ∠
,AB EF≅ ,BC FG≅ ,CD GH≅ and
;DA HE≅ Sample answer: BCDA FGHE≅
2. 8; 19x y= =
3. From the figure, ,J N∠ ≅ ∠ ,K P∠ ≅ ∠
,Q M∠ ≅ ∠ ,JK NP≅ ,KL PL≅ ,LQ LM≅
and .QJ MN≅ Also, KLQ PLM∠ ≅ ∠ by the
Vertical Angles Congruence Theorem (Thm. 2.6). Because all pairs of corresponding angles and all pairs of corresponding sides are congruent,
.JKLQ NPLM≅
4. The figure shows that A R∠ ≅ ∠ and .B S∠ ≅ ∠
So, C T∠ ≅ ∠ by the Third Angles Theorem
(Thm. 5.4). Using the Triangle Sum Theorem (Thm. 5.1), 65 .m T∠ = °
5. ABC and DEF are both equilateral and equiangular.
5.2 Practice B
1. ,A H∠ ≅ ∠ ,B I∠ ≅ ∠ ,C J∠ ≅ ∠ ,D F∠ ≅ ∠,E G∠ ≅ ∠ ,AB HI≅ ,BC IJ≅ ,CD JF≅
,DE FG≅ ;EA GH≅ ABCDE GFJIH≅
2. 65; 6; 146x y z= = =
3. From the figure, ,LP ON≅ ,PM NM≅ and
.LM OM≅ P N∠ ≅ ∠ because all right angles
are congruent to each other. By the Vertical Angles Congruence Theorem (Thm. 2.6),
.LMP OMN∠ ≅ ∠ L O∠ ≅ ∠ by the Third
Angles Theorem (Thm. 5.4). Because corresponding sides and angles are congruent,
.LPM ONM≅
A
D
EF
B C2 cm51°
51°
39°
39°
2.5 cm 3.2 cm5 cm6.4 cm
4 cm
x
y8
4
−8
−4
8 12−4
X
Y
X′
Y′
Answers
Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers
A47
4. Because RSTU UVQR≅ and VUR∠ is a right
angle, SUR∠ is a right angle. Using vertical
angles, 47 ,m UPV∠ = ° and by the Triangle Sum
Theorem (Thm. 5.1), 2 5 43x y+ = and
43 .m RSV∠ = ° Because it forms a linear pair
with a 47° angle, 133 .m UPS∠ = ° Using angle
addition and corresponding parts of congruent figures, VST∠ is a right angle. Then
( )360 90 90 133
47
4 3 .
m PUT
x y
∠ = − + +
== −
Solving the system 2 5 43
4 3 47
x y
x y
+ =− =
gives 14x = and 3.y = Finally,
90 43 133 .m RST∠ = + = °
5.
yes; To prove the triangles congruent, use the facts that AC CA= and opposite sides of a rectangle are congruent to show that 3 pairs of corresponding sides are congruent. Then use the Alternate Interior Angles Theorem (Thm. 3.2) to show two pairs of angles congruent. The third pair of corresponding angles are congruent right angles.
5.2 Enrichment and Extension
1.
2.
3. a. yes; You are given .ADB CDA CDB≅ ≅
So, .AB BC CA≅ ≅ Because all three sides of ABC are congruent, it is an equilateral triangle.
b. 120°
c. 30 , 30° °
d. The angle measures are equal because CDB is isosceles.
e. The measure of each of the congruent angles of each small triangle is 30 .° By the Angle Addition Postulate (Post. 1.4), the measure of each angle of ABC is 60 .°
5.2 Puzzle Time
CRAB CAKES
5.3 Start Thinking
Sample answer:
ABC ADE≅ because all angles and sides are
congruent. There is no further information needed because the given directions make it impossible to
construct ADE ≅ .ABC
5.3 Warm Up
1. AC 2. FH
A
B
D
C
A B
C
D
E
x
y
4
6
2
−2
4−2−4 (0, 0)(3, 0)(−3, 0)
(−3, 4) (3, 4)
STATEMENTS REASONS
1. ABD CDB∠ ≅ ∠ 1. Given
2. ADB CBD∠ ≅ ∠ 2. Given
3. AD BC≅ 3. Given
4. AB DC≅ 4. Given
5. BD BD≅ 5. Reflexive Property of Equality
6. BAD BCD∠ ≅ ∠ 6. Triangle Sum Theorem
7. ABD CDB≅ 7. All corresponding parts are congruent.
STATEMENTS REASONS
1. ||AB DC 1. Given
2. AB DC≅ 2. Given
3. is the midpoint
of and .
E
AC BD 3. Given
4. AE EC≅ 4. Definition of midpoint
5. BE ED≅ 5. Definition of midpoint
6. EAB ECD∠ ≅ ∠ 6. Alternate interior angles
7. ABD BDC∠ ≅ ∠ 7. Alternate interior angles
8. AEB CED∠ ≅ ∠ 8. Vertical Angles
Theorem
9. AEB CED≅ 9. All corresponding parts are congruent.
Answers
Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A48
5.3 Cumulative Review Warm Up
1. ED 2. EC
5.3 Practice A
1. yes; You know that two sides and the included angle are congruent, so you can use the SAS Congruence Theorem (Thm. 5.5).
2. no; The congruent vertical angles are not the included angles, so SAS cannot be used.
3. ;ABO CBO≅ ,OB OB≅ and because they are
radii, .AO CO≅ It is given that .AOB COB∠ ≅
So, SAS is satisfied.
4. ;ABE CBE≅ Because ,AB BC≅
,ABE CBE∠ ≅ ∠ and ,BE BE≅ SAS is satisfied.
5. no; The SAS Congruence Theorem (Thm. 5.5) applies after a translation, reflection, or rotation, because those are congruence transformations, but not after a dilation, which changes the size of the figure.
6.
5.3 Practice B
1. yes; You know that two sides are congruent, and using the Third Angles Theorem (Thm. 5.4) you can find that the included angles are congruent, so SAS applies.
