Download Geometry Final Assessment 11

Geometr y Final Assessment 11-12, 1st semester
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. Name three collinear points.
a. P, G, and N
b. R, P, and N
____
c. R, P, and G
d. R, G, and N
2. Draw and label a pair of opposite rays
a.
b.
____

→
FH
c.
3. Name a plane that contains .
←→
AC
c. plane WRT
d. plane RCA
4. Find the length of .
BC
a.
b.
____
and .
d.
a. plane ACR
b. plane WCT
____

→
FG
BC
BC
= –7
= –9
c.
d.
BC
BC
=7
=8
5. Find the best sketch, drawing, or construction of a segment congruent to .
JK
a.
b.
c.
d.
____
6. D is between C and E.
a. CE = 17.5
b. CE = 78
____
CE
= ,
6x
CD
= , and DE = 27. Find CE.
4x + 8
c. CE = 105
d. CE = 57
7. The map shows a linear section of Highway 35. Today, the Ybarras plan to drive the 360 miles from
Springfield to Junction City. They will stop for lunch in Roseburg, which is at the midpoint of the trip. If
they have already traveled 55 miles this morning, how much farther must they travel before they stop for
lunch?
a. 125 mi
b. 145 mi
____
____
c. 180 mi
d. 305 mi
8. K is the midpoint of . and
a. JK = 1, KL = 1, JL = 2
b. JK = 6, KL = 6, JL = 12
9. m and m . Find m .
JL
∠IJK = 57°
∠IJL = 20°
JK = 6x
. Find JK, KL, and JL.
c. JK = 12, KL = 12, JL = 6
d. JK = 18, KL = 18, JL = 36
KL = 3x + 3
∠LJK
a. m
b. m
c. m
d. m
∠LJK = −37°
∠LJK = 77°
____
10.
bisects , m
a. m = 22°
b. m = 3°

→
∠ABC
BD
, and m
∠ABD = (7x − 1)°
∠ABD
c. m = 40°
d. m = 20°
∠ABD
∠ABD
____
____
∠ABD
11. Find the measure of the complement of , where m
a.
c.
b.
d.
12. Find the measure of the supplement of , where m
a.
c.
b.
d.
13. A billiard ball bounces off the sides of a rectangular billiards table in such a way that
∠M
∠M = 31.1°
58.9°
31.1°
148.9°
121.1°
∠R
44.5°
(190 − 8z)°
(80 − 8z)°
, find m , m , and m .
∠1 = 26.5°
a.
b.
c.
d.
∠R = (8z + 10)°
(170 − 8z)°
are complementary. If m
____
∠LJK = 40°
. Find m .
∠DBC = (4x + 8)°
∠ABD
____
∠LJK = 37°
m
m
m
m
;m
;m
;m
;m
∠3
∠4
;m
;m
;m
;m
∠3 = 26.5°
∠4 = 63.5°
∠5 = 63.5°
∠3 = 26.5°
∠4 = 63.5°
∠5 = 53°
∠3 = 63.5°
∠4 = 26.5°
∠5 = 53°
∠3 = 26.5°
∠4 = 153.5°
∠5 = 26.5°
14. Find the perimeter and area of the figure.
∠5
,
∠1 ≅ ∠3
, and
∠4 ≅ ∠6
∠3
and
∠4
a. perimeter =
;
c. perimeter = ;
area =
area =
b. perimeter = ;
d. perimeter = ;
area =
area =
15. Find the circumference and area of the circle. Use 3.14 for , and round your answer to the nearest tenth.
6x 2 + 14
7x + 14
3x + 24
6x + 48
7x + 14
7x + 14
6x 2 + 14
3x + 24
____
π
a.
b.
____
C = 201.0
C = 50.2
ft;
ft;
A = 50.2
A = 25.1
ft 2
ft 2
c.
d.
C = 25.1
C = 50.2
ft;
ft;
A = 50.2
A = 201.0
ft 2
ft 2
16. The width of a rectangular mirror is the measure of the length of the mirror. If the area is 192 in , what are
2
3
4
the length and width of the mirror?
a. length = 24 in., width = 8 in.
b. length = 16 in., width = 12 in.
____
17. Find the coordinates of the midpoint of
c. length = 48 in., width = 4 in.
d. length = 25 in., width = 71 in.
CM
with endpoints C(1, –6) and M(7, 5).
a. (3, -1)
b. (8, –1)
____
c. (4, 0/-1/2)
d. (4/1/2, 0/1/2)
18. Use the Distance Formula and the Pythagorean Theorem to find the distance, to the nearest tenth, from T(4,
–2) to U(–2, 3).
a. –1.0 units
b. 3.4 units
____
c. 0.0 units
d. 7.8 units
19. R is the midpoint of . T is the midpoint of . S is the midpoint of . Use the diagram to find the coordinates
AB
AC
BC
of T, the area of RST, and . Round your answers to the nearest tenth.
∆
AB
T(3, 1); area of ∆RST = 8;
T(3, 1); area of ∆RST = 32;
T(3, 1); area of ∆RST = 16;
T(3, 1); area of ∆RST = 8;
20. Determine if the conjecture is valid by the Law of Syllogism.
a.
b.
c.
d.
____
AB ≈ 17.9
AB ≈ 17.9
AB ≈ 8.9
AB ≈ 8.9
Given: If you are in California, then you are in the west coast. If you are in Los Angeles, then you are in
California.
Conjecture: If you are in Los Angeles, then you are in the west coast.
a. No, the conjecture is not valid.
b. Yes, the conjecture is valid.
____
21. Solve the equation
4x − 6
. Write a justification for each step.
Given equation
[1]
Simplify.
[2]
Simplify.
4x − 6 = 34
= 34
+6
+6
= 40
4x
4x
4
= 40
4
x
= 10
a. [1] Substitution Property of Equality;
____
c. [1] Division Property of Equality;
[2] Division Property of Equality
[2] Subtraction Property of Equality
[1]
Addition
Property
of
Equality;
b.
d. [1] Addition Property of Equality;
[2] Division Property of Equality
[2] Reflexive Property of Equality
22. Use the given plan to write a two-column proof.
Given: m + m = 90 , m + m = 90 , m = m
∠1
°
∠2
∠3
∠4
°
∠2
∠3
Prove: m = m
Plan: Since both pairs of angle measures add to 90 , use substitution to show that the sums of both pairs are
equal. Since m = m , use substitution again to show that sums of the other pairs are equal. Use the
Subtraction Property of Equality to conclude that m = m .
