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Brooklyn Technical High School
Review #2 (Answers)
Geometry Term I
Name ____________________________
Instructor: Mr. Rodriguez Period ____________________________
1)
Answer: If two angles are vertical angles, then they are congruent. Choice 4.
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

Supplementary angles add up to 180o but are not necessarily congruent.
A linear pair of angles add up to a straight angle but are not necessarily congruent
Adjacent angles share a common vertex and a common side and have no interior
points in common but are not necessarily congruent
Ans.
4
2)
Answer - In choice 4 you have a disjunction which is true as long as one of the parts is true.
Since a triangle has 3 sides, that part is true so the entire disjunction is true.
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

In choice 1, you have a conjunction and both parts have to be true. 2nd part is not true
In choice 2, you have a biconditional and both parts have to be false or both parts have to be true for it to be true.
Since the 1st part is true and the second part is false, the biconditional is false
In choice 3, you have a conditional. A conditional is true unless the first part is true and the 2nd is false.
In choice 3 the first part is true and the 2nd part is false, so the conditional is incorrect.
Ans.
Know your truth tables for conjunctions, biconditionals, conditionals and conjunctions
4
3)
Mark the diagram knowing that the triangle congruency statement is listed in corresponding
parts order. After doing that you can see that the answer is choice 4.
A
ns
4
.
4)
To prove by SAS postulate, you will need another pair of sides that will include the
̅̅̅̅ ≅ 𝑂𝐿
̅̅̅̅
angle. You will need 𝐴𝐺
2
Ans
.
5)
12
All 3 medians are concurrent, in other words, they intersect at the same point. That point is called the centroid.
The centroid divides the medians in the ratio of 2:1 where the longer part is closer to the
vertex that is connecting the median. Median ̅̅̅̅
𝐹𝐶 is divided into two parts ̅̅̅̅
𝐺𝐶 and ̅̅̅̅
𝐹𝐺 that are in the ratio Ans
̅̅̅̅
̅̅̅̅
.
2 to 1 respectively. Since 𝐹𝐺 is 12 then 𝐺𝐶 is 24.
24
6)
The statement “The medians of a triangle are concurrent” is true because the medians of a triangle all intersect
at one point. The negation of the statement is “The medians of the triangle are not concurrent”. This
negation is false.
7)
The angle bisectors are concurrent, in other words, they intersect
at one point. That point is called the incenter and in this case the
̅̅̅̅ is
incenter is point G. Since point B is drawn to the incenter, 𝐵𝐺
Ans.
also an angle bisector. Since <B is bisected, then <DBG≅<EBG. An additional note is that the incenter
is equidistant from the sides of the triangle (meaning the line segment perpendicular from the incenter to the
side of the triangle
4
8)
If you have a conditional pq, the inverse is
formed by negating each part of the
conditional, namely ~p~q. In this case the
answer should be “If two triangles are
similar, their corresponding angles are
congruent. You should know what is the
converse, inverse, and contrapositive of
the conditional.
(3)If two triangles are similar, their corresponding
angles are congruent
9)
The medians of a triangle intersect at a single point, meaning they are
concurrent. The point of intersection is called the centroid. The centroid divides
the medians into a ratio of 2:1 where the longer segment is closer to the vertex.
In this case, the entire median is 9. They are asking for the length of the longer
segment. Two numbers in the ratio 2:1 that add up to 9 are 3 and 6, or you can
set up the equation 2x + x = 9 and solve for x. x=3, so the longer segment ̅̅̅̅
𝑇𝐷 is 6.
6
10)
Ans.
4
The circle that goes around the triangle in this way (circumscribed) where the
vertices of that triangle are on the circle is called the circumcircle. The center
of the circumcircle is called the circumcenter. The circumcenter is the
intersection of the perpendicular bisectors of the sides of the triangle. Know
that the circumcenter can lie inside the triangle or outside the triangle if the
triangle is obtuse.
Ans
.
11)
Ans
33.3
After marking the given, it is clear that the two triangles can be proven congruent via the ASA
postulate
12)
The medians of a triangle intersect at a single point, meaning they are concurrent. The point
of intersection is called the centroid. The centroid divides the medians into a ratio of 2:1
where the longer segment is closer to the vertex. In this case, median ̅̅̅̅
𝐶𝐹 is measured to be 6.