2. yes; One pair of sides is marked congruent, you can use segment addition to show a second pair of sides congruent, and you can use the Third Angles Theorem (Thm. 5.4) to show their included angles congruent. So, SAS applies.
3. , ,PLM PMN and ;PNL The sides ,PL ,PM
and PN are congruent because they are radii of the same circle. The three angles at P are congruent, so you can use SAS with these angles and the sides including them.
4. ,RVU ,SXW and ;TZY Show that each
obtuse angle measures ( )360 60 90 90− + +
and that the sides including these angles are all congruent because they are sides of squares that border an equilateral triangle. So, SAS applies.
5. 7; 24x y= =
6. AED is equilateral and equiangular, so
,AE DE≅ and .CAD BDA∠ ≅ ∠ ,EB EC≅ so
EB DE EC DE+ = + by the Addition Property of Equality. Substituting AE for DE gives
.EB DE EC AE+ = + Using the Segment Addition Postulate (Post. 1.2), BD EB DE= +and .CA EC AE= + So, BD CA= by
substitution. Also, AD DA≅ by the Reflexive Property of Segment Congruence (Thm. 2.1).
ACD DBA≅ by the SAS Congruence
Theorem (Thm. 5.5).
5.3 Enrichment and Extension
1. NP OP≅ because they have an equal length of 6;
MP MP≅ by the Reflexive Property of Segment
Congruence (Thm. 2.1); MO MN≅ because they
have an equal length of 3 2. So, PMO PMN≅ by the SSS Congruence
Theorem (Thm. 5.8). It is possible to prove PMO PMN≅ by the SAS Congruence
Theorem (Thm. 5.5) because you already know that
MP MP≅ and .MO MN≅ You also know that
NO MP⊥ because the slopes are opposite reciprocals. So, ,OMP NMP∠ ≅ ∠ leading to
PMO PMN≅ by the SAS Congruence
Theorem (Thm. 5.5).
2. yes; HL Congruence Theorem (Thm. 5.9)
STATEMENTS REASONS
1. and ABD CBD∠ ∠ are right angles.
1. Given
2. bisects .BD AC 2. Given
3. AB BC≅ 3. Definition of a segment bisector
4. BD BD≅ 4. Reflexive Property of Segment Congruence (Thm. 2.1)
5. ABD CBD≅ 5. SAS Congruence Theorem (Thm. 5.5)
Answers
Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers
A49
3.
4.
5. ( ) ( )4, 10 , 15, 3X Y
5.3 Puzzle Time
OBTUSE
5.4 Start Thinking
Sample answer:
.ABD CBD≅ The length of 8.BD = EGH EGF≅ because all side lengths and angles
are congruent.
5.4 Warm Up
1. 89x = 2. 83x =
3. 21x = 4. 80x =
5.4 Cumulative Review Warm Up
1. A triangle is an isosceles triangle if and only if the legs are of equal length.
2. A Chinese puzzle is a tangram if and only if it is made up of seven pieces.
3. A parallelogram is a rectangle if and only if it has four right angles.
5.4 Practice A
1. 25x = 2. 6x =
3. 20; 30x y= = 4. 30; 4x y= =
5. Because they are vertical angles, .DCE BCA∠ ≅ ∠ Because they both measure
70 ,° .B D∠ ≅ ∠ So, A E∠ ≅ ∠ by the Third
Angles Theorem (Thm 5.4). But E D∠ ≅ ∠
because they are base angles of an isosceles triangle, so 70 .m A∠ = ° Because they have equal
measures, A∠ and B∠ are base angles of an
isosceles triangle.
6. yes; An isosceles triangle is obtuse when the vertex angle is obtuse and the base angles are acute.
x
y
4
−2
−2 2−4 DC
A
GH
E F
B
STATEMENTS REASONS
1. AC BD⊥ 1. Given
2. is the midpoint
of .
D
AC 2. Given
3. is a rightangle.
ADB∠ 3. Definition of perpendicular lines
4. 90m ADB∠ = ° 4. Definition of right angle
5. is a right angle.
CDB∠ 5. Definition of perpendicular lines
6. 90m CDB∠ = ° 6. Definition of right angle
7. m ADB m CDB∠ = ∠ 7. Substitution
8. AD DC≅ 8. Definition of midpoint
9. BD BD≅ 9. Reflexive Property of Segment Congruence (Thm. 2.1)
10. ADB CDB≅ 10. SAS Congruence Theorem (Thm. 5.5)
STATEMENTS REASONS
1. DE BF= 1. Given
2. AE CF= 2. Given
3. AE DB⊥ 3. Given
4. CF BD⊥ 4. Given
5. is a right angle.
AEB∠ 5. Definition of perpendicular lines
6. is a right angle.
CFD∠ 6. Definition of perpendicular lines
7. AEB CFD∠ ≅ ∠ 7. Right Angles Congruence Theorem (Thm. 2.3)
8. BE BF FE= + 8. Segment Addition Postulate (Post. 1.2)
9. FD FE ED= + 9. Segment Addition Postulate (Post. 1.2)
10. FD FE BF= + 10. Substitution
11. FD BE= 11. Substitution
12. AE CF≅ 12. Definition of congruent segments
13. FD BE≅ 13. Definition of congruent segments
14. AEB CFDΔ ≅ Δ 14. SAS Congruence Theorem (Thm. 5.5)
Answers
Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A50
5.4 Practice B
1. 36x = 2. 90x =
3. 14; 7x y= = 4. 29; 3x y= =
5. It is given that CBD CDB∠ ≅ ∠ and
.CAE CEA∠ ≅ ∠ By the Converse of the Base
Angles Theorem (Thm. 5.7), BC DC≅ and
.AC EC≅ By the Reflexive Property of Angle Congruence (Thm. 2.2), .C C∠ ≅ ∠
ACD ECB≅ by the SAS Congruence
Theorem (Thm. 5.5). Because congruent parts of
congruent triangles are congruent, .AD EB≅
6.