∠1
∠4
°
∠2
∠3
∠1
∠4
Complete the proof.
Proof:
Statements
1. m
2. [1]
3. m
4. m
5. m
6. m
∠1
∠1
∠2
∠1
∠1
+ m = 90
°
∠2
+m =m +m
=m
+m =m +m
=m
∠2
∠3
∠4
∠2
∠4
∠3
∠2
∠4
a. [1] m + m = 90
∠3
∠4
°
[2] Substitution Property
[3] Subtraction Property of Equality
Reasons
1. Given
2. Given
3. Substitution Property
4. Given
5. [2]
6. [3]
b. [1] m + m = 90
∠5
____
∠6
°
[2] Substitution Property
[3] Subtraction Property of Equality
c. [1] m + m = 90
[2] Subtraction Property of Equality
[3] Substitution Property
d. [1] m + m = 90
[2] Addition Property of Equality
[3] Substitution Property
23. Use the given flowchart proof to write a two-column proof of the statement
∠3
∠4
°
∠5
∠6
°
.
AF ≅ FD
Flowchart proof:
;
AB = CD
AB + BF = AF
FC + CD = FD
BF = FC
Segment Addition
Postulate
Given
AB + BF =
FC + CD
Addition
Property of
Equality
AF = FD
AF ≅ FD
Substitution
Definition of
congruent segments
Complete the proof.
Two-column proof:
Statements
1. ;
2. [1]
3. [2]
4.
5.
AB = CD
BF = FC
AF = FD
AF ≅ FD
a. [1]
Reasons
1. Given
2. Addition Property of Equality
3. Segment Addition Postulate
4. Substitution
5. Definition of congruent segments
AB + BF = AF ; FC + CD = FD
[2]
b. [1]
[2]
c. [1] ;
[2]
d. [1]
[2]
24. Identify the transversal and classify the angle pair
AF = FD
AF = FD
AB + BF = FC + CD
AB = CD
BF = FC
AB + BF = FC + CD
AB + BF = FC + CD
AB + BF = AF ;FC + CD = FD
____
∠11
and .
∠7
a.
b.
c.
d.
____
The transversal is line l. The angles are corresponding angles.
The transversal is line l. The angles are alternate interior angles.
The transversal is line n. The angles are alternate exterior angles.
The transversal is line m. The angles are corresponding angles.
25. Find m .
∠RST
a. m =
b. m =
____
∠RST
108°
∠RST
24°
26. Classify
∆DBC
c. m =
d. m =
by its angle measures, given m
a. obtuse triangle
b. acute triangle
____
27.
∆ABC
a.
b.
____
is an isosceles triangle.
AB
AB
= 110
= 24
,m
∠DAB = 60°
∠RST
156°
∠RST
72°
, and m
∠ABD = 75°
.
∠BDC = 25°
c. right triangle
d. equiangular triangle
AB
is the longest side with length .
4x + 4
c.
d.
AB
AB
BC
=
8x + 3
and
CA
=
. Find .
7x + 8
AB
= 43
=5
28. One of the acute angles in a right triangle has a measure of . What is the measure of the other acute angle?
34.6°
a.
b.
____
c.
d.
145.4°
34.6°
29. Find m , given
∠DCB
,
∠A ≅ ∠F
, and m
∠B ≅ ∠E
.
c. m
d. m
∠DCB = 134°
∠DCB = 67°
30. Given: P is the midpoint of
Prove:
90°
∠CDE = 46°
a. m
b. m
____
55.4°
∠DCB = 44°
∠DCB = 46°
and .
RS
TQ
∆TPR ≅ ∆QPS
Complete the proof.
Proof:
Statements
1. P is the midpoint of and .
2. ,
3. [2]
4.
Reasons
1. Given
2. [1]
3. Vertical Angles Theorem
4. [3]
a. [1]. Definition of midpoint
c. [1] Definition of midpoint
TQ
TP ≅ QP
RS
RP ≅ SP
∆TPR ≅ ∆QPS
[2]
[3] SAS
b. [1] Definition of midpoint
d.
[2]
[3] SSS
31. What additional information do you need to prove
∠TPR ≅ ∠QPS
RT ≅ SQ
____
[2]
[3] SAS
[1] Definition of midpoint
[2]
[3] SSS
by the SAS Postulate?
∠PRT ≅ ∠PSQ
∠TPR ≅ ∠QPS
∆ABC ≅ ∆ADC
a.
b.
____
c.
d.
AB ≅ AD
∠ACB ≅ ∠ACD
c. x = 2
d. x = 8
33. Which of the following is not a positioning of a right triangle with leg lengths of 4 units and 5 units?
a.
c.
b.
____
BC ≅ DC
32. Find the value of x.
a. x = 6
b. x = 4
____
∠ABC ≅ ∠ADC
d.
34. Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints
.
B(5, 4)
a.
b.
____
y−3 =
y−3 =
35. In ,
∆ABC
-7/2
2/7
c.
d.
(x − 1.5)
(x − 1.5)
.
BY = 26.4 and CO = 24
AX BY, and CZ
are medians. Find .
BO
-2/7
7/2
y−1 =
(x − 3.5)
y−1 =
(x − 3.5)
A(−2, 2)
and
a.
b.
____
BO
BO
= 17.6
= 26.4
36. Given
∆ABC
c.
d.
with
,
AB = 3
, and
37. In ,
∆ABC
a.
b.
____
,
m∠ADC > m∠BDC
____
____
XY
c. XY = 2.5
d. XY = 2
, and
AC = 3x + 32
. Find the range of values for x.
BC = 7x + 16
c.
d.
0<x<4
− 167 < x < 4
x>4
− 78 < x < 4
38. Tell if the measures 6, 14, and 13 can be side lengths of a triangle. If so, classify the triangle as acute, right,
c. Yes; right triangle
d. No.
39. Tell whether the polygon is regular or irregular. Tell whether it is concave or convex.
a. regular and concave
b. irregular and concave
____
= 24
= 8.8
CA = 6
or obtuse.
a. Yes; acute triangle
b. Yes; obtuse triangle
____
BO
, find the length of midsegment .
BC = 5
a. XY = 3
b. XY = 1.5
____
BO
c. regular and convex
d. irregular and convex
40. Find the measure of each interior angle of a regular 45-gon.
a. 176
c. 172
b. 164
d. 188
41. Find the measure of each exterior angle of a regular decagon.
a. 45°
c. 18°
b. 22.5°
d. 36°
42.
and
. Determine if the quadrilateral must be a parallelogram. Justify your answer.
