75.
1.
2.
3.
4.
5.
Statements
̶̶
​ PX​ ⊥ ℓ; Y is any point
on ℓ other than X.
m∠1 = 90°
∠1 is a rt. ∠.
△XPY is a rt. △.
∠2 and ∠P are comp.
6. 90° = m∠2 + m∠P
7. 90° > m∠2
8. m∠1 > m∠2
9. PY > PX
3.
Reasons
B
2. Def. of ⊥
3. Def. of rt. ∠
4. Def. of rt. △
5. Acute  of rt. △
are comp.
6. Def. of comp. 
7. Comparison
Prop. of Inequal.
8. Subst.
9. In △, longer side
is opp. larger ∠
A
Hinge Theorem:
If AB XY, AC XZ,
and m∠A > m∠X, then
BC > YZ.
5. AB > ED
b.
Reasons
1. Given
2. Def. of midpoint
3. Def. of ≅ 
4. Con. of Isosc. △
Thm.
5. Hinge Thm.
Statements
1. ∠
SRT ≅ ∠STR,
TU > RU
̶̶ ̶̶
2. ​ ST​ ≅ ​ SR​ 
̶̶ ̶̶
3. ​ SU​ ≅ ​ SU​ 
4. m∠TSU > m∠RSU
Reasons
1. Given
2. C
on. of Isosc. △
Thm.
3. Reflex. Prop. of ≅
4. Con. of the Hinge
Thm.
think and discuss
1. Possible answer: kitchen tongs
2. No; in this case, 2 sides of the 1st △ are ≅ to 2
sides of the 2nd △, but the given ∠ measures are
not the measures of  included between the ≅
sides. Thus you cannot apply the Hinge Theorem.
Z
Converse of Hinge Theorem:
If AB XY, AC XZ,
and BC > YZ, then
m∠A > m∠X.
2. Compare the side lengths in △SRT and △QRT.
RT = RT RS = RQ ST > QT
By the Converse of Hinge Theorem, m∠SRT >
m∠QRT.
b. Compare the sides and the  in △ABD and △CBD.
AD = CD BD = BD m∠CDB > m∠ADB
By the Hinge Theorem, BC > AB.
Statements
X
1. Compare the sides and the  in △ABC and △XYZ.
AB = YZ BC = XY m∠B < m∠Y
By the Hinge Theorem, AC < XZ.
1a. Compare the side lengths in △EFG and △EHG.
EF > EH EG = EG FG = HG
By the Converse of the Hinge Theorem, m∠EGF >
m∠EGH.
1. C
is the midpoint
̶̶
of ​ BD​; 
m∠1 = m∠2 m∠3 > m∠4
̶̶ ̶̶
2. ​ BC​ ≅ ​ DC​ 
3. ∠1 ≅ ∠2
̶̶ ̶̶
4. ​ AC​ ≅ ​ EC​ 
C
guided Practice
check it out!
2. The ∠ of swing at full speed is greater than the ∠ of
swing at low speed.
Y
exercises
5-6 inequalities in two triangles
3a.
Inequalities in Two Triangles
1. Given
3. Compare the sides and  in △KLM and △KNM.
KM = KM LM = NM m∠KML > m∠KMN
By the Hinge Theorem, KL > KN.
4. Step 1 Compare the side lengths in . By the
Converse of the Hinge Theorem,
2x + 8 < 25
2x< 17
x< 8.5.
Step 2 Since (2x + 8)° is an angle in a △,
2x + 8 > 0
2x> -8
x> -4.
Step 3 Combine the inequals.
The range of values is -4 < x < 8.5.
5. Step 1 Compare the sides and the  in . By the
Hinge Theorem,
5x - 6 < 9
5x< 15
x< 3.
Step 2 Since 5x - 6 is a length,
5x - 6 > 0
5x> 6
x> 1.2.
Step 3 Combine the inequals.
The range of values is 1.2 < x < 3.
6. Step 1 Compare the sides and the  in . By the
Hinge Theorem,
2x - 5 < x + 7
x< 12.
Step 2 Since 2x - 5 is a length,
2x - 5 > 0
2x> 5
x> 2.5.
Step 3 Combine the inequals.
The range of values is 2.5 < x < 12.
108 Holt McDougal Geometry
7. The 2nd position; the lengths of the upper and lower
arm are the same in both positions, but the distance
from the shoulder to the wrist is greater in the 2nd
position. So the included ∠ measure is greater by
the Converse of the Hinge Theorem.
8.
Statements
̶̶
1. FH​
​   is a median of △DFG;
m∠DHF > m∠GHF
̶̶
2. H is midpoint of ​ DG​. 
̶̶ ̶̶
3. ​ DH​ ≅ ​ GH​ 
̶̶ ̶̶
4. ​ FH​ ≅ ​ FH​ 
5. DF > GF
Reasons
1. Given
2.
3.
4.
5.
Def. of median
Def. of midpoint
Reflex. Prop. of ≅
Hinge Thm.
practice and problem solving
9. BC = CD, CA = CA, AD > AB; by Converse of
Hinge Theorem, m∠DCA > m∠BCA.
10. GH = KL, HJ = LM, GJ < KM; by Converse of
Hinge Theorem, m∠GHJ < m∠KLM.
11. ST = UV, SU = SU, m∠UST > m∠SUV; by Hinge
Theorem, TU > SV.
4z - 12 > 0
12. 4z - 12 < 16
4z> 12
4z< 28
z > 3
z< 7
Combining, 3 < z < 7.
2z + 7 > 0
13. 2z + 7 < 72
2z> -7
2z< 65
z> -3.5
z< 32.5