5.4 Enrichment and Extension
1. ( )12
r p q= +
2.
3. 12.9; 51.4x y= = 4. 29; 64x y= =
5.
It is possible to partition a right triangle into two isosceles triangles. Suppose ABC is a right triangle with right angle C. Because m CAD∠ is
less than ,m ACB∠ it is possible to construct D
such that ,m ACD m CAD∠ = ∠ as shown above.
Because 90m ACD m DCB m CAD∠ + ∠ = ° = ∠
m DBC+ ∠ and ,m CAD m ACD∠ = ∠ it follows
that .m DCB m DBC∠ = ∠ So, DBC is an
isosceles triangle.
5.4 Puzzle Time
NEITHER IT’S BEST TO WRITE WITH A PEN
5.5 Start Thinking
In ,JKL 20 , 80 ,m J m K∠ = ° ∠ = ° 80 .m L∠ = °
JKL is an isosceles triangle. The angles in PQR
are congruent. It is not possible to create two triangles with the same side lengths but different angles. As discussed in Section 5.4, the side lengths and angles are related.
5.5 Warm Up
1. RUT∠ 2. STR∠ 3. TRS∠
4. UTR∠ 5. SRT∠ 6. RST∠
5.5 Cumulative Review Warm Up
1. 1 54 ;m∠ = ° straight angle
2 54 ;m∠ = ° Corresponding Angles Theorem
(Thm. 3.1)
2. 2 131 ;m∠ = ° Corresponding Angles Theorem
(Thm. 3.1) 1 49 ;m∠ = ° straight angle
5.5 Practice A
1. congruent; Two pairs of sides are marked congruent and the third pair of sides is shared, so the SSS Congruence Theorem (Thm. 5.8) applies.
a° a°
3a° 3a°
4a°
2a°2a°
4a°6a°
6a°
5a° 5a°
7a°7a°
(180 − 2a)°
(180 − 6a)°
(180 − 4a)° (180 − 8a)°
(180 − 10a)°
(180 − 14a)°
(180 − 12a)°
STATEMENTS REASONS
1. , EBC ECB
AE DE
∠ ≅ ∠
≅
1. Given
2. EB EC≅ 2. Converse of the Base Angles Theorem (Thm. 5.7)
3. AEB DEC∠ ≅ ∠ 3. Vertical Angles Congruence Theorem (Thm. 2.6)
4. AEB DEC≅ 4. SAS Congruence Theorem (Thm. 5.5)
5. AB DC≅ 5. Corresponding parts of congruent triangles are congruent.
1 in.
0.5 in.K L
J
1 in.
75.5° 75.5°
29°1 in.
0.5 in.Q R
P
1 in.
75.5° 75.5°
29°
A
C B
D
Answers
Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers
A51
2. not congruent; The hypotenuse and a leg of one triangle are congruent to the legs of the other triangle, so the triangles cannot be congruent.
3. congruent; The hypotenuses are shared and a pair of legs is congruent, so the HL Congruence Theorem (Thm. 5.9) applies.
4.
5.
6. not congruent
7. a. You know the shared sides are congruent, so you need to measure each of the other sides of the two triangular faces to determine whether the SSS Congruence Theorem (Thm. 5.8) applies.
b. a regular hexagon
5.5 Practice B
1. The triangles share one pair of sides and two other pairs of sides are marked congruent, so the SSS Congruence Theorem (Thm. 5.8) can be used.
2. The triangles are right triangles and the hypotenuses and a pair of legs are shown to be congruent, so the HL Congruence Theorem (Thm. 5.9) can be used.
3. The triangles have three pairs of congruent sides and each triangle has a right angle, so they can be proven congruent using the SSS Congruence Theorem (Thm. 5.8), the HL Congruence Theorem (Thm. 5.9), or the SAS Congruence Theorem (Thm. 5.5).
4.
5.
6. Sample answer: Sketch the lines in a coordinate plane. The slopes of lines a and b show that they form vertical right angles of the triangles. The Distance Formula shows that the hypotenuses have different lengths, so the triangles are not congruent.
STATEMENTS REASONS
1. ;
bisects .
AB AD
AC BD
≅ 1. Given
2. AC AC≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)
3. BC DC≅ 3. Definition of segment bisector
4. ABC ADC≅ 4. SSS Congruence Theorem (Thm. 5.8)
STATEMENTS REASONS
1. and JL GF
KL HF
≅≅
1. Given
2. and
are right angles.J G∠ ∠
2. Given
3. and JKL GHF are right triangles.
3. Definition of a right triangle
4. JKL GHF≅ 4. HL Congruence Theorem (Thm. 5.9)
STATEMENTS REASONS
1. , PS RS
SQ PR
≅
⊥
1. Given
2. and are right angles.
PQS RQS∠ ∠ 2. Definition of perpendicular lines
3. and are right triangles.
PQS RQS 3. Definition of right triangles
4. QS QS≅ 4. Reflexive Property of Segment Congruence (Thm. 2.1)
5. PSQ SQR≅ 5. HL Congruence Theorem (Thm. 5.9)
STATEMENTS REASONS
1. ,
,
BC ED
AB FE
AD FC
≅
≅
≅
1. Given
2. BC ED= 2. Definition of congruent segments
3. ,BD BC CD
EC ED CD
= += +
3. Segment Addition Postulate (Post. 1.2)
4. BD ED CD= + 4. Substitution Property of Equality
5. BD EC= 5. Substitution Property of Equality
6. BD EC≅ 6. Definition of congruent segments
7. ABD FEC≅ 7. SSS Congruence Theorem (Thm. 5.8)
Answers
Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A52
5.5 Enrichment and Extension
1. a. 7x = b. 4x = c. 4 or 4x = −
2.
3.
4.
If one side of RSTΔ is congruent to one side of
,DEF such that ,RS MN≅ then you know that the triangles are congruent because equilateral triangles have three congruent sides. So, all sides in
RST would be equal to all sides in ,MNO making them congruent by the SSS Congruence Postulate.