KL ≅ MN
°
°
°
°
∠KLM ≅ ∠MNK
a. No. Only one set of angles and sides are given as congruent. The conditions for a
parallelogram are not met.
b. Yes. Opposite angles are congruent to each other. This is sufficient evidence to prove that
the quadrilateral is a parallelogram.
c. Yes. Opposite sides are congruent to each other. This is sufficient evidence to prove that
the quadrilateral is a parallelogram.
d. Yes. One set of opposite sides are congruent, and one set of opposite angles are congruent.
This is sufficient evidence to prove that the quadrilateral is a parallelogram.
Numer ic Response
43. Find the measure of the angle formed by the hands of a clock when it is 7:00.
44. The supplement of an angle is 26 more than five times its complement. Find the measure of the angle.
45. A satellite completely orbits Earth in 216 days. Determine the angle through which the satellite travels over
a period of 12 days.
46. In the quadrilateral,
. If
m∠1 + m∠2 + m∠3 + m∠4 = 360°
,
m∠2 = 3m∠1
, and
m∠3 = m∠1 + 6
, find
m∠4 = m∠1
m∠3
in degrees.
47. An isosceles triangle has a perimeter of 50 in. The congruent sides measure (2x + 3) cm. The length of the
third side is 4x cm. What is the value of x?
48. Find the value of x.
49. Find the value of n in the triangle.
50. In parallelogram LMNO,
, and
NO = 10.2
. What is the perimeter of parallelogram LMNO?
LO = 14.7
51. Find the value of x in the rhombus.
Matching
Match each vocabulary term with its definition.
a. bisect
e.
b. midpoint
f.
c. angle bisector
g.
d. angle
h.
____
____
____
____
____
____
52.
53.
54.
55.
56.
57.
perpendicular bisector
measure
degree
segment bisector
a unit of angle measure
the point that divides a segment into two congruent segments
to divide into two congruent parts
a figure formed by two rays with a common endpoint
the use of units to find a size or quantity
a line, ray, or segment that divides a segment into two congruent segments
Match each vocabulary term with its definition.
a. conclusion
e. hypothesis
converse
b.
f. truth value
c. inverse
g. contrapositive
d. negation
____
____
____
____
____
58.
59.
60.
61.
62.
for a statement, either true (T) or false (F)
operations that undo each other
the contradiction of a statement by using “not,” written as ∼
the statement formed by exchanging the hypothesis and conclusion of a conditional statement
the statement formed by both exchanging and negating the hypothesis and conclusion
Geometr y Final Assessment 11-12, 1st semester
Answer Section
MULTIPLE CHOICE
1. ANS: D
Collinear points are points that lie on the same line.
R, G, and N are three collinear points.
Feedback
A
B
C
D
Collinear points are points that lie on the same line.
Collinear points are points that lie on the same line.
Points R, P, and G are noncollinear.
Correct!
PTS:
OBJ:
STA:
KEY:
2. ANS:
1
DIF: Basic
REF: Page 7
1-1.1 Naming Points, Lines, and Planes
NAT: 12.3.1.c
GE1.0
TOP: 1-1 Understanding Points Lines and Planes
points | lines | planes
B
In the diagram, rays

→
FG
and

→
FH
share a common endpoint F and form the line .
←

→
GH
Feedback
A
B
C
D
Opposite rays form a line.
Correct!
Opposite rays form a line.
Opposite rays are two rays that have a common endpoint and form a line.
PTS:
OBJ:
TOP:
3. ANS:
1
DIF: Basic
REF: Page 7
1-1.2 Drawing Segments and Rays NAT: 12.3.1.d
1-1 Understanding Points Lines and Planes
C
STA: GE1.0
KEY: opposite rays
A plane can be described by any three noncollinear points. Of the choices given, only points W, R, and T are
noncollinear. Thus, lies in plane WRT.
←→
AC
Feedback
A
B
C
D
Points A, C, and R are collinear. A plane can be described by any three noncollinear
points.
Points W, C, and T are collinear. A plane can be described by any three noncollinear
points.
Correct!
A plane can be described by any three noncollinear points.
PTS:
OBJ:
STA:
KEY:
4. ANS:
1
DIF: Basic
REF: Page 7
1-1.3 Identifying Points and Lines in a Plane
NAT: 12.3.4.b
GE1.0
TOP: 1-1 Understanding Points Lines and Planes
points | lines | planes
C
= | −8 − (−1) |
BC
= | −8 + 1 |
= | −7 |
=7
Feedback
A
B
C
D
The length of a segment is always positive.
Find the absolute value of the difference of the coordinates.
Correct!
Find the absolute value of the difference of the coordinates.
PTS:
OBJ:
STA:
KEY:
5. ANS:
1
DIF: Basic
REF: Page 13
1-2.1 Finding the Length of a Segment
NAT: 12.2.1.e
GE1.0
TOP: 1-2 Measuring and Constructing Segments
segments | length
B
Use a ruler to measure the lengths of , , , , and .
JK
LM
PQ
GH
RS
Feedback
A
B
C
D
Use a ruler to measure the lengths of all the segments.
Correct!
Use a ruler to measure the lengths of all the segments.
Use a ruler to measure the lengths of all the segments.
PTS:
OBJ:
TOP:
KEY:
6. ANS:
1
DIF: Average
REF: Page 14
1-2.2 Copying a Segment
STA: GE16.0
1-2 Measuring and Constructing Segments
copying a segment | constructions
C
CE = CD + DE
6x
6x = ( 4x + 8 ) + 27
6x = 4x + 35
CE
4x + 8
CD
4x
2x = 35
2x = 35
2
2
or 17.5
x = 35
2
CE
Segment Addition Postulate
Substitute for and for .
Simplify.
Subtract from both sides.
Divide both sides by 2.
Simplify.
= 6x
= 6( 17.5 ) = 105
Feedback
A
You found the value of x. Find the length of the specified segment.
B
C
D
You found the length of a different segment.
Correct!
Check your equation. Make sure you are not subtracting instead of adding.