Combining, -3.5 < z < 32.5.
4z - 6 > 0
14. 4z - 6 < z + 11
4z> 6
3z< 17
3  ​ 
17
_
z> ​ _
z< ​   ​ 
2
3
3
17
_
_
Combining, ​   ​  < z < ​   ​ . 
2
3
15. The lengths of the arms are the same in both
positions, but the included ∠ measure is greater in
the 2nd position. Therefore, by the Hinge Theorem,
the distance from the cab to the bucket is greater in
the 2nd position.
16.
1.
2.
3.
4.
Statements
̶̶ ̶̶̶ ̶̶ ̶̶̶
​ JK​ ≅ ​ NM​,  ​ KP​ ≅ ​ MQ​, 
JQ > NP
̶̶ ̶̶
​ QP​ ≅ ​ QP​ 
QP = QP
JQ + QP > NP + QP
5. J Q + QP = JP,
NP + QP = NQ
6. JP > NQ
7. m∠K > m∠M
7. BC = YZ
1
19. m∠QPR > m∠QRP
21. m∠RSP = m∠RPS
Reasons
1. Given
2. Reflex. Prop. of ≅
3. Def. of ≅ segs.
4. Add. Prop. of
Inequal.
5. Segment Add.
Post.
6. Subst.
7. Con. of the Hinge
Thm.
24. Corr. sides are ≅, and the included  are ∠B and
∠E. By the Hinge Theorem, m∠B > m∠E → AC >
DF.
̶̶ ̶̶
̶̶ ̶̶
25.​ SR​ ≅ ST​
​   by definition, and SV​
​   ≅ SV​
​  .  So by the
Converse of the Hinge Theorem, RV < TV →
m∠RSV < m∠TSV.
26. Corr. sides are ≅, and the included  are ∠G and
∠K. m∠G = 90° > m∠K, so by the Hinge Theorem,
HJ > LM.
̶̶ ̶̶
̶̶ ̶̶
27.​ YM​ ≅ MZ​
​   by definition, and XM​
​   ≅ XM​
​  . So by the
Converse of the Hinge Theorem, YX > ZX →
m∠YMX > m∠ZMX.
28. Possible answer: As the angle made by a door
hinge gets larger, the width of the door opening
increases. As the angle made by the hinge gets
smaller, the width of the door opening decreases.
This is like the side opposite an angle in a triangle
getting larger as the measure of the angle increases
or getting smaller as the angle decreases.
29. Possible answer:
Similarities: Both the SAS ≅ Post. and the Hinge
Theorem concern the relationship between 2 .
Both involve 2 sides and the included ∠ of each △.
Differences: To apply the SAS ≅ Post., you must
know that 2 sides and the included ∠ of one △ are
≅ to 2 sides and the included ∠ of the 2nd △. To
apply the Hinge Theorem, you must know that 2
sides of one △ are ≅ to 2 sides of the 2nd △, but
the included  are ≠ in measure. The SAS ≅ Post.
allows you to conclude that the 2  are ≅; then by
CPCTC, you can show that the sides opposite the ≅
 are ≅. The Hinge Theorem involves 2  that are
≇; in this case, the sides opposite the included 
are ≠ in length, and the exact relationship between
the lengths is determined by the sizes of the
included .
̶̶ ̶̶ ̶̶ ̶̶
30a. Newton Springs; ​ NS​ ≅ HS​
​  ,  SJ​
​   ≅ SJ​
​  ,  and
m∠NSJ < m∠HSJ, so NJ < JH by the Hinge
Theorem.
b. By △ Inequal. Theorem,
NJ + SJ > SN
NJ + 182 > 300
NJ > 118 mi
Min. distance =
SN + NJ
> 300 + 118 = 418 mi
test prep
31. D;
0 < 3x - 9 < 2x + 1
9 < 3x or 3 < x, and x < 10
3 < x < 10
32. H;
̶̶
D lies on AB​
​  ;  AD = DB by the definition of median.
8. m∠QRP < m∠SRP
1
20.m∠PRS < m∠RSP
22. m∠QPR > m∠RPS
23. m∠PSR < m∠PQR
109 Holt McDougal Geometry
2. Let y be the distance in ft from the foot of the ladder
to the base of the wall. Then 4y is the distance in ft
from the top of the ladder to the base of the wall.
33. Group A is closer to the camp.
Possible answer: The 6.5-mi and 4-mi paths
together with the distance lines back to the camp
form 2 . 2 sides of 1 △ are ≅ to 2 sides of the
other △. In the △ for Group A, the measure of the
included ∠ is 90° + 35° = 125°. In the △ for Group
B, the measure of the included ∠ is 90° + 45° =
135°. By the Hinge Theorem, the side opposite the
125° ∠ is shorter than the side opposite the 135° ∠.
So Group A is closer to the camp.
​a  2​+ ​b  2​= ​c  2​
(4y​) 2​+ ​y  2​= ​30 2​
17​y  2​= 900
2
900 ​ 
 