5. ( ) ( )2 , 2 , 2 , 2J e c f d K e a f b− − − −
5.5 Puzzle Time
A TEENAGER
5.6 Start Thinking
Sample answer:
;XYZ ABC≅/ Any size triangle can have the angle
measures indicated. One example would be to draw the triangle on a coordinate plane and then dilate it with the origin as the center. This example shows that knowing two triangles have the same angle measures is not enough to prove congruence.
5.6 Warm Up
1. SAS Congruence Theorem (Thm. 5.5)
2. none
3. SAS Congruence Theorem (Thm. 5.5)
4. none
5.6 Cumulative Review Warm Up
1.
5.6 Practice A 1. yes; AAS Congruence Theorem (Thm. 5.11)
X Z
Y
70° 60°
50°
A C
B
70° 60°
50°
x
y
8
−4
84−2
X
X
Z
Y
X′
X″
Y″
Z″
Y′
Z′
R T
S
M O
N
STATEMENTS REASONS
1. WA WT≅ 1. Given
2. is the midpoint of .S AT 2. Given
3. AS ST≅ 3. Definition of midpoint
4. WS WS≅ 4. Reflexive Property of Segment Congruence (Thm. 2.1)
5. WSA WSTΔ ≅ Δ 5. SSS Congruence Theorem (Thm. 5.8)
6. 1180
m WAS m∠ + ∠= °
6. Definition of perpendicular lines
7. AEB CFD∠ ≅ ∠ 7. Definition of linear pair
8. 2180
m WTS m∠ + ∠= °
8. Definition of linear pair
9. WAS WTS∠ ≅ ∠ 9. Corresponding parts of congruent triangles are congruent.
10. m WAS m WTS∠ = ∠ 10. Definition of congruence
11. 12
m WAS mm WTS m
∠ + ∠= ∠ + ∠
11. Substitution
12. 12
m WAS mm WAS m
∠ + ∠= ∠ + ∠
12. Substitution
13. 1 2m m∠ = ∠ 13. Subtraction
14. 1 2∠ ≅ ∠ 14. Definition of congruent angles
STATEMENTS REASONS
1. GR GT≅ 1. Given
2. RS ST≅ 2. Given
3. GS GS≅ 3. Reflexive Property of Segment Congruence (Thm. 2.1)
4. GRS GST≅ 4. SSS Congruence Theorem (Thm. 5.8)
Answers
Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers
A53
2. no; Although two pairs of sides and one pair of angles are congruent, they are not the corresponding parts needed for the SAS Congruence Theorem (Thm. 5.5).
3. yes; ASA Congruence Theorem (Thm. 5.10)
4.
5.
6. Given: , ;AB CD ABD CDB BD DB≅ ∠ ≅ ∠ ≅
by the Reflexive Property of Segment Congruence (Thm. 2.1); So, ABD CDB≅ by the SAS
Congruence Theorem (Thm. 5.5).
Given: , ;AB EF ABD F ABD EDF≅ ∠ ≅ ∠ ∠ ≅ ∠by the Vertical Angles Congruence Theorem (Thm. 2.6); So, ABD EFD≅ by the AAS
Congruence Theorem (Thm. 5.11).
Given: , ;CD EF BDC F BD DF≅ ∠ ≅ ∠ ≅
because congruent parts of congruent triangles are congruent; So, CDB EFD≅ by the SAS
Congruence Theorem (Thm. 5.5).
There is not enough information to show that AGD is congruent to any other triangle.
5.6 Practice B
1. yes; ASA Congruence Theorem (Thm. 5.10)
2. no; The sides and angles marked congruent do not correspond to be used for either ASA or AAS.
3. yes; AAS Congruence Theorem (Thm. 5.11)
4.
5.
STATEMENTS REASONS
1. , PS RT PQ TQ≅ 1. Given
2. and S R
P T
∠ ≅ ∠∠ ≅ ∠
2. Alternate Interior Angles Theorem (Thm. 3.2)
3. PSQ TRQ≅ 3. AAS Congruence Theorem (Thm. 5.11)
STATEMENTS REASONS
1. bisects , BD ADC
BD AC
∠
⊥
1. Given
2. BD BD≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)
3. ADB CDB∠ ≅ ∠ 3. Definition of angle bisector
4. and are right angles.
ABD CBD∠ ∠ 4. Definition of perpendicular lines
5. ABD CBD∠ ≅ ∠ 5. Right Angles Congruence Theorem (Thm. 2.3)
6. ABD BCD≅ 6. ASA Congruence Theorem (Thm. 5.10)
STATEMENTS REASONS
1. bisects ,BD AE
A E∠ ≅ ∠ 1. Given
2. AC CE≅ 2. Definition of segment bisector
3. ACB ECD∠ ≅ ∠ 3. Vertical Angles Congruence Theorem (Thm. 2.6)
4. ABC EDC≅ 4. ASA Congruence Theorem (Thm. 5.10)
STATEMENTS REASONS
1. ,
,
I J
IM JN
KL MN
∠ ≅ ∠
≅
1. Given
2. M N∠ ≅ ∠ 2. Corresponding Angles Theorem (Thm. 3.1)
3. KL MN= 3. Definition of congruent segments
4. KL LMLM MN
+= +
4. Addition Property of Equality
5. and KM KL LM
LN LM MN
= +
= +
5. Segment Addition Postulate (Post. 1.2)
6. KM LN= 6. Substitution Property of Equality
7. KM LN≅ 7. Definition of segment congruence
8. IKM LJN≅ 8. AAS Congruence Theorem (Thm. 5.11)
Answers
Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A54
6. Draw RT . By the definition of a parallelogram,
|| and || .RS QT QR TS QTR SRT∠ ≅ ∠ and
QRT STR∠ ≅ ∠ by the Alternate Interior Angles
Theorem (Thm. 3.2). RT TR≅ by the Reflexive Property of Segment Congruence (Thm. 2.1).