PTS:
OBJ:
STA:
KEY:
7. ANS:
1
DIF: Average
REF: Page 15
1-2.3 Using the Segment Addition Postulate
NAT: 12.3.5.a
GE1.0
TOP: 1-2 Measuring and Constructing Segments
segment addition postulate
A
If the Ybarra’s current position is represented by X, then the distance they must travel before they stop for
lunch is XR.
Segment Addition Postulate
Solve for XR.
Substitute known values. R is the midpoint of , so
Simplify.
SX + XR = SR
XR = SR − SX
XR =
1
2
SJ
( 360 ) − 55
XR = 125
SR =
1
2
.
SJ
Feedback
A
B
C
D
Correct!
Use the definition of midpoint and the Segment Addition Postulate to find the distance
to Roseburg.
This is the distance from Springfield to Roseburg. You must subtract the distance they
have already traveled.
This is the distance to Junction City. Use the definition of midpoint and the Segment
Addition Postulate to find the distance to Roseburg.
PTS:
NAT:
KEY:
8. ANS:
1
DIF: Average
REF: Page 15
OBJ: 1-2.4 Application
12.2.1.e
STA: GE1.0
TOP: 1-2 Measuring and Constructing Segments
application | segment addition postulate
B
Step 1 Write an equation and solve.
K is the midpoint of .
Substitute for and for .
Subtract from both sides.
Divide both sides by 3.
JL
JK = KL
6x
6x = 3x + 3
JK
3x + 3
KL
3x
3x = 3
x=1
Step 2 Find JK, KL, and JL.
JK = 6x = 6( 1 ) = 6
KL = 3x + 3 = 3(1) + 3 = 6
JL = JK + KL = 6 + 6 = 12
Feedback
A
B
This is the value of x. Substitute this value for x to solve for the segment lengths.
Correct!
C
D
Reverse your answers. The first two segments are half as long as the last segment.
Check your simplification methods when solving for x. Use division for the last step.
PTS:
OBJ:
STA:
KEY:
9. ANS:
m
m
m
m
∠IJK =
∠IJL +
m
Angle Addition Postulate
Substitute for m and 20° for m .
Subtract ° from both sides.
∠LJK
∠LJK
57° = 20° +
37° =
1
DIF: Average
REF: Page 16
1-2.5 Using Midpoints to Find Lengths
NAT: 12.2.1.e
GE1.0
TOP: 1-2 Measuring and Constructing Segments
midpoints | length
C
∠IJK
57°
∠LJK
∠IJL
20
Feedback
A
B
C
D
Use the Angle Addition Postulate.
Subtract the smaller angle measure from the larger angle measure.
Correct!
Subtract the smaller angle measure from the larger angle measure.
PTS:
OBJ:
STA:
KEY:
10. ANS:
1
DIF: Basic
REF: Page 22
1-3.3 Using the Angle Addition Postulate
NAT: 12.2.1.f
GE1.0
TOP: 1-3 Measuring and Constructing Angles
angle addition postulate
D
Step 1 Solve for x.
m m
Definition of angle bisector.
(7x − 1)° = (4x + 8)°
Substitute
∠ABD =
∠DBC
7x − 1
for
∠ABD
and
4x + 8
for
.
∠DBC
Add 1 to both sides.
Subtract from both sides.
Divide both sides by 3.
7x = 4x + 9
4x
3x = 9
x=3
Step 2 Find m .
m
∠ABD
∠ABD = 7x − 1 = 7(3) − 1 = 20°
Feedback
A
B
C
D
Check your simplification technique.
Substitute this value of x into the expression for the angle.
This answer is the entire angle. Divide by two.
Correct!
PTS:
OBJ:
STA:
KEY:
11. ANS:
1
DIF: Average
REF: Page 23
1-3.4 Finding the Measure of an Angle
NAT: 12.2.1.f
GE1.0
TOP: 1-3 Measuring and Constructing Angles
angle bisectors | angle measures
A
Subtract from 90º and simplify.
90° − 31.1°=
58.9
°
Feedback
A
B
C
D
Correct!
Find the measure of a complementary angle, not a supplementary angle.
Complementary angles are angles whose measures have a sum of 90 degrees.
The measures of complementary angles add to 90 degrees.
PTS:
OBJ:
NAT:
KEY:
12. ANS:
1
DIF: Basic
REF: Page 29
1-4.2 Finding the Measures of Complements and Supplements
12.3.3.g
STA: 6MG2.2
TOP: 1-4 Pairs of Angles
complementary angles | supplementary angles
A
Subtract from 180º and simplify.
180° − (8z + 10)°
= 180 − 8z − 10
= (170 − 8z)°
Feedback
A
B
C
D
Correct!
The measures of supplementary angles add to 180 degrees.
Supplementary angles are angles whose measures have a sum of 180 degrees.
Find the measure of a supplementary angle, not a complementary angle.
PTS:
OBJ:
NAT:
KEY:
13. ANS:
1
DIF: Average
REF: Page 29
1-4.2 Finding the Measures of Complements and Supplements
12.3.3.g
STA: 6MG2.2
TOP: 1-4 Pairs of Angles
complementary angles | supplementary angles
B
Since ,
Thus m .
∠1 ≅ ∠3
.
m∠1 ≅ m∠3
∠3 = 26.5°
Since
∠3
and
Since ,
Thus m .
∠4 ≅ ∠6
∠4
are complementary, m
.
∠4 = 90° − 26.5° = 63.5°
.
m∠4 ≅ m∠6
∠6 = 63.5°
By the Angle Addition Postulate,
180° = m∠4 + m∠5 + m∠6
= 63.5° + m∠5 + 63.5°
Thus, m .
∠5 = 53°
Feedback
A
B
C
D
The measure of angle 5 is 180 degrees minus the sum of the measure of angle 4 and the
measure of angle 6.
Correct!
Angle 1 and angle 3 are congruent. Congruent angles have the same measure.
Angle 3 and angle 4 are complementary, not supplementary.
PTS:
OBJ:
TOP:
KEY:
14. ANS:
1
DIF: Average
REF: Page 30
1-4.4 Problem-Solving Application NAT: 12.3.3.g
STA: 6MG2.2
1-4 Pairs of Angles
application | complementary angles | supplementary angles
B
Solve for the perimeter of the triangle.
P=
=
=
Solve for the area of the triangle.
A=
=
=
a+b+c
6 + (x + 8) + 6x
7x + 14
1
2
bh
1
2
(x + 8)(6)
3x + 24
Feedback
A
B
C
D
Check your algebra when adding like terms.
Correct!
The triangle's area is half of its base times its height.
