Download Investigating Geometry Activity: The Transitive Property of Segments

Name———————————————————————— lesson
2.6
Date —————————————
Investigating Geometry Activity: The
Transitive Property of Segments and Angles
For use before the lesson “Prove Statements about Segments and Angles”
Materials: Question
explore
metric ruler and protractor
What is the transitive property of segments and angles?
Measuring segments and angles
Measure each segment to the nearest centimeter and each angle to the nearest degree.
A
F
B
D
C
E
Y
draw
conclusions
m∠ X 5    ?   
m∠ Y 5    ?   
m∠ Z 5    ?   
Z
Use your observations above to complete the following.
}
}
}
}
1. Because ​AB​ and CD​
​  are the same length, we can say that AB​
​  is    ?    to CD​
​  .
}
}
}
}
2. Because ​CD​ and EF​
​  are the same length, we can say that CD​
​  is    ?    to EF​
​  .
} }
} }
}
}
3. If ​AB​ > CD​
​  and CD​
​  > EF​
​  , what must be true about AB​
​  and EF​
​  ? Use the
Lesson 2.6
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
X
AB 5    ?   
CD 5    ?   
EF 5    ?   
results of the Explore to support your answer.
4. Use your answers to Exercises 1–3 to complete the transitive property of
congruent segments.
}
}
}
}
}
}
If AB​
​     ?    CD​
​  and CD​
​     ?    EF​
​  , then AB​
​     ?    EF​
​  .
5. Because ∠ X and ∠ Y have the same measure, we can say that ∠ X is    ?    to ∠ Y.
6. Because ∠ Y and ∠ Z have the same measure, we can say that ∠ Y is    ?    to ∠ Z.
7. If ∠ X > ∠ Y and ∠ Y > ∠ Z, what must be true about ∠ X and ∠ Z? Use the
results of the Explore to support your answer.
8. Use your answers to Exercises 5–7 to complete the transitive property of
congruent angles.
If ∠ X    ?    ∠ Y and ∠ Y    ?    ∠ Z, then ∠ X    ?    ∠ Z.
Geometry
Chapter Resource Book
CS10_CC_G_MECR710761_C2L06IG.indd 75
2-75
4/27/11 5:30:26 PM
L 2 a 1 d 5 dn Addition Prop. of Equality
L2a1d
Division Prop. of Equality
d
n
3. S 5 }
​ 2 ​ [2a 1 (n 2 1)d] Given
2
Sp}
​ n ​5 2a 1 (n 2 1)d Mult. Prop. of Equality
2S
  ​2 (n 2 1)d 5 2a Subtract. Prop. of Equality
​ }
n
(n 2 1)d
S
​  2   
​ 5 a Division Prop. of Equality
​ }n ​2 }
1
4.
V5}
​ 3 ​ πh2(3r 2 h) Given
}
​ 
 
 
 
​5 n
3 p V 5 πh2(3r 2 h) Mult. Prop. of Equality
3V
πh
​ }2  ​ 5 3r 2 h
3V
πh
​ }2  ​ 1 h 5 3r
V
πh
h
​ 3 ​5 r
​ } 2 ​ 1 }
ivision Prop. of
D
Equality
ddition Prop. of
A
Equality
ivision Prop. of
D
Equality
5. a. n 5 c(r 1 1)
b. n 5 c(r 1 1)
n 5 cr 1 c
n 2 c 5 cr n2c
   