It is divided into two parts in the ratio 2:1. Therefore the equation to find ̅̅̅̅
𝑃𝐹 , labeled as x is
2x + x = 6
13)
Ans.
2
A median goes from a vertex to the
opposite side. Median ̅̅̅̅
𝐵𝐹 starts at
angle B and ends at point F.
Therefore point F is the midpoint of
̅̅̅̅
𝐴𝐶 . The midpoint of a segment
divides the segment into two
congruent segments. As a result
̅̅̅̅
𝐶𝐹 ≅ ̅̅̅̅
𝐴𝐹
Ans
.
4
16)
Ans.
3
The angle bisectors are
concurrent.
The point of
concurrency is called the
incenter which is the center of
the incircle which is the circle
inscribed in the triangle. The
incenter is always inside the
triangle.
The incenter is
equidistant from the sides of
the triangle.
17)
Statements
̅̅̅̅ ⊥ 𝐴𝐷
̅̅̅̅
1. 𝐵𝐶
2. ̅̅̅̅
𝐸𝐷 ⊥ ̅̅̅̅
𝐴𝐷
3. <ACB and < CDE are right angles
4. <ACB≅<CDE
5. <A≅<EFD
6. ̅̅̅̅
𝐴𝐹 ≅ ̅̅̅̅
𝐶𝐷
̅̅̅̅ ≅ 𝐹𝐶
̅̅̅̅
7. 𝐹𝐶
8. ̅̅̅̅
𝐴𝐹 + ̅̅̅̅
𝐹𝐶 ≅ ̅̅̅̅
𝐶𝐷 + ̅̅̅̅
𝐹𝐶
Or
̅̅̅̅̅
̅̅̅̅
𝐴𝐶 ≅
𝐷𝐹
9. ΔACB≅ΔFDE
(a.≅a.)
(a.≅a.)
Reasons
1. Given
2. Given
3. Definition of perpendicular lines (1,2)
4. If 2 angles are right angles, they are congruent (3)
5. Given
6. Given
7. Reflexive postulate of congruence
8. Addition Postulate (6,7)
(s.≅ 𝑠. )
9. a.s.a≅a.s.a. (4,5,8)
[1
]
18)
Of the choices listed, the one that is false
about the centroid of a triangle is that is can
lie on the outside or on the triangle. The
centroid of a triangle always lies inside
the triangle.
Ans.
2
19)
A conditional an its contrapositive
are logically equivalent. To form
the contrapositive switch
conditional and hypothesis and
negate each one
Ans.
2
Because of vertical angles,
<AED≅<BEC, giving us a.s.a≅a.s.a
20)
Ans.
1
21)
Since AB=DE, set 2x+10=4x20 and solve for x. Once
you find x, substitute into
3x-16, which is same length
as AC. Solving for x gives
you x=15. Therefore AC =
29
Ans.
29
22)
23) Given the two intersecting lines, find x
Since the angles with the expressions are vertical angles, they are equal to each
other.
119 − 𝑥 = 3𝑥 + 11
Solving for x gives us 27.
Ans.
27
24) Given: BD is the perpendicular bisector of AC
Prove: <A
<C
B
A
D
C
Statements
̅̅̅̅
̅̅̅̅ is the perpendicular bisector of 𝐴𝐶
1.𝐵𝐷
̅̅̅̅
2.D is the midpoint of 𝐴𝐶
̅̅̅̅ ≅ ̅̅̅̅
3.𝐴𝐷
𝐶𝐷
(s.≅ 𝑠. )
4.<ADB and <CDB are right angles
5)<ADB ≅ <CDB
(a.≅a.)
̅̅̅̅ ≅ ̅̅̅̅
6)𝐵𝐷
𝐵𝐷
7)ΔADB ≅ 𝛥CDB
8)<A≅<C
(s.≅ 𝑠. )
Reasons
1. Given
2 Definition of a line segment bisector (1)
3. Definition of a midpoint (2)
4. Definition of perpendicular lines (1)
5. If 2 angles are right angles, then they are
congruent. (4)
6. Reflexive property of congruent
7. s.a.s. ≅s.a.s (3,5,6)
8. Corresponding parts of congruent triangles
are congruent