The triangle's area is half of its base times its height.
PTS:
OBJ:
STA:
KEY:
15. ANS:
C = 2πr = 2π ( 4 ) ≈ 25.1
A = πr2 = π ( 4 )
2
1
DIF: Average
REF: Page 36
1-5.1 Finding the Perimeter and Area
GE8.0
TOP: 1-5 Using Formulas in Geometry
perimeter | area | triangles
C
NAT: 12.2.1.h
ft
≈ 50.2
ft 2
Feedback
A
B
Use the radius, not the diameter, in your calculations.
The circumference of a circle is 2 times pi times the radius. The area of a circle is pi
times the radius squared.
Correct!
Use the radius, not the diameter, in your calculations.
C
D
PTS:
OBJ:
STA:
KEY:
16. ANS:
1
DIF: Average
REF: Page 37
1-5.3 Finding the Circumference and Area of a Circle
GE8.0
TOP: 1-5 Using Formulas in Geometry
circles | circumference | area
B
NAT: 12.2.1.h
The area of a rectangle is found by multiplying the length and width. Let l represent the length of the
mirror. Then the width of the mirror is .
3
4
l
A = lw
192 = l( 34 l)
192 =
3 2
l
4
256 = l2
16 = l
The length of the mirror is 16 inches. The width of the mirror is
Feedback
3
4
(16) = 12
inches.
A
B
C
D
First, find the length. Then, use substitution to find the width.
Correct!
First, find the length. Then, use substitution to find the width.
The formula for the area of a rectangle is length times width.
PTS: 1
DIF: Advanced
TOP: 1-5 Using Formulas in Geometry
17. ANS: C
=
NAT: 12.2.1.h
STA: GE8.0
KEY: area | rectangles | application
(4, 0/-1/2)
Êx + x
Á
y + y ˆ˜ Ê
Á 1 + 7 −6 + 5 ˜˜ˆ
1
2
Á
MÁ
, 1 2 ˜˜˜˜ = Á
,
Á
˜
Á
Á
Á
2 ˜¯ Á
2 ˜¯
Ë 2
Ë 2
Feedback
A
The x- and y-coordinates of the midpoint are the averages of the x- and y-coordinates of
the endpoints.
The x- and y-coordinates of the midpoint are the averages of the x- and y-coordinates of
the endpoints.
Correct!
The x- and y-coordinates of the midpoint are the averages of the x- and y-coordinates of
the endpoints.
B
C
D
PTS:
OBJ:
STA:
KEY:
18. ANS:
1
DIF: Basic
REF: Page 43
1-6.1 Finding the Coordinates of a Midpoint
NAT: 12.2.1.e
GE17.0
TOP: 1-6 Midpoint and Distance in the Coordinate Plane
midpoint formula | coordinates
D
Method 1 Substitute the values for the
coordinates of T and U into the Distance
Formula.
TU =
Method 2 Use the Pythagorean Theorem.
Plot the points on a coordinate plane. Then
draw a right triangle.
ˆ˜ 2
Ê
ˆ˜ 2 Ê
Á
Á
Ëx 2 − x 1 ¯ + Ëy 2 − y 1 ¯
=
( −2 − 4 ) + ( 3 − −2 )
=
2
( −6 ) + ( 5 )
=
61
2
2
2
≈ 7.8 units
Count the units for sides a and b. and .
Then apply the Pythagorean Theorem.
a=6
c 2 = a2 + b2 = 62 + 52 = 36 + 25 = 61
c ≈ 7.8 units
Feedback
A
B
C
D
The distance is the square root of the quantity
The distance is the square root of the quantity
The distance is the square root of the quantity
Correct!
PTS:
1
DIF:
Average
.
.
.
(x2 − x1)^2 + (y2 − y1)^2
(x2 − x1)^2 + (y2 − y1)^2
(x2 − x1)^2 + (y2 − y1)^2
REF: Page 45
b=5
OBJ:
STA:
KEY:
19. ANS:
1-6.4 Finding Distances in the Coordinate Plane
NAT: 12.2.1.e
GE15.0
TOP: 1-6 Midpoint and Distance in the Coordinate Plane
congruent segments | distance formula | Pythagorean Theorem
D
Using the given diagram, the coordinates of T are (3, 1).
The area of a triangle is given by .
From the diagram, the base of the triangle is .
From the diagram, the height of the triangle is .
Therefore the area is
.
To find , use the Distance Formula with points A(1,5) and B .
≈ 8.9
A=
1
2
bh
b = RT = 4
h=4
A=
1
2
(4)(4) = 8
AB
AB =
(x 2 − x 1) 2 + (y 2 − y 1) 2
=
(−3, −3)
(−3 − 1) 2 + (−3 − 5) 2
=
16 + 64
=
80
Feedback
A
B
C
D
Use the distance formula to find the measurement of AB.
The area of a triangle is one half the measure of its base times the measure of its height.
The area of a triangle is one half times the measure of its base times the measure of its
height.
Correct!
PTS:
TOP:
KEY:
20. ANS:
1
DIF: Advanced
NAT: 12.2.1.e
1-6 Midpoint and Distance in the Coordinate Plane
area | distance formula | triangles
B
STA: GE17.0
Let p, q, and r represent the following.
p: You are in California.
q: You are in the west coast.
r: You are in Los Angeles.
You are given that
p→q
and
r→p
Since p is the conclusion of the second statement and the hypothesis of the first statement, reorder the
statements like this and .
r→p
p→q
By the Law of Syllogism, if and are true, then is true.
is the statement, If you are in Los Angeles, then you are in the west coast.
r→p
p→q
r→q
r→q
Feedback
A
B
The Law of Syllogism states that if (if p, then q) and (if q, then r) are true statements,
then (if p, then r) is a true statement.
Correct!
PTS:
OBJ:
NAT:
TOP:
21. ANS:
1
DIF: Average
REF: Page 89
2-3.3 Verifying Conjectures by Using the Law of Syllogism
12.3.5.a
STA: GE3.0
2-3 Using Deductive Reasoning to Verify Conjectures
B
4x − 6
= 34
Given equation
+6
[1] Addition Property of Equality
Simplify.
[2] Division Property of Equality
Simplify.
+6
= 40
4x
4x
4
= 40
4
x
= 10
Feedback
A
B
C
D
Check the properties.
Correct!
Check the properties.
Check the properties.