 ​5 r
​ }
c
Given
Distributive Property
Subtract. Prop. of Equality
Division Prop. of Equality
c. 2%
d. To find the co-worker’s old wage, solve the
formula n 5 c(r 1 1) for c.
n 5 c(r 1 1) Given
n
   ​ 5 c
​ }
r11
10.24
  ​ 
5c
​ }
0.04 1 1
Division Prop. of Equality
Substitution Prop. of Equality
$9.85 < c Simplify.
6. You are given that AB 5 CD. By the Addition
Property of Equality, you can write
AB 1 BC 5 BC 1 CD. You know that
AC 5 AB 1 BC and BD 5 BC 1 CD by the
Segment Addition Postulate. By the Substitution
Property of Equality, you have AC 5 BC 1 CD.
A26
You are given AC 5 6x 2 12, BC 5 4, and
CD 5 3x 2 2. Substitute these expressions into
the equation AC 5 BC 1 CD to obtain
6x 2 12 5 4 1 3x 2 2. Simplify the right side of
the equation to obtain 6x 2 12 5 3x 1 2. By the
Subtraction Property of Equality you have
3x 2 12 5 2. Next, by the Addition Property of
Equality you have 3x 5 14. Finally, by the
14
Division Property of Equality, x 5 }
​ 3  ​. Substitute
this value of x into the expression for CD to obtain
CD 5 12. Because you are given that AB 5 CD,
you know that AB 5 12 also.
7. m∠ RPQ 5 m∠ RPS Given
m∠ SPQ 5 m∠ RPQ 1 m∠ RPS
Segment Addition Postulate
m∠ SPQ 5 m∠ RPQ 1 m∠ RPQ
Substitution Prop. of Equality
m∠ SPQ 5 2(m∠ RPQ) Simplify.
8. a 5 b Given
ac 5 bc Multiplication Prop. of Equality
c5d Given
bc 5 bd Multiplication Prop. of Equality
ac 5 bd Substitution Prop. of Equality
9. You are given that a is a positive integer.
Assume a is even. Then a 5 2k, where k is a
positive integer. Substitute 2k for a in a 1 1 to
obtain 2k 1 1. Because 2k is even, adding 1 to this
expression produces an odd number. Therefore,
a 1 1 is odd.
Lesson 2.6 Prove Statements
about Segments and Angles
Teaching Guide
Statements
2. 81 is between 80 and 89.
Reasoning
3. Definition of a B 4. 80 and 81 are the least
two percents in the interval 80–89.
Investigating Geometry Activity
1. congruent 2. congruent
}
}
3. AB​
​  must be congruent to EF​
​  . Both segments
have a measure of 5 centimeters.
} }
} }
} }
4. If AB​
​  > CD​
​  and CD​
​  > EF​
​  , then AB​
​  > EF​
​  .
5. congruent 6. congruent 7. ∠ X must be
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
answers
Lesson 2.5 Reason Using
Properties from Algebra,
continued
Geometry
Chapter Resource Book
CS10_CC_G_MECR710761_C2AK.indd 26
4/27/11 6:42:31 PM
Lesson 2.6 Prove Statements
about Segments and Angles,
continued
3. 1. Given 2. Reflexive Property of Equality
Practice Level A
1. Transitive Property of Equality; ∠ A > ∠ C
2. Given; DE 5 DF; Symmetric Property of
} }
​   3. ∠ 1 and ∠ 2 are a linear
Equality; DF​
​  > DE​
pair; ∠ 1 and ∠ 2 are supplementary; Definition of
Supplementary Angles; m∠ 1 5 1808 2 m∠ 2
} }
4. ∠ 4 5. ​DX​ 
; CD​
​   6. Transitive Property of
Congruence 7. Reflexive Property of
Congruence 8. Symmetric Property of
Congruence 9. Symmetric Property of
Congruence
10. Sample sketch:
D
C
A
B
E
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
11. 1. 2m∠ ABC 5 m∠ ABD (Given)
2. m∠ ABC 1 m∠ CBD 5 m∠ ABD
(Angle Addition Postulate)
3. 2m∠ ABC 5 m∠ ABC 1 m∠ CBD
(Transitive Property of Equality)
4. m∠ ABC 5 m∠ CBD
(Subtraction Property of Equality)
5. ∠ ABC > ∠ CBD
(Definition of congruent angles)
B
12. Sample answer: a. A
95 mi
C
D
95 mi
b. Given: AB 5 95, CD 5 95 Prove: AC 5 BD
c. 1. AB 5 95, CD 5 95 (Given)
2. AB 1 BC 5 AC, CD 1 BC 5 BD (Segment
Addition Postulate) 3. 95 1 BC 5 AC,
95 1 BC 5 BD (Substitution Property of
Equality) 4. AC 5 95 1 BC (Symmetric Property
of Equality) 5. AC 5 BD (Transitive Property of
Equality)
Practice Level B
1. 1. Given 2. Given 3. Substitution Property
} }
​   5. Given 6. Transitive
of Equality 4. ​HI​ > IJ​
Property of Congruence
2. 1. Given 2. Given 3. Definition of
complementary angles 4. Transitive Property of
Equality 5. Subtraction Property of Equality
6. Definition of congruent angles
Practice Level C
1. Given; m∠ CBD 1 m∠ DBE; Substitution
Property of Equality; Subtraction Property of
Equality; m∠ DBE; ∠ CBD > ∠ DBE; Transitive
Property of Equality 2. Given; definition of
congruent segments; Transitive Property of
Equality; definition of perimeter;
P(ABCD) 5 AB 1 AB 1 AB 1 AB;
P(ABCD) 5 4AB
3. ∠ 5 > ∠ 7 4. ∠ 2 > ∠ 1 and ∠ 4 > ∠ 3
5. Reflexive Property of Congruence
6. Symmetric Property of Congruence
7. Transitive Property of Congruence
} }
} }
8. RS​
​  > ST​
​  and ST​
​  > TU​
​  by the definition of
} }
​  by the Transitive
midpoint. Then ​RS​ > TU​
} }
​ .  Then
Property of Congruence, so RS​
​  5 RT​
5x 1 7 5 7x 2 3 by the Substitution Property of
Geometry
Chapter Resource Book
CS10_CC_G_MECR710761_C2AK.indd 27
answers
congruent to ∠ Z. Both angles have a measure
of 758. 8. If ∠ X > ∠ Y and ∠ Y > ∠ Z, then
∠ X > ∠ Z.
3. Addition Property of Equality 4. Segment
Addition Postulate 5. Segment Addition Postulate
6. Substitution Property of Equality
4. 1. Given 2. Transitive Property of Angle
Congruence 3. m∠ 2 5 m∠ 4 4. Substitution
Property of Equality 5. x 5 6; Because the
angles are congruent, the measures of the angles
are congruent by the definition of congruent
angles. Set the measure of the angles equal to
each other to find x. 6. x 5 3; By the transitive
} }
​  . Set the lengths of the
property, ​FG​ > JH​
segments equal to each other to find x.
7. x 5 5; By the transitive property,
∠ ABD > ∠ EBC. Because the angles are
congruent, the measures of the angles are
congruent by the definition of congruent angles.
Set the measures of the angles equal to each other
to find x. 8. x 5 4; Because the segments are
congruent, the lengths of the segments are
congruent by the definition of congruent
segments. Set the lengths of the segments equal
to each other to find x.
} } } }
9. UV​
​  > ZY​
​ ,  UW​
​  > ZX
​  ​(Given)
UV 5 ZY, UW 5 ZX (Def. of >)
VW 5 UW 2 UV (Segment Addition Postulate)
YX 5 ZX 2 ZY (Segment Addition Postulate)
YX 5 UW 2 UV ( Substitution Property of
Equality)
VW 5 YX ( Transitive Property of
Equality)
} }
​  (Def. of >)
​VW​ > YX​
A27
4/27/11 6:42:31 PM