QRT STR≅ by the ASA Congruence Theorem
(Thm. 5.10). So, QR TS≅ and RS QT≅
because corresponding parts of congruent triangles are congruent.
5.6 Enrichment and Extension
1. 12
any real value except 2; or 0;m m m= = − =
The triangles are congruent by AAS or ASA.
2. Sample answer I: Label point ( )1, 6 as A, point
(5, 4) as B, point (4, 3) as C, point (1, 0) as D, and point (5, 2) as E. You can calculate
4.472, 1.414,AB DE CE CB= ≈ = ≈ and
4.242.AC CD= ≈ So, ABC DEC≅ by
the SSS Congruence Theorem (Thm. 5.8).
Sample answer II:
Label point (1,6) as A, point (5,4) as B, point (4,3) as C, point (1,0) as D, and point (5,2) as E. 1.414CE CB= ≈ and 4.242.AC CD= ≈ Additionally, the slope of ,BD AE⊥ so, 90 .m ACB m DCE∠ = ∠ = ° By the SAS Congruence Theorem (Thm. 5.5), .ABC DEC≅
3. BC and AD are parallel, because their slopes are
equal, with AC being a transversal. The Alternate Interior Angles Theorem (Thm. 3.2) applies to
make .BCA CAD∠ ≅ ∠ AB and CD are parallel
because their slopes are equal, with AC being a transversal. The Alternate Interior Angles Theorem (Thm. 3.2) applies to make .BAC ACD∠ ≅ ∠
AC AC≅ by the Reflexive Property of Segment Congruence (Thm. 2.1). So, ABC CDA≅ by
the ASA Congruence Theorem (Thm. 5.10).
4.
5.
x
y
2
−2
2−2
x
y
4
6
2
4 6
(1, 6)
(5, 4)
(4, 3)
(1, 0)
(5, 2)
STATEMENTS REASONS
1. HB EB
BHG BEA
HGJ EAD
JGB DAB
≅∠ ≅ ∠∠ ≅ ∠∠ ≅ ∠
1. Given
2. m BHG m BEA
m HGJ m EAD
m JGB m DAB
∠ = ∠∠ = ∠∠ = ∠
2. Definition of congruency
3. m HGJ m JGB
m HGB
∠ + ∠= ∠
3. Angle Addition Postulate (Post. 1.4)
4. m EAD m DAB
m EAB
∠ + ∠= ∠
4. Angle Addition Postulate (Post. 1.4)
5. m EAD m DAB
m HGB
∠ + ∠= ∠
5. Substitution
6. m HGB m EAB∠ = ∠ 6. Substitution
7. HGB EAB∠ ≅ ∠ 7. Definition of congruency
8. BHG BEA≅ 8. AAS Congruence Theorem (Thm. 5.11)
STATEMENTS REASONS
1. ABC ABDFCA EDA
≅∠ ≅ ∠ 1. Given
2. CA DA≅ 2. Corresponding parts of congruent triangles are congruent.
3. CAF DAE∠ ≅ ∠ 3. Vertical Angles Congruence Theorem (Thm. 2.6)
4. CAF DAE≅ 4. ASA Congruence Theorem (Thm. 5.10)
Answers
Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers
A55
6.
5.6 Puzzle Time
A WATCH
5.7 Start Thinking
Answer should include, but is not limited to: Any construction of a small triangle (preferably small enough to fit on a notebook page) with labels for side lengths and angle measurements.
5.7 Warm Up
1. Transitive Property of Segment Congruence
2. Reflexive Property of Angle Congruence
3. Symmetric Property of Angle Congruence
4. Reflexive Property of Segment Congruence
5. Symmetric Property of Segment Congruence
6. Transitive Property of Angle Congruence
5.7 Cumulative Review Warm Up
1. , ,ABC CBA B∠ ∠ ∠ 2. , , 2JKL LKJ∠ ∠ ∠
3. , ,HMN NMH M∠ ∠ ∠ 4. , ,MPR RPM P∠ ∠ ∠
5.7 Practice A
1. The figure shows that , ,DAC E B C∠ ≅ ∠ ∠ ≅ ∠
and .AB DC≅ So, ACD EBA≅ by the AAS
Congruence Theorem (Thm. 5.11) and EB AC≅ because corresponding parts of congruent triangles are congruent.
2. The figure shows that ACB DCB∠ ≅ ∠ and
.AC DC≅ Use the Reflexive Property of Segment Congruence (Thm. 2.1) to show
.BC BC≅ Then ABC DBC≅ by the SAS
Congruence Theorem (Thm. 5.5). So, A D∠ ≅ ∠because corresponding parts of congruent triangles are congruent.
3. Show that PQRS is a parallelogram by definition, then show QPS RSP∠ ≅ ∠ by the Alternate
Interior Angles Theorem (Thm. 3.2). Show PS SP≅
by the Reflexive Property of Segment Congruence
(Thm. 2.1). The figure shows .PQ SR≅ So,
PQS SRP≅ by the SAS Congruence Theorem
(Thm. 5.5), and PR SQ≅ because corresponding
parts of congruent triangles are congruent.
4. The figure shows that HI JI≅ and .HK JK≅ Use the Reflexive Property of Segment Congruence
(Thm. 2.1) to show that .IK IK≅ Then HIK JIK≡ by the SSS Congruence Theorem
(Thm. 5.8), and H J∠ ≅ ∠ because corresponding parts of congruent triangles are congruent.
5. Place a stake at L so that .LM MN⊥ Find K, the
midpoint of .ML Locate the point J so that
,ML JL⊥ and J, K, and N are collinear. Then find JL; Given: M∠ and L∠ are right angles, K is the
midpoint of .ML Paragraph proof: Because they are right angles, .M L∠ ≅ ∠ MK LK= by the definition of the midpoint of a segment.
JKL NKM∠ ≅ ∠ by the Vertical Angles
Congruence Theorem (Thm. 2.6). JKL NKM≅
by the ASA Congruence Theorem (Thm. 5.10). LJ MN= because corresponding parts of congruent triangles are congruent.