PTS:
OBJ:
STA:
22. ANS:
1
DIF: Basic
REF: Page 104
2-5.1 Solving an Equation in Algebra
1A5.0
TOP: 2-5 Algebraic Proof
A
NAT: 12.5.2.e
Proof:
1. m
2. m
3. m
4. m
5. m
6. m
+m
+m
+m
=m
+m
=m
∠1
∠2
∠3
∠4
∠1
∠2
∠2
Statements
= 90
= 90
=m +m
Reasons
1. Given
2. Given
3. Substitution Property
4. Given
5. Substitution Property
6. Subtraction Property of Equality
°
°
∠3
∠4
∠3
∠1
∠2
∠1
=m +m
∠2
∠4
∠4
Feedback
A
B
C
D
Correct!
Check the given information.
To get from Step 4 to Step 5, use substitution, not subtraction.
To get from Step 4 to Step 5, use substitution, not addition.
PTS:
OBJ:
STA:
23. ANS:
1
DIF: Average
REF: Page 112
2-6.3 Writing a Two-Column Proof from a Plan
GE2.0
TOP: 2-6 Geometric Proof
D
NAT: 12.3.5.a
In a flowchart, reasons flow from the statement above. The statement above Reason 2 is
statement above Reason 3 is
.
AB + BF = AF ; FC + CD = FD
Feedback
A
B
C
D
Reasons flow from the statement above.
Reasons flow from the statement above.
Reasons flow from the statement above.
Correct!
PTS: 1
DIF: Average
REF: Page 118
OBJ: 2-7.1 Reading a Flowchart Proof
NAT: 12.3.5.a
TOP: 2-7 Flowchart and Paragraph Proofs
STA: GE2.0
. The
AB + BF = FC + CD
24. ANS: A
To determine which line is the transversal for a given angle pair, locate the line that connects the vertices.
Corresponding angles lie on the same side of the transversal l, on the same sides of lines n and m.
Feedback
A
B
C
Correct!
Alternate interior angles lie on opposite sides of the transversal, between two lines.
To find which line is the transversal for a given angle pair, locate the line that connects
the vertices.
To find which line is the transversal for a given angle pair, locate the line that connects
the vertices.
D
PTS:
OBJ:
STA:
25. ANS:
1
DIF: Average
REF: Page 147
3-1.3 Identifying Angle Pairs and Transversals
GE7.0
TOP: 3-1 Lines and Angles
D
NAT: 12.3.3.g
Alternate Exterior Angles Theorem
Subtract from both sides.
Divide both sides by .
(3x)° = (4x − 24)°
4x
−x = −24
−1
x = 24
m
∠RST = 3x = 3(24) = 72°
Substitute 24 for .
x
Feedback
A
B
C
D
Find the measure of angle RST, not the supplement.
Find the measure of angle RST, not the value of x.
After finding x, substitute to find the angle measure.
Correct!
PTS:
OBJ:
TOP:
26. ANS:
∠ABD
∆DBC
1
DIF: Average
REF: Page 156
3-2.2 Finding Angle Measures
NAT: 12.3.3.g
3-2 Angles Formed by Parallel Lines and Transversals
A
and form a linear pair, so they are supplementary. Therefore
is an obtuse triangle by definition.
∠DBC
STA: GE7.0
. By substitution,
m∠ABD + m∠DBC = 180°
Feedback
A
B
C
D
Correct!
An acute triangle has three acute angles.
A right triangle has one right angle.
An equiangular triangle has three congruent acute angles.
PTS:
OBJ:
STA:
27. ANS:
1
DIF: Average
REF: Page 216
4-1.1 Classifying Triangles by Angle Measures
GE12.0
TOP: 4-1 Classifying Triangles
B
Step 1 Find the value of x.
Step 2 Find .
AB
NAT: 12.3.3.f
. So
75° + m∠DBC = 180°
.
m∠DBC = 105°
BC
=
CA
=
4x + 4
8x + 3
=
7x + 8
=
4(5) + 4
x
=
5
=
24
AB
Feedback
A
B
C
D
First, find the value of x. Then, use substitution to find AB.
Correct!
This is equals BC and CA. Now find AB.
This is the value of x. Now find AB.
PTS:
OBJ:
TOP:
28. ANS:
1
DIF: Average
REF: Page 217
4-1.3 Using Triangle Classification NAT: 12.3.3.f
4-1 Classifying Triangles
C
Let the acute angles be
m m
m
m
∠M
34.6°
+
∠N
∠N
+
∠N
STA: GE12.0
and , with m .
The acute angles of a right triangle are complementary.
Substitute for m .
Subtract from both sides.
∠M
∠N
= 90°
= 90°
∠M = 34.6°
∠M
34.6°
= 55.4°
34.6°
Feedback
A
B
C
D
The two acute angles in a right triangle are complementary.
This is the measure of the given angle. Find the measure of the other acute angle.
Correct!
The measure of the other acute angle is less than 90 degrees.
PTS:
OBJ:
STA:
29. ANS:
1
DIF: Basic
REF: Page 225
4-2.2 Finding Angle Measures in Right Triangles
NAT: 12.3.3.f
GE12.0
TOP: 4-2 Angle Relationships in Triangles
D
The Third Angles Theorem states that if two angles of one triangle are congruent to two angles of another
triangle, then the third pair of angles are congruent.
It is given that
∠A ≅ ∠F
and
. Therefore,
∠B ≅ ∠E
. So, m
∠CDE ≅ ∠DCB
.
∠DCB = 46°
Feedback
A
B
C
D
This is the supplement. Use the Third Angles Theorem.
The Third Angles Theorem states that if two angles of one triangle are congruent to two
angles of another triangle, then the third pair of angles are congruent.
This is the complement. Use the Third Angles Theorem.
Correct!
PTS: 1
DIF: Advanced
NAT: 12.3.3.f
TOP: 4-2 Angle Relationships in Triangles
30. ANS: A
STA: GE12.0
Proof:
Statements
Reasons
1. P is the midpoint of
2. ,
3.
4.
TQ
1. Given
2. Definition of midpoint
3. Vertical Angles Theorem
4. SAS
and .
RS
RP ≅ SP
TP ≅ QP
∠TPR ≅ ∠QPS
∆TPR ≅ ∆QPS
Feedback
A
B
C
D
Correct!
There is not enough information to show that segment RT is congruent to segment SQ.
Angle PRT and angle PSQ are not vertical angles.
Use the correct postulate to prove the triangles congruent.
