Geometry (G)
1st Quarter Exam - Review Answers
For extra practice refer to your textbook/CD/Workbook. The chapter and sections are listed below.
Pg1 – Definitions
You may find these in your glossary or on QuizLet. There will not be any vocabulary or matching on the exam. However, you will need to know what these words mean in order to be able to work some of the problems.
Pg1 – Ch1
#1) 5x + 5 +x – 5 = 180 solve for x
x = 30
7 = 155˚
8 = 25˚
#2) 5x + 4x + 6 + 10x + 12x -12 = 180 solve for x
x = 6
5 = 30˚6 = 30˚
7 = 60˚8 = 60˚
#3) 11x = 10x + 10 solve for x
x = 10
11 = 110˚12 = 110˚
13 = 70˚
Pg2 – Ch2.1
#1) Valid.
#2) Invalid. Just because 2 angles are congruent does not make them complementary to a third angle.
Counterexample: If A= 100˚ and C= 100˚ then they are congruent but they cannot be complementary to B because the combined measure would exceed 90˚.
#3) Invalid. E and F are congruent because they are complementary to the same angle; however, that does not require them to be vertical.
#10) Conditional: If 89 is divisible by 2, then 89 is an even number.TRUE
Converse:If 89 is an even number, then it is divisible by 2.TRUE
Inverse:If 89 is not divisible by 2, then 89 is not an even number.TRUE
Contrapositive:If 89 is not an even number, then it is not divisible by 2.TRUE
Biconditional:89 is divisible by 2 iff it is an even number.
Based on this series of statements is zero an even number?
Pg2 – Ch2.4
For the proofs you may want to read through the Postulates and Theorems beginning on page 827.
Pg2 – Ch2.4
a. 8x – 5 = 2x + 1 / a. Given
b. 8x – 5 -2x = 2x + 1 – 2x / b. Subtraction Property of Equality
c. 6x – 5 = 1 / c. Simplify (I changed this because this is more accurate)
d. 6x – 5 + 5 = 1 + 5 / d. Addition Property
e. 6x = 6 / e. Simplify
f. 6x = 6
6 6 / f. Division Property
g. x = 1 / g. Simplify
For the proofs the blue text indicates the answers that you must supply.
#6)
#7)
Statements / Reasonsa. and / a. Given
b. / b. Transitive
c. / c. Definition of Segments
Pg3 – Ch2.5
#5)
Statements / Reasonsa. / a. Given
b. SU=LR and TU=LN / b. Definition of segments
c. SU=ST+TU
LR=LN+NR / c. Segment Addition Postulate
d. ST + TU = LN + NR / d. Substitution
e. ST + LN = LN + NR / e. Substitution of TU=LN
f. ST+LN –LN=LN+NR - LN / f. Subtraction Property of Equality
g. ST=NR / g. Simplify(I changed this because this is more accurate)
h. / h. Definition of segments
Pg3 – Ch2.6
#1)
Statements / Reasonsa. / a. Given
b. is a right angle / b. Definition of
c. =90˚ / c. Definition of right angle
d. = / d. Angle addition postulate
e. 90= / e. Substitution
f. are complementary / f. Definition of complementary angles
g. are complementary / g. Given
h. / h. complementary to the sameare
Statements / Reasons
a. and form a linear pair.
/ a. Given
b. and are supplementary / b. Supplementary Angle Theorem
c. is supplementary to / c. Definition of Supplementary
d. / d. supplementary to the sameare
Pg3 – Ch2.6
#2)
Pg4 – Ch2.6
#9)
Statements / Reasonsa. / a. Given
b. / b. Definition of congruence
c.
/ c. Angle addition postulate
d. / d. Substitution
e. / e. Subtraction Property of Equality
f. / f. Definition of congruence
Pg4 – Ch3.3
Exercises #1) 5x – 5 = 6x – 20 alternate exterior angles are = Solve for x
x = 15
Exercises #2) 4x + 20 = 6x alternate interior angles are = Solve for x
x = 10
Exercises #3) 3x + 15 = 90 alternate interior angles are = Solve for x
x = 25
#1) 5x – 5 = 4x + 10corresponding angles are = Solve for x
x = 15
Now set up another equation with the y
5x – 5 + 6y – 4 = 180linear pair angles = 180
5(15) -5 + 6y – 4 = 180substitute x=15 and solve for y
y = 19
#2) 3y + 18 = 90same side interior angles =180 since one in 90 I know the other =90 also
y = 24
Now set up another equation with the x
15x + 30 + 10x = 180same side interior angles =180 Solve for x
x = 6
#3) 5y + 5 +13y – 5 = 180 same side interior angles =180 Solve for y
y = 10
Now set up another equation with the x
11x + 4 +5x = 180same side interior angles =180 Solve for x
x = 11
Pg4 – Ch3.3
#4) 3x = 5x – 20alternate interior angles are = Solve for x
x = 10
Now set up another equation with the y
3x + 2y + 4y = 180same side interior = 180You could also add the angles of the ∆ = 180
3(10) + 2y + 4y = 180substitute x=10 and solve for y
y = 25
#5) 2y + 106 = 180linear pair = 180Solve for y
y = 37
x +106 = 180same side interior = 180Solve for x
x = 74
x +4z + 6 = 180vertical angles are = THEN same side interior = 180
(74) +4z + 6 = 180 Substitute x=74Solve for z
z = 25
#6) 2x + 90 +x = 180linear pairs = 180Solve for x
x = 30
x = 2yalternate interior anglesSolve for x
(30) = 2ySubstitute x=30Solve for y
y = 15
z+2y = 180linear pairs = 180
z +2(15) = 180Substitute y=15Solve for z
z = 150
Pg5 – Ch3.4
Example 1r ll s if corresponding angles are then lines are parallel
Example 23x + 10 = 6x – 20alternate interior angles are = Solve for x
x= 10
= 40˚
#1) First Proof
Statements / Reasons1. / 1. Given
2. / 2. vertical are
3. / 3. Transitive
4. r ll s / 4. if corresponding are then lines are ll
5. / 5. Given
6. l ll m / 6. if corresponding are then lines are ll
Pg5 – Ch3.4
#a) Second Proof
Statements / Reasons1.
/ 1. Given
2. / 2. Transitive
3. ll / 3. if alternate interior are then lines are ll
Pg6 – Ch7.3
For the reflections I will give the coordinates. On the exam you will need to draw the pre-image in one color and the image in a different color so that it is easy to grade.
#4) D’ (-2,1) E’ (-1,-3) F’ (3,1)
#5) A’ (-1,4) B’ (-3,2) C’ (-2,-2) D’ (3,1)
Rotations –here are the coordinates. You will need to graph it on the exam.
#1) P’ (2,-1) Q’ (-3,1)
#2) P’ (-3,4) Q’ (-1,0) R’ (2,-1) To rotate around Point T you must identify the relationship each point has in location to T. See how far left/right and up/down the preimage is from T then go the same at a 90˚to T.
Pg6 – Ch7.4
#3)#4)
Translations- here are the coordinates. You will need to graph it on the exam.
#1) P’ (-3,4) Q’ (0,3)
#2) P’ (0,-6) Q’ (1,0) R’ (4,-1)
#3) Q’ (6,2) R’ (2,0) S’ (3,3) U’ (5,-1)