6. 5DE = ; Sample answer: 44m C∠ = ° by the Triangle Sum Theorem (Thm. 5.1), so you have
, , andA D C F AC DF∠ ≅ ∠ ∠ ≅ ∠ ≅ .
ABC DEF≅ by the ASA Congruence
Theorem (Thm. 5.10). Therefore, DE AB= because corresponding parts of congruent triangles are congruent.
STATEMENTS REASONS
1. AE BF
CE DF
AB CD≅
1. Given
2. EAC FBD∠ ≅ ∠ 2. Corresponding Angles Theorem (Thm. 3.1)
3. ECA FDB∠ ≅ ∠ 3. Corresponding Angles Theorem (Thm. 3.1)
4. BC BC≅ 4. Reflexive Property of Segment Congruence (Thm. 2.1)
5. AB BC AC+ = 5. Segment Addition Postulate (Post. 1.2)
6. CD BC DB+ = 6. Segment Addition Postulate (Post. 1.2)
7. AB BC
CD BC
+
= +
7. Addition Property of Equality
8. AC DB= 8. Substitution
9. AC DB≅ 9. Definition of congruent segments
Answers
Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A56
5.7 Practice B
1. The figure shows that ,G J∠ ≅ ∠ and .GI JH=Because they are vertical angles, .GKI JKH∠ ≅ ∠
So, GKI JKH≅ by the AAS Congruence
Theorem (Thm. 5.11) and GK JK≅ because corresponding parts of congruent triangles are congruent.
2. The figure shows that BE CE= and .AD CB⊥Because they are right angles, .AEB AEC∠ ≅ ∠
AE AE≅ by the Reflexive Property of Segment Congruence (Thm. 2.1). AEB AEC≅ by the
SAS Congruence Theorem (Thm. 5.5). Because corresponding parts of congruent triangles are congruent, .BA CA=
3. The figure shows that ,BA FG≅ ,A G∠ ≅ ∠ and
B∠ and F∠ are right angles. So, show that
ABC GFE≅ by the ASA Congruence
Theorem (Thm. 5.10). Then BCA FEG∠ ≅ ∠because corresponding parts of congruent triangles are congruent, which leads to congruence of their respective vertical angles, .DCH DEH∠ ≅ ∠
Then DH DH≅ by the Reflexive Property of Segment Congruence (Thm. 2.1) and right angles
are formed by ,DH CE⊥ so CDH EDH≅ by the AAS Congruence Theorem (Thm. 5.11).
So, DC DE≅ because corresponding parts of congruent triangles are congruent.
4. Use the Converse of the Base Angles Theorem
(Thm. 5.7) to show that .SV SU≅ Show that RVS TUS∠ ≅ ∠ because they form linear pairs
with congruent angles. Then RVS TUS≅ by
the SAS Congruence Theorem (Thm. 5.5), which leads to 1 2∠ ≅ ∠ because corresponding parts of
congruent triangles are congruent.
5. a. Sample answer: Mark point C along the edge of
your roof. Then find the midpoint D of .AC
Locate point E so that AC CE⊥ and E, D, and
B are collinear. Measure .CE This is the same as AB.
b. Sample answer: The method works because you have right angles, congruent segments, and vertical angles which lead to ABD CED≅ by the ASA Congruence Theorem (Thm. 5.10),
and CE corresponds to AB in these congruent triangles.
5.7 Enrichment and Extension
1. 2; 5x x= = 2. 0; 4x x= =
3. 20, 120, 6x y z= = = ±
4. In ABC and ,DEF you can use the Distance
Formula to find that 5 units,AB DE= =5.385,BC EF= ≈ and 10 units.AC FD= =
So, you can conclude that ABC DEF≅ by
the SSS Congruence Theorem (Thm. 5.8). So, A D∠ ≅ ∠ because corresponding parts of
congruent triangles are congruent.
5.
STATEMENTS REASONS
1. is the midpoint
of .
L
JN
PJ QN
PL QL
≅
≅
1. Given
2. JL LN≅ 2. Definition of midpoint
3. PLJ PLN≅ 3. SSS Congruence Theorem (Thm. 5.8)
4. and are right angles.
PKJ QMN∠ ∠ 4. Given
5. PKJ QMN∠ ≅ ∠ 5. Right Angles Congruence Theorem (Thm. 2.3)
6. KJP MNQ∠ ≅ ∠ 6. Corresponding parts of congruent triangles are congruent.
7. PKJ QMN≅ 7. AAS Congruence Theorem (Thm. 5.11)
8. MQN KPJ∠ ≅ ∠ 8. Corresponding parts of congruent triangles are congruent.
A B
EC
D
Answers
Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers
A57
6.
5.7 Puzzle Time
TO KEEP THEIR INSIDES IN
5.8 Start Thinking
Answer should include, but is not limited to: Any construction using dynamic geometry software consisting of ABC with whole-number degree angle measures, centered at the origin. A second triangle, larger or smaller, should also be centered at the origin and contain the same angle measures. The two triangles both have the same angle measures but are not the same size, so that they are not congruent.
5.8 Warm Up
1. 11.7 2. 3.2 3. 18
4. 7.3 5. 12.6 6. 21.8
5.8 Cumulative Review Warm Up
1. 30 EF+ 2. 4 GH• 3. AB JK−
5.8 Practice A
1.
It is easy to find the width and length of the rectangle by placing a vertex at the origin.
2.
It is easy to see the length of each leg by placing the vertex with the right angle at the origin.
3.
It is easy to see the width and length of the rectangle by placing a vertex at the origin.
4.
It is easy to see the length of the base and the height, and to determine that the triangle is isosceles by placing one vertex at the origin.
STATEMENTS REASONS
1.