PTS:
OBJ:
TOP:
31. ANS:
1
DIF: Average
REF: Page 244
4-4.4 Proving Triangles Congruent NAT: 12.3.5.a
4-4 Triangle Congruence: SSS and SAS
B
STA: GE5.0
The SAS Postulate is used when two sides and an included angle of one triangle are congruent to the
corresponding sides and included angle of a second triangle.
From the given, .
From the figure,
by the Reflexive Property of Congruence.
You have two pair of congruent sides, so you need information about the included angles.
BC ≅ DC
AC ≅ AC
Use these pairs of sides to determine the included angles.
The angle between sides
is .
The angle between sides
is .
You need to know
to prove
by the SAS Postulate.
∠ACB ≅ ∠ACD
AC and BC
∠ACB
AC and DC
∠ACD
∆ABC ≅ ∆ADC
Feedback
A
B
C
D
This information is needed to use the SSS Postulate.
Correct!
You need the included angle between the two sides.
This information is already given. Find information that you need that is not given or
true in the figure.
PTS: 1
DIF: Advanced
NAT: 12.3.5.a
TOP: 4-4 Triangle Congruence: SSS and SAS
32. ANS: A
The triangles can be proved congruent by the SAS Postulate.
By CPCTC,
.
3x − 5 = 2x + 1
Solve the equation for x.
3x − 5 = 2x + 1
3x = 2x + 6
x=6
Feedback
A
B
Correct!
When solving, you can either add 5 or subtract 1 from each side.
STA: GE5.0
C
D
Remember to combine the like terms when solving.
These two triangles have SAS congruence, so the two expressions are equal by CPCTC.
PTS: 1
DIF: Advanced
TOP: 4-6 Triangle Congruence: CPCTC
33. ANS: B
NAT: 12.3.2.e
STA: GE5.0
These graphs show
right triangles with
leg lengths of 4
units and 5 units
with one vertex at
the origin.
This graph shows a right triangle with leg
lengths of 4 units and 5 units with the base
centered at the origin.
This graph shows an isosceles triangle that is
not a right triangle.
Feedback
A
B
C
D
This is a right triangle with a vertex at the origin.
Correct!
This is a right triangle with a vertex at the origin.
This is a right triangle with a vertex at the origin.
PTS:
OBJ:
STA:
34. ANS:
1
DIF: Basic
REF: Page 267
4-7.1 Positioning a Figure in the Coordinate Plane
NAT: 12.3.4.d
GE17.0
TOP: 4-7 Introduction to Coordinate Proof
A
Step 1 Plot .
AB
The perpendicular bisector of
AB
is perpendicular to
AB
at its midpoint.
Step 2 Find the midpoint of .
Midpoint of
AB
Ê −2 + 5 2 + 4 ˆ˜
Á
˜
AB = Á
Á
Á 2 , 2 ˜˜ = (1.5, 3)
Ë
¯
Step 3 Find the slope of the perpendicular bisector.
2/7
Slope of
AB =
(4) − (2)
=
(5) − (−2)
Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is
-7/2.
Step 4 Use point-slope form to write the equation.
y − y 1 = m(x − x 1)
y−3 =
-7/2
(x − 1.5)
Feedback
A
B
C
Correct!
The perpendicular bisector has a slope that is perpendicular to the given segment.
To find the midpoint, add the x-coordinates and divide by 2, and add the y-coordinates
and divide by 2. The perpendicular bisector has a slope that is perpendicular to the given
segment.
To find the midpoint, add the x-coordinates and divide by 2, and add the y-coordinates
and divide by 2.
D
PTS:
OBJ:
NAT:
35. ANS:
BO =
2
3
BY
BO =
2
3
(26.4) = 17.6
1
DIF: Average
REF: Page 303
5-1.4 Writing Equations of Bisectors in the Coordinate Plane
12.5.2.c
STA: 1A8.0
TOP: 5-1 Perpendicular and Angle Bisectors
A
Centroid Theorem
Substitute 3.3 for and simplify.
BY
Feedback
A
B
C
Correct!
Segment BO is two-thirds the length of segment BY.
Segment BO is two-thirds the length of segment BY.
D
Segment BO is two-thirds the length of segment BY.
PTS:
OBJ:
STA:
36. ANS:
1
DIF: Average
REF: Page 315
5-3.1 Using the Centroid to Find Segment Lengths
NAT: 12.3.3.f
GE1.0
TOP: 5-3 Medians and Altitudes of Triangles
B
The length of a midsegment is half the length of the corresponding base.
XY =
1
2
AB =
1
2
(3) = 1.5
Feedback
A
B
C
D
Find the base corresponding to segment XY.
Correct!
Find the base corresponding to segment XY.
The length of the midsegment is half the length of the corresponding base.
PTS:
OBJ:
NAT:
37. ANS:
1
DIF: Average
REF: Page 323
5-4.2 Developing and Using the Triangle Midsegment Theorem
12.3.3.f
STA: GE1.0
TOP: 5-4 The Triangle Midsegment Theorem
B
From the diagram,
.m
AD ≅ BD
∠ADC
>m
is given, and
∠BDC
CD ≅ CD
by the Reflexive Property. By the Hinge Theorem,
3x + 32 > 7x + 16
16 > 4x
4>x
Since ,
which means .
So the range of values for x is
BC > 0
7x + 16 > 0
x > − 167
.
− 167 < x < 4
Feedback
A
B
C
D
By the Hinge Theorem, AC > BC. Set up an inequality and solve for x.
Correct!
By the Hinge Theorem, AC > BC. Set up an inequality and solve for x.
By the Hinge Theorem, AC > BC. Set up an inequality and solve for x.
PTS: 1
38. ANS: A
DIF:
Advanced
TOP: 5-6 Inequalities in Two Triangles
Step 1 Determine if the sides form a triangle.
By the Triangle Inequality Theorem, 6, 14, and 13 can be the sides of a triangle.
Step 2 Classify the triangle.
c2 ? a2 + b2
Compare c2 to a2 + b2.
Substitute the longest side length for c.
Multiply.
Add and compare.
142 ? 62 + 132
196 ? 36 + 169
196 < 205
Since
142 < 62 + 132
, the triangle is acute.
Feedback
A
Correct!
.
AC > BC
B
C
D
Compare the square of the longest side to the squares of the other two sides to determine
the type of triangle.
Compare the square of the longest side to the squares of the other two sides to determine
the type of triangle.