​y  ​= ​ _
17

​  900 ​ ​ 
 
y= ​  _
17

​  900 ​ ​ 
 ≈ 29 ft 1 in.
4y= 4​  _
17
√
√
challenge and extend
34. Step 1 Apply Hinge Theorem.
̶̶ ̶̶ ̶̶
By Converse of Isosc. △ Theorem, VZ​
​   ≅ VY​
​  ;  VX​
​   
̶̶
≅ VX​
​  ;  m∠XVZ > m∠XVY. So XZ > XY.
Step 2 Write and solve 2 inequals.
5x + 15 > 8x - 6
8x - 6 > 0
21> 3x
8x> 6
7> x
x> 0.75
Step 3 Combine the inequals.
0.75 < x < 7
2
35a. Locate point P outside △ABC so that ∠ABP ≅
̶̶ ̶̶
̶̶
∠DEF and BP​
​   ≅ EF​
​  .  It is given that AB​
​   ≅
̶̶ ̶̶
̶̶
​ DE​,  so △ABP ≅ △DEF by SAS. Thus ​ AP​ ≅ DF​
​   
by CPCTC.
̶̶
̶̶
b. Locate point Q on AC​
​   so that BQ​
​   bisects ∠PBC.
By the definition of ∠ bisector, ∠QBC ≅ ∠QBP. It
̶̶ ̶̶
̶̶ ̶̶
is given that ​ BC​ ≅ EF.​
​   Since ​ BP​ ≅ EF​
​   from part a,
̶̶ ̶̶
​   ≅ BP​
BC​
​   by the Trans. Prop. of ≅. By the Reflex.
̶̶ ̶̶̶
Prop. of ≅, BQ​
​   ≅ BQ.​
​ 
 So △BQP ≅ △BQC by SAS,
̶̶ ̶̶
and QP​
​   ≅ QC​
​   by CPCTC.
c. AQ + QP > AP by the △ Inequal. Theorem in
△AQP. AQ + QC = AC by the Segment Add.
̶̶ ̶̶
Post. From part b, QP​
​   ≅ QC​
​  , so QP = QC by
the definition of ≅ segs. Thus AQ + QC > AP by
̶̶
subst., and so AC > AP by subst. From part a, AP​
​   
̶̶
≅ ​ DF​.  So by the definition of ≅ segs., AP = DF.
Therefore AC > DF by subst.
5-7 the pythagorean theorem
check it out!
1a.​a 2​+ ​b 2​= ​c  2​
​4 2​+ ​8 2​= ​x  2​
80 = ​x  2​

​ √80 ​
    = x
x= ​ √
(16)(5) ​
   
= 4​ √
5 ​
  
b. ​ a  2​+ ​b  2​= ​c  2​
​x  2​+ ​12  2​= (x + 4​) 2​
​x  2​+ 144 = ​x  2​+ 8x + 16
128= 8x
x= 16
2
2
3a. ​a  ​+ ​b  ​= ​c  ​
​8 2​+ ​10 2​= ​c 2​
164= ​c 2​


c= ​ √164 ​
   = 2​ √41 ​
   
The side lengths do not form a Pythagorean triple

because 2​ √41 ​
   is not a whole number.
b. ​a 2​+ ​b 2​= ​c 2​
​24 2​+ ​b 2​= ​26 2​
​b 2​= 100
b = 10
The side lengths are nonzero whole numbers that
satisfy the equation a
​  2​+ ​b 2​= ​c 2​, so they form a
Pythagorean triple.
2
2
2
c. ​a  ​+ ​b  ​= ​c  ​
​1 2​+ ​2.4 2​= ​c 2​
6.76= ​c 2​
c = 2.6
The side lengths do not form a Pythagorean triple
because 2.4 and 2.6 are not whole numbers.
d.​ a 2​+ ​b 2​= ​c 2​
​16 2​+ ​30 2​= ​c 2​
1156= ​c 2​
c= 34
The side lengths are nonzero whole numbers that
satisfy the equation a
​  2​+ ​b 2​= ​c 2​, so they form a
Pythagorean triple.
4a. Step 1 Determine if the measures form a △.
By the △ Inequal. Theorem, 7, 12, and 16 can be
the side lengths of a △.
Step 2 Classify the △.
​c 2​≟ ​a 2​+ ​b 2​
​16 2​≟ ​7 2​+ ​12 2​
256 ≟ 49 + 144
256 > 193
Since ​c 2​> ​a 2​+ ​b 2​, △ is obtuse.
b. Step 1 Determine if the measures form a △.
Since 11 + 18 = 29 ≯ 34, these cannot be the side
lengths of a △.
c. Step 1 Determine if the measures form a △.
By the △ Inequal. Theorem, 3.8, 4.1, and 5.2 can
be the side lengths of a △.
Step 2 Classify the △.
​c 2​≟ ​a 2​+ ​b 2​
​5.2 2​≟ ​3.8 2​+ ​4.1 2​
27.04 ≟ 14.44 + 16.81
27.04 < 31.25
Since ​c 2​< ​a 2​+ ​b 2​, △ is acute.
110 Holt McDougal Geometry
41. z + z ⩼ 3z
2z≯ 3z
No; possible answer: when z = 5, the value of 3z
is 15. So the 3 lengths are 5, 5, and 15. the sum of
5 and 5 is 10, which is not greater than 15. By the
△ Inequality Thm., a △ cannot have these side
lengths.
42. Possible answer:
Given: △ABC
Prove: △ABC cannot have 2 obtuse .
Proof: Assume that △ABC has 2 obtuse . Let ∠A
and ∠B be the obtuse . By the definition of obtuse,
m∠A > 90° and m∠B > 90°. If the 2 inequalities
are added, m∠A + m∠B > 180°. However, by the
△ Sum Theorem, m∠A + m∠B + m∠C = 180°.
So m∠A + m∠B = 180° - m∠C. But then 180° m∠C > 180° by subst., and thus m∠C < 0°. A △
cannot have an ∠ with a measure less than 0°. So
the assumption that △ABC has 2 obtuse  is false.
Therefore a △ cannot have 2 obtuse .
5-6 Inequalities in Two Triangles
̶̶ ̶̶ ̶̶ ̶̶
43.​ PQ​ ≅ QR​
​  , QS​
​   ≅ QS​
​  , and m∠PQS < m∠RQS.
By the Hinge Theorem, PS < RS.
̶̶ ̶̶ ̶̶ ̶̶
44.​ BC​ ≅ DC​
​  , AC​
​   ≅ AC​
​  ,  and AB < AD.
By the Converse of the Hinge Theorem,
m∠BCA < m∠DCA.
45. m∠GFH < m∠EFH
5n + 7< 22
5n< 15
n< 3
-1.4 < n < 3
46. XZ < JK
4n - 11 < 39
4n< 50
n< 12.5
2.75 < n < 12.5
2
2
5-8 Applying Special Right Triangles
55. 45°-45°-90° △
x = 26​ √
2 ​
  
56. 45°-45°-90° △

12= x​ √2 ​
  

12​ √2 ​
    = 2x

x= 6​ √2 ​
  
57. 45°-45°-90° △
x = (​ 16​ √
2 ​
  )  ​​ √
2 ​
    = 32
58. 30°-60°-90° △
48 = 2x
x= 24


y = x​ √3 ​
    = 24​ √3 ​
  
59. 30°-60°-90° △
x = 6​ √
3 ​
  
y = 2(6) = 12
60. 30°-60°-90° △

14= x​ √3 ​
  

14​ √3 ​
    = 3x

  
14​ √3 ​
 
x= _
​   ​ 
3
y = 2x
3 ​
  
3 ​
  
14​ √
28​ √
​   ​ 
 ​= _
​   ​ 
 
= 2​ _
3
3
(
)
XZ > 0
4n - 11 > 0
4n > 11
n > 2.75
62. The altitude forms two 30°-60°-90° . The shorter
legs measure 9 ft.

h = 9​ √3 ​
    ≈ 15 ft 7 in.
chaPter test
48.​a  2​+ ​b  2​= ​c  2​
​x  2​+ ​8 2​= ​14 2​
​x  2​= 132

x= 2​ √33 ​
   
2
50. ​a  2​+ ​b  2​= ​c  2​
​24 2​+ ​32 2​= ​x  2​
1600= ​x  2​
x= 40
The side lengths form a Pythagorean triple because
they are nonzero whole numbers that satisfy
53. 1.5 + 3.6 = 5.1 > 3.9 ✓ 54. 2 + 3.7 = 5.7 > 4.1 ✓
​3.9 2​≟ ​1.5 2​+ ​3.6 2​
​4.1 2​≟ ​2 2​+ ​3.7 2​
15.21 = 15.21
16.81 < 17.69
The side lengths can
The side lengths can
form a rt. △.
form an acute △.
61. The diagonal forms two 45°-45°-90° .
2 ​
  
30= s​ √

30​ √2 ​
    = 2s
s= 15​ √
2 ​
    ≈ 21 ft 3 in.
49. ​a  ​+ ​b  ​= ​c  ​
​x  2​+ (4.5​) 2​= (7.5​) 2​
​x  2​= 36
x= 6
The side lengths do not form a Pythagorean triple
because 4.5 and 7.5 are not whole numbers.
​
52. 11 + 14 = 25 ≯ 27
The side lengths cannot
form a △.
m∠GFH > 0
5n + 7> 0
5n> -7
n> -1.4
5-7 The Pythagorean Theorem
47.​a  2​+ ​b  2​= ​c  2​
​2 2​+ ​6 2​= ​x  2​
40= ​x  2​

x= 2​ √10 ​
   
51. 9 + 12 = 21 > 16 ✓
​16 2​≟ ​9 2​+ ​12 2​
256 > 225
The side lengths can
form an obtuse △.
1. KL = JK = 9.8
2. m∠WXY = 2m∠WXZ = 2(17) = 34°
3. AC= BC
2n + 9 = 5n - 9
18= 3n
n= 6
BC = 5(6) - 9 = 21
4. RS = 2MS
= 2(3.4) = 6.8
RQ = SQ = 4.9
5. m∠DEF + m∠EFD + m∠FDE = 180
2m∠GEF + 2(25) + 42= 180
2m∠GEF= 88
m∠GEF= 44°
̶̶
̶̶
distance from G to DF​
​   = distance from G to DE​
​   
= 3.7
a  2​+ b
​   2​= c​   2​.
119 Holt McDougal Geometry