2 3
R S∠ ≅ ∠∠ ≅ ∠
1. Given
2. TV TV≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)
3. RTV SVT≅ 3. AAS Congruence Theorem (Thm. 5.11)
4. RT SV≅ 4. Corresponding parts of congruent triangles are congruent.
5. 5 6∠ ≅ ∠ 5. Vertical Angles Congruence Theorem (Thm. 2.6)
6. RTU SVU≅ 6. AAS Congruence Theorem (Thm. 5.11)
7. RU SU≅ 7. Corresponding parts of congruent triangles are congruent.
x
y4
−2
42 6
(0, 2) (6, 2)
(6, 0)(0, 0)
x
y
(0, w)
( , 0)(0, 0)
( , w)
x
y
4
2
42 6
(0, 4)
(0, 0)(4, 0)
x
y
(b, 0)
, h)(
(0, 0)
b2
Answers
Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A58
5.
( )
2 24, , midpoint of :
, 12
2, undefined, midpoint of :
, 1
, 0, midpoint of : , 02
AB a mAB ABa
a
BC mAB AB
a
aAC a mAB AB
= + =
= =
= =
ABC is a right triangle because AC is horizontal
and BC is vertical.
When 2,a ABC= is also isosceles because
then 2.AC BC= =
6.
2 2
, undefined, midpoint of :
0, 2
, , midpoint of :
, 2 2
, 0, midpoint of : , 02
JK a mJK JK
a
aKL a b mKL KL
b
b a
bJL b mJL JL
= =
= + = −
= =
JKL is a right triangle because JL is horizontal
and JK is vertical. JKL is not isosceles because .a b≠
7. ( ) ( ) 2 20, 0 , , 0 ;O C h OB h k= +
8. ( ) ( ) ( ) 2
2 2
2 , 0 , 0, 0 , 2, 4 ; 4 4 1,
2 4 2 1
G h O D k FD k
DE k h h
− = +
= + − +
9.
( )
( ) ( )
( ) ( )
2 2 2 2
2 2
Coordinate proof:
Segments and have the same length.
0 ; 0
Segments and have the same length.
0 ; 0
Segments and have the same length.
0 0 ;
0 0
CO DO
CO h h DO h h
AC BD
AC k k BD k k
AO BO
AO h k h k
DO h k h
= − − = = − =
= − = = − =
= − − + − = +
= − + − = 2 2
by the SSS Congruence
Theorem (Thm. 5.8).
k
ACO BDO
+
≅
10. Sample answer: yes; In this position, it is relatively easy to find the lengths of the horizontal base and the side adjacent to the base with its vertex at ( )0, 0 .
11. Sample answer: ( ) ( ) ( )0, 0 , 0, 3 , 4 , 0 ; 5R S a T a a
5.8 Practice B
1.
It is easy to find the width and length of the rectangle by placing a vertex at the origin with one side on each axis.
2.
It is easy to see the length of one leg by placing a vertex at the origin with a side on the horizontal axis.
3.
It is easy to see the length of one side by placing a vertex at the origin with a side on the horizontal axis.
x
y
C(a, 0)
B(a, 2)
A(0, 0)
2
4
x
y
L(b, 0)J(0, 0)
K(0, a)
x
y
(0, k) (2k, k)
(0, 0) (2k, 0)
x
y
(3, 0)
(3, k)
(0, 0)
2 4 6
x
y
(0, 0) (a, 0)
(b, c)
Answers
Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers
A59
4.
( )2 2
2 2
2 , 0, midpoint of : , 0
, , midpoint of :
, 2 2
, , midpoint of :
3,
2 2
JL a mJL JL a
bJK a b mJK JK
a
a b
bKL a b mKL KL
a
a b
= =
= + =
= + = −
JKL is isosceles, because .JK KL≅
5.
( )
( )
( )
5 , 0, midpoint of : 2.5 , 0
14 5, , midpoint of : 4 , 2
24
5 , , midpoint of : 6.5 , 23
PQ a mPQ PQ a
PR a mPR PR a a
QR a mQR QR a a
= =
= =
= =
PQR is isosceles, because .PQ QR≅
6. ( )0, 0 ,F ( )0, ,G k ( )2 , ,H k k− ( ), 0 ;J k
2 2,GH k= 5FH k=
7. ( )0, 0 , , 2 ,2
kA B k
3, 2 ,
2
kC k
( ), 0 ;E k ,BC k=
117
2CD k=
8. parallelogram; 5
,2
mWX = − 1,
2mXY =
5,
2mYZ = − and
1.
2mZW = Because both pairs
of opposite sides are parallel, WZYZ is a parallelogram by definition. It is not a trapezoid because it has two pairs of parallel sides instead of one. It is not a rectangle because the slopes are not negative reciprocals of each other, which indicates that the sides are not perpendicular.
9. Given: Vertex C of ABC is on the line the perpendicular bisector of .AB
Prove: ABC is isosceles.
AC and BC have the lengths
( )2 2 2 2 ,AC a b a b= − + = +
2 2 .BC a b= +
So, ,AC BC= and by the definition of congruent
segments, .AC BC≅ Therefore, ABC is isosceles by the definition of isosceles triangle.
5.8 Enrichment and Extension
1. Call the point S the midpoint of ,RP with the
coordinates , .2 2
b a a bS
− + + =
The slope of
( ),
b aRP
b a
− −=
− − and the slope of ,
a bSQ
b a
− −=−
making them opposite reciprocals and forming the
90° angles RSQ and PSQ. SQ SQ≅ by the
Reflexive Property of Segment Congruence
(Thm. 2.1). The length of both RS and SP is equal
to 2 2
,2 2
a b b a+ − +
so RSQ PSQ≅
by the SAS Congruence Theorem (Thm. 5.5). The
slope of a
RQb
= − and the slope of .b
QPa
=
They are opposite reciprocals and form the right angle PQR.
0,x =
x
y
J(0, 0) L(2a, 0)
K(a, b)
x
y
P(0, 0) Q(5a, 0)
R(8a, 4a)
x
y
A(−a, 0) B(a, 0)
C(0, b)
Answers
Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A60
2. The slope of CD is equal to the slope of ,AB but .AD BC≠ So, it is a trapezoid, but not an
isosceles trapezoid.
