C Challenge
-
Proof Construction In the figure below, point D is constructed by drawing two arcs. One has center C and radius AB. The other has center B and radius AC. Prove that
A
M
¯
eh m bar is a median of
Δ
A
B
C
.
cap delta eh b c .
-
Probability If two opposite angles of a quadrilateral measure 120 and the measures of the other angles are multiples of 10, what is the probability that the quadrilateral is a parallelogram?
Standardized Test Prep
SAT/ACT
- From which set of information can you conclude that RSTW is a parallelogram?
-
R
S
¯
∥
W
T
¯
,
R
S
¯
≅
S
T
¯
r s bar , parallel to , w t bar , comma , r s bar , approximately equal to , s t bar
-
R
S
¯
∥
W
T
¯
,
S
T
¯
≅
R
W
¯
r s bar , parallel to , w t bar , comma , s t bar , approximately equal to , r w bar
-
R
S
¯
≅
S
T
¯
,
R
W
¯
≅
W
T
¯
r s bar , approximately equal to , s t bar , comma , r w bar , approximately equal to , w t bar
-
R
Z
¯
≅
T
Z
¯
,
S
Z
¯
≅
W
Z
¯
r z bar , approximately equal to , t z bar , comma , s z bar , approximately equal to , w z bar
Short Response
-
Write a proof using the diagram.
Given:
Δ
N
R
J
≅
Δ
C
P
T
,
J
N
¯
∥
C
T
¯
cap delta n r j approximately equal to cap delta c p t comma , j n bar , parallel to , c t bar
Prove:
JNTC is a parallelogram.
Extended Response
-
Use the figure below.
Image Long Description
- Write an equation and solve for x.
- Is
A
F
¯
∥
D
E
¯
?
eh f bar , parallel to , d e bar , question mark Explain.
- Is BDEF a parallelogram? Explain.
Mixed Review
See Lesson 6-2.
Algebra Find the value of each variable in each parallelogram.
-
-
-
See Lessons 4-4 and 4-7.
- Explain how you can use overlapping congruent triangles to prove
A
C
¯
≅
B
D
¯
.
eh c bar , approximately equal to , b d bar , .
Get Ready! To prepare for Lesson 6-4, do Exercises 36–44.
See Lessons 5-2 and 6-2.
PACE is a parallelogram and
m
∠
P
A
C
=
124
.
m angle p eh c , equals 124 , . Complete the following.
-
A
C
=
□
eh c equals white square
-
C
E
=
□
c e equals white square
-
P
A
=
□
p eh equals white square
-
R
E
=
□
r e equals white square
-
C
P
=
□
c p equals white square
-
m
∠
C
E
P
=
□
m angle c e p equals white square
-
m
∠
E
P
A
=
□
m angle e p eh equals white square
-
m
∠
E
C
A
=
□
m angle e c eh equals white square
-
m
∠
A
C
R
=
□
m angle eh c r equals white square