The largest side length is greater than the sum of the other two sides, so this can be the
side lengths of a triangle.
PTS:
OBJ:
TOP:
39. ANS:
1
DIF: Average
5-7.4 Classifying Triangles
5-7 The Pythagorean Theorem
B
REF: Page 351
NAT: 12.3.3.d
STA: GE6.0
If the shape were regular all the sides would be congruent and all the angles would be congruent. Since
neither the sides nor the angles are congruent, the shape is irregular. Because some of the diagonals of this
shape contain points in the exterior of the polygon, the shape is concave.
Feedback
A
B
C
D
In regular shapes, all sides and all angles are congruent.
Correct!
In regular shapes, all sides and all angles are congruent.
In convex shapes, no diagonals contain points in the exterior.
PTS:
OBJ:
TOP:
40. ANS:
1
DIF: Basic
REF: Page 383
6-1.2 Classifying Polygons
NAT: 12.3.3.f
6-1 Properties and Attributes of Polygons
C
STA: GE12.0
Step 1 Find the sum of the interior angle measures.
(n – 2)180°
Polygon Angle Sum Theorem
= (45 – 2)180°
A 45-gon has 45 sides, so substitute 45 for n.
= 7740
Simplify.
Step 2 Find the measure of one interior angle.
= 172
The interior angles are , so divide by 45.
≅
7740
45
Feedback
A
B
C
D
Subtract 2, not 1, from the number of sides.
According to the Polygon Angle Sum Theorem, the sum of the interior angle measures
is the product of 180 and the number of sides minus 2.
Correct!
Subtract, not add, 2 from the number of sides.
PTS:
OBJ:
NAT:
41. ANS:
1
DIF: Average
REF: Page 384
6-1.3 Finding Interior Angle Measures and Sums in Polygons
12.3.3.f
STA: GE12.0
TOP: 6-1 Properties and Attributes of Polygons
D
A decagon has 10 sides and 10 vertices.
sum of exterior angle measures = 360°
Polygon Exterior Angle Sum Theorem
measure of one exterior angle =
°
360 = 36
10
A regular decagon has 10 congruent exterior
angles, so divide the sum by 10.
The measure of each exterior angle of a regular decagon is 36°.
Feedback
A
B
C
D
Divide by the number of sides the polygon has.
Divide 360 by the number of sides the polygon has.
Divide 360 by the number of sides.
Correct!
PTS:
OBJ:
STA:
42. ANS:
1
DIF: Average
REF: Page 384
6-1.4 Finding Exterior Angle Measures in Polygons
NAT: 12.3.3.f
GE12.0
TOP: 6-1 Properties and Attributes of Polygons
A
One set of opposite sides are congruent and one set of opposite angles are congruent. This is insufficient
information to prove that the quadrilateral is a parallelogram.
Feedback
A
B
C
D
Correct!
Both sets of opposite angles must be congruent in order for the quadrilateral to be a
parallelogram.
Both sets of opposite sides must be congruent in order for the quadrilateral to be a
parallelogram.
If one set of opposite sides are congruent and parallel, then the quadrilateral is a
parallelogram.
PTS: 1
DIF: Average
REF: Page 400
OBJ: 6-3.2 Applying Conditions for Parallelograms
STA: GE7.0
TOP: 6-3 Conditions for Parallelograms
NAT: 12.3.3.f
NUMERIC RESPONSE
43. ANS: 150
PTS:
TOP:
KEY:
44. ANS:
1
DIF: Average
NAT: 12.2.1.f
1-3 Measuring and Constructing Angles
application | angle measures
74
PTS: 1
DIF:
TOP: 1-4 Pairs of Angles
45. ANS: 20
Average
NAT: 12.2.1.f
STA: 6MG2.2
KEY: supplementary angles | complementary angles
PTS: 1
DIF: Advanced
NAT: 12.2.1.f
TOP: 1-7 Transformations in the Coordinate Plane
KEY: transformations | application | rotations
46. ANS: 65
PTS: 1
47. ANS: 5.5
DIF:
Average
PTS: 1
DIF: Advanced
TOP: 4-1 Classifying Triangles
48. ANS: 21.6
NAT: 12.2.1.f
TOP: 2-5 Algebraic Proof
NAT: 12.3.2.e
STA: GE8.0
KEY: perimeter | isosceles
PTS: 1
DIF: Average
NAT: 12.2.1.f
TOP: 4-8 Isosceles and Equilateral Triangles
49. ANS: 11
STA: GE12.0
KEY: equilateral | equiangular
PTS: 1
DIF: Average
NAT: 12.3.3.f
TOP: 5-4 The Triangle Midsegment Theorem
50. ANS: 49.8
PTS: 1
DIF: Average
TOP: 6-2 Properties of Parallelograms
51. ANS: 0.5
NAT: 12.2.1.h
STA: GE7.0
PTS: 1
DIF: Advanced
NAT: 12.3.3.f
TOP: 6-4 Properties of Special Parallelograms
STA: GE7.0
MATCHING
52. ANS:
TOP:
53. ANS:
TOP:
54. ANS:
TOP:
55. ANS:
TOP:
56. ANS:
TOP:
57. ANS:
TOP:
G
PTS: 1
DIF:
1-3 Measuring and Constructing Angles
B
PTS: 1
DIF:
1-2 Measuring and Constructing Segments
A
PTS: 1
DIF:
1-2 Measuring and Constructing Segments
D
PTS: 1
DIF:
1-3 Measuring and Constructing Angles
F
PTS: 1
DIF:
1-3 Measuring and Constructing Angles
H
PTS: 1
DIF:
1-1 Understanding Points Lines and Planes
Basic
REF: Page 20
Basic
REF: Page 15
Basic
REF: Page 15
Basic
REF: Page 20
Basic
REF: Page 20
Basic
REF: Page 16
58. ANS:
TOP:
59. ANS:
TOP:
60. ANS:
TOP:
F
PTS: 1
2-2 Conditional Statements
C
PTS: 1
2-2 Conditional Statements
D
PTS: 1
2-2 Conditional Statements
DIF:
Basic
REF: Page 82
DIF:
Basic
REF: Page 83
DIF:
Basic
REF: Page 82
61. ANS:
TOP:
62. ANS:
TOP:
B
PTS: 1
2-2 Conditional Statements
G
PTS: 1
2-2 Conditional Statements
DIF:
Basic
REF: Page 83
DIF:
Basic
REF: Page 83