3. a. coordinates of : , ;2 2
s tP
coordinates of
: ,2 2
s r tQ
+
b. equation of ( );2
tPR y x r
s r= = −
−
equation of t
QO y xr s
= =+
4. The trapezoid is isosceles because .TR PA≅ The base is on the x-axis, and the y-axis bisects the bases. Label point ( )0, 2 ,E c= point ( )0, 0 ,G =
point ( ), ,D b a c= − − and point ( ), .F b a c= +You can then show that
( )22 .DE EF FG DG c b a= = = = + +
All sides are equal, making DEFG a rhombus.
5. ( ),b b− 6. ( )0, a
7. ( ) ( ), or ,a b b a b b− − −
8. ( ),b a b−
5.8 Puzzle Time
AIR BAG
Cumulative Review
1. 3
, 32
M −
2. 5 3,
2 2M −
3. 5, 2
2M
4. ( )2, 0M 5. ( ) 5, 4M 6. 11, 1
2M
7. 7
, 12
M −
8. 31,
2M −
9. 9, 1
2M −
10. 1
0,2
M −
11. ( )1, 3M − 12. 5, 2
2M
13. 5 , 0
2M −
14. 15, 4
2M −
15. 3, 2
2M −
16. 15 9,
2 2M
17. 1
7,2
M − −
18. ( )5, 6M
19. ( )5, 3M − 20. 3 3
,2 2
M
21. 8.54
22. 12.37 23. 10.44 24. 13.34
25. 7.21 26. 6.32 27. 15.03
28. 6.32 29. 12.53 30. 5
31. 7 32. 10 33. 8.25
34. 6.40 35. 13.60 36. 7.62
37. 6.71 38. 7.62 39. 12.53
40. 15.81 41. 324 in. 42. 3234 in.
43. 21 44. 11 45. 18
46. 9 47. 10 48. 20
49. 8 50. 34 51. 2750 ft
52. a. 2216 ft
b. 2 containers
c. $31.98
53. 87° 54. 69° 55. 74°
56. 61° 57. 115° 58. 78°
59. 20x = 60. 6x = − 61. 5x = −
62. 9x = − 63. 8x = 64. 4x =
65. 2x = − 66. 7x = 67. 9x =
68. 11x = 69. 6x = 70. 5x =
71. 5x = 72. 4x = 73. 3x = −
74. 14x = 75. 1x = 76. 10x =
77. 6x = 78. 4x = 79. 4x = −
80. 6x = − 81. 32x = − 82. 8x = −
83. 3 9y x= − + 84. 7 11y x= −
85. 8 13y x= − 86. 3 6y x= −
87. 2 6y x= − + 88. 5 7y x= −
89. 0.125 3y x= − + 90. 12 8y x= +
91. 0.125 8y x= − 92. 2 24y x= − −
Answers
Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers
A61
93. 94
24y x= − 94. 8 2y x= −
95. a. 2224 ft b. 12 ft by 14 ft c. 2168 ft
96. a. 10 ft b. 24 ft c. 224 ft
Chapter 6 6.1 Start Thinking
The roof lines become steeper; The two top chords will get longer as the king post gets longer, but the two top chords will always be congruent to each other. The angles formed by the top chords and the king post are congruent. The angles formed by the top chords and the bottom chord are congruent.
6.1 Warm Up
1. 5 2. 8 3. 2 4. 1−
6.1 Cumulative Review Warm Up
1.
2.
6.1 Practice A
1. P lies on the perpendicular bisector of ;RS The
markings show that TP
satisfies the definition of a perpendicular bisector.
2. P lies on the perpendicular bisector of ;RS Because
T is equidistant from the endpoints R and S, T lies
on the perpendicular bisector of RS by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2). Because only one line can be
perpendicular to RS at U, TU
must be the
perpendicular bisector of ,RS and P is on .TU
3. P lies on the angle bisector of ;DEF∠ P is
equidistant from sides ED
and EF
of angle ,DEF∠ so P is on the angle bisector of DEF∠ by
the Converse of the Angle Bisector Theorem (Thm. 6.4).
4. 20; Because D is on the perpendicular bisector of
,AC D is equidistant from A and C.
5. 17; By the Perpendicular Bisector Theorem (Thm. 6.1), .GJ GH= Solving 4 5 2 11x x+ = + gives 3,x = so 2 11 17.x + =
6. 14; Q is on the angle bisector of ,PSR∠ so Q is
equidistant from and .SP SR
7. 76 ;° By the Converse of the Angle Bisector
Theorem (Thm. 6.4), GE
bisects .DGF∠
Solving 3 8 5 12x x+ = − gives 10.x = So,
( ) ( )3 10 8 5 10 12 76 .m DGF∠ = + + − ° = °
8. 4 7y x= − +
9. Because any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment, you can draw an isosceles triangle by drawing segments from each endpoint of the segment to the same point on the perpendicular bisector.
10. yes; In a right triangle, the bisector of the right angle is also the perpendicular bisector of the hypotenuse when the right angle is isosceles. The figure shows that when right of C ABC∠
is bisected by , CD ACD BCD≅ by either
the ASA Congruence Theorem (Thm. 5.10) or the SAS Congruence Theorem (Thm. 5.5). Then the corresponding sides
andAC BC must be
congruent, so ABC is isosceles.
STATEMENTS REASONS
1. P is the midpoint
of MN and .TQ
1. Given
2. MP NP≅ 2. Definition of segment midpoint
3. PT PQ≅ 3. Definition of segment midpoint
4. MPQ NPT∠ ≅ ∠ 4. Vertical Angles Congruence Theorem (Thm. 2.6)
5. MQP NTP≅ 5. SAS Congruence Theorem (Thm. 5.5)
STATEMENTS REASONS
1. ,AB DC≅
AC DB≅
1. Given
2. BC BC≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)
3. ABC DCB≅ 3. SSS Congruence Theorem (Thm. 5.8)
A
C B
D