If lines s and t are parallel, which pair of angles is supplementary and why?
(1 point)
Responses
∠B
and ∠G
because they are consecutive interior angles
∠B
and ∠G
because they are consecutive interior angles
∠G
and ∠D
because they are a linear pair
∠G
and ∠D
because they are a linear pair
∠A
and ∠B
because they are corresponding angles
∠A
and ∠B
because they are corresponding angles
∠A
and ∠D
because they are consecutive exterior angles
∠G and ∠D because they are a linear pair. When two parallel lines are intersected by a transversal, alternate interior angles are congruent. Since ∠G and ∠D are alternate interior angles, they are congruent and form a linear pair, which means they are supplementary.
Use the image to answer the question.
A horizontal line is intersected by two parallel lines lower s and lower t that slant from left to right forming eight angles.
If lines s and t are parallel, which pair of angles is supplementary and why?
(1 point)
Responses
∠B
and ∠G
because they are consecutive interior angles
∠B
and ∠G
because they are consecutive interior angles
∠G
and ∠D
because they are a linear pair
∠G
and ∠D
because they are a linear pair
∠A
and ∠B
because they are corresponding angles
∠A
and ∠B
because they are corresponding angles
∠A
and ∠D
because they are consecutive exterior angles
∠A
and ∠D
because they are consecutive exterior angles
Based on the given information and the image, the correct answer is:
∠G and ∠D because they are a linear pair.
Use the image to answer the question.
A vertical line is intersected by two parallel lines lower m and lower n forming eight angles.
Lines m and n are parallel. Angle E
measures (3x+18)
degrees and angle B
measures (14x−8)
degrees. What is the value of x?
(1 point)
Responses
132
48
Start Fraction 26 over 11 End Fraction
10
To find the value of x, we can set the measures of the two angles equal to each other and solve for x.
3x + 18 = 14x - 8
Combine like terms:
8 + 18 = 14x - 3x
26 = 11x
Divide both sides by 11:
x = 26/11
Therefore, the value of x is Start Fraction 26 over 11 End Fraction.
Line segment AC¯¯¯¯¯¯¯¯
is a perpendicular bisector of segment BD¯¯¯¯¯¯¯¯
, with the two segments meeting at point E
. What is true of segment BE¯¯¯¯¯¯¯¯
? (1 point)
Responses
It must be the same length as segment BC¯¯¯¯¯¯¯¯
.
It must be the same length as segment upper B upper C .
It must be the same length as segment DE¯¯¯¯¯¯¯¯
.
It must be the same length as segment upper D upper E .
It must be the same length as segment EA¯¯¯¯¯¯¯¯
.
It must be the same length as segment upper C upper D .
It must be the same length as segment AB¯¯¯¯¯¯¯¯
.
It must be the same length as segment AB¯¯¯¯¯¯¯¯.
Since segment AC¯¯¯¯¯¯¯¯ is a perpendicular bisector of segment BD¯¯¯¯¯¯¯¯, point E is the midpoint of segment BD¯¯¯¯¯¯¯¯. This means that segment BE¯¯¯¯¯¯¯¯ is equal in length to segment AB¯¯¯¯¯¯¯¯, as they both connect to the midpoint E.
Lenny wrote a paragraph proof of the Perpendicular Bisector Theorem. What mistake did Lenny make in his proof?
HK¯¯¯¯¯¯¯¯¯
is a perpendicular bisector of IJ¯¯¯¯¯¯
, and L is the midpoint of IJ¯¯¯¯¯¯
. M is a point on the perpendicular bisector, HK¯¯¯¯¯¯¯¯¯
. By the definition of a perpendicular bisector, I know that IM¯¯¯¯¯¯¯¯≅JM¯¯¯¯¯¯¯¯
. By the definition of a perpendicular bisector, I also know that ∠MLI
and ∠MLJ
are right angles. ∠MLI≅∠MLJ
because of the Right Angle Congruence Theorem. I can also say that ML¯¯¯¯¯¯¯¯¯≅ML¯¯¯¯¯¯¯¯¯
by the Reflexive Property of Congruence. With this information, I know that △MLI≅△MLJ
by the SAS Congruence Theorem. Since the triangles are congruent, the CPCTC Theorem allows me to know that IL¯¯¯¯¯¯≅JL¯¯¯¯¯¯¯
. Knowing that these segments are congruent proves the Perpendicular Bisector Theorem.
(1 point)
Responses
The definition of a perpendicular bisector tells you that∠LMI
and ∠LMJ
are right angles, not that ∠MLI
and ∠MLJ
are right angles.
