Name ________________________________________ Date __________________ Class__________________
LESSON
5-6
Reteach
Inequalities in Two Triangles
Theorem
Example
Hinge Theorem
If two sides of one triangle are congruent to
two sides of another triangle and the included
angles are not congruent, then the included
angle that is larger has the longer third side
If ∠K is larger than ∠G, then side LM is
across from it.
longer than side HJ.
The Converse of the Hinge Theorem is also true. In the example above, if side LM is longer
than side HJ, then you can conclude that ∠K is larger than ∠G. You can use both of these
theorems to compare various measures of triangles.
Compare NR and PQ in the figure at right.
PN = QR
PR = PR
m∠NPR < m∠QRP
Since two sides are congruent and ∠NPR is smaller
than ∠QRP, the side across from it is shorter than
the side across from ∠QRP.
So NR < PQ by the Hinge Theorem.
Compare the given measures.
2. m∠G and m∠L
1. TV and XY
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4. m∠FHE and m∠HFG
3. AB and AD
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5-46
Holt Geometry
Name ________________________________________ Date __________________ Class__________________
LESSON
5-6
Reteach
Inequalities in Two Triangles continued
You can use the Hinge Theorem and its converse to find a range of values in triangles.
Use UMNP and UQRS to find the range of values for x.
Step 1 Compare the side lengths in the triangles.
NM = SR
NP = SQ
m∠N < m∠S
Since two sides of UMNP are congruent to two sides
of UQRS and m∠N < m∠S, then MP < QR by the
Hinge Theorem.
MP < QR
3x − 6 < 24
3x < 30
x < 10
Hinge Thm.
Substitute the given values.
Add 6 to each side.
Divide each side by 3.
Step 2 Check that the measures are possible for a triangle.
Since MP is in a triangle, its length must be greater than 0.
MP > 0
3x − 6 > 0
x>2
Def. of U
Substitute 3x − 6 for MP.
Simplify.
Step 3 Combine the inequalities.
A range of values for x is 2 < x < 10.
Find a range of values for x.
5.
6.
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7.
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8.
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5-47
Holt Geometry
8. Possible answer: The legs of a compass
and the length spanned by it form a
triangle, but the lengths of the legs
cannot change. Therefore any two
settings of the compass are subject to the
Hinge Theorem. To draw a largerdiameter circle, the measure of the hinge
angle must be made larger. To draw a
smaller-diameter circle, the measure of
the hinge angle must be made smaller.
4.
Statements
1. JK || HL , JK ≅ HL ,
m∠KML > m∠HML
Practice C
1. The length of BD increases, and the
length of AC decreases.
2. Vert. ∠s Thm.
3. ∠JKH ≅ ∠LHK
3. Alt. Int. ∠s Thm.
4. UJKM ≅ ULHM
4. AAS
5. MK ≅ MH
5. CPCTC
6. ML ≅ ML
6. Reflex. Prop. of ≅
7. KL > HL
7. Hinge Thm.
2. the second cyclist
6. No, DC cannot be longer than BC ;
possible answer: the inequalities lead to
the contradiction that x must be both less
4
and greater than 2.
than
3
3. The ∠ formed by the compass when
drawing the first circle is smaller. So the
distance between the points of the
compass is greater for the second circle.
4. B
7. AB, CD, DE, FA, EF , BC
5. G
Reading Strategies
8. DE
1. m∠ADB < m∠DBC
Reteach
2. ZY > WZ
2. m∠G > m∠L
3. m∠ABC < m∠EFD
3. AB > AD
4. QR < RS
4. m∠FHE < m∠HFG
5. −2 < x < 7
6. 3 < x < 21
7. 3 < x < 57
8. −0.6 < x < 7
LESSON 5-7
Practice A
Challenge
1. AB ≅ AD and AC ≅ AE because radii of
the same circles are congruent. Since
m∠BAC > m∠DAE, then by the Hinge
Thm., BC > DE.
2. 16.5 < y < 34
2. ∠JMK ≅ ∠LMH
1. Greatest at relaxed position; least at
writing position; the length of his leg and
the length of his body are the same in all
three triangles. So, by the Converse of
the Hinge Thm., the larger included ∠ is
across from the longer third side.
1
4. 6 < x < (y + z)
3
5. Yes, BD can be longer than DC 2 < x <
188
5
1. TV < XY
1. Given
Problem Solving
2. between zero and (a + b)
3. 10 < x < 58
Reasons
1. 26
2. 16
3. 8.9
4. 48 in.
5. whole numbers
6. 7.2; no
7. 11.5; no
8. 12; yes
9. The segments can form a triangle.
3. 0.25 < z < 3
10. 11
11. 130; 121
12. acute
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A51
Holt Geometry