The definition of a perpendicular bisector tells you that ∠LMI
and ∠LMJ
are right angles, not that ∠MLI
and ∠MLJ
are right angles.
The definition of a perpendicular bisector tells you that IL¯¯¯¯¯¯≅JL¯¯¯¯¯¯¯
, not that IM¯¯¯¯¯¯¯¯≅JM¯¯¯¯¯¯¯¯
.
The definition of a perpendicular bisector tells you that line segment cap i liters is congruent to line segment cap j liters, not that line segment cap i cap m is congruent to line segment cap j cap m.
Lenny did not make a mistake. His proof is correct.
Lenny did not make a mistake. His proof is correct.
The SAS Congruence Theorem tells you that △MLI≅△JLM
, not that △MLI≅△MLJ
.
The SAS Congruence Theorem tells you that △MLI≅△JLM
, not that △MLI≅△MLJ
.
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The mistake that Lenny made in his proof is:
The definition of a perpendicular bisector tells you that ∠LMI and ∠LMJ are right angles, not that ∠MLI and ∠MLJ are right angles.
Lenny mistakenly wrote ∠MLI and ∠MLJ as the right angles formed by the perpendicular bisector, while it should be ∠LMI and ∠LMJ.
¯¯¯¯¯
is a perpendicular bisector of KM¯¯¯¯¯¯¯¯¯¯
. Hallie locates point H
along line segment JL¯¯¯¯¯¯¯
. If the length of KH¯¯¯¯¯¯¯¯¯
is given by 6y+3
and the length of HM¯¯¯¯¯¯¯¯¯¯
is given by 2y+7
, what is the length in units of segment HM¯¯¯¯¯¯¯¯¯¯
? (1 point)
Responses
1
1
8
8
9
9
18
To find the length of segment HM¯¯¯¯¯¯¯¯¯¯, we can substitute the given expressions for KH¯¯¯¯¯¯¯¯¯ and HM¯¯¯¯¯¯¯¯¯¯ into the expression for segment KM¯¯¯¯¯¯¯¯¯¯:
KM¯¯¯¯¯¯¯¯¯¯ = KH¯¯¯¯¯¯¯¯¯ + HM¯¯¯¯¯¯¯¯¯¯
6y + 3 + 2y + 7 = HM¯¯¯¯¯¯¯¯¯¯
Combine like terms:
8y + 10 = HM¯¯¯¯¯¯¯¯¯¯
Therefore, the length of segment HM¯¯¯¯¯¯¯¯¯¯ is given by the expression 8y + 10.
JL¯¯¯¯¯¯¯
is a perpendicular bisector of KM¯¯¯¯¯¯¯¯¯¯
. Hallie locates point H
along line segment JL¯¯¯¯¯¯¯
. If the length of KH¯¯¯¯¯¯¯¯¯
is given by 6y+3
and the length of HM¯¯¯¯¯¯¯¯¯¯
is given by 2y+7
, what is the length in units of segment HM¯¯¯¯¯¯¯¯¯¯
? (1 point)
Responses
1
8
9
18
To find the length of segment HM¯¯¯¯¯¯¯¯¯¯, we need to find the value of y and then substitute it into the expression 2y + 7.
Since JL¯¯¯¯¯¯¯ is a perpendicular bisector of KM¯¯¯¯¯¯¯¯¯¯, we know that KH¯¯¯¯¯¯¯¯¯ and HM¯¯¯¯¯¯¯¯¯¯ are congruent. Therefore, we can set the expressions for their lengths equal to each other:
6y + 3 = 2y + 7
Subtract 2y from both sides:
4y + 3 = 7
Subtract 3 from both sides:
4y = 4
Divide both sides by 4:
y = 1
Now we can substitute y = 1 into the expression 2y + 7 to find the length of segment HM¯¯¯¯¯¯¯¯¯¯:
2(1) + 7 = 2 + 7 = 9
Therefore, the length of segment HM¯¯¯¯¯¯¯¯¯¯ is 9 units.
Use the image to answer the question.
An image is shown of perpendicular bisector VZ intersecting line segment WY at point X. Point U is a point on the perpendicular bisector.
Harriet has written a proof to prove that point U is equidistant from points W and Y. She knows that ZV¯¯¯¯¯¯¯¯
is the perpendicular bisector of WY¯¯¯¯¯¯¯¯¯
.
Statement Reason
1. ZV¯¯¯¯¯¯¯¯
is the perpendicular bisector of WY¯¯¯¯¯¯¯¯¯
. given
2. WX¯¯¯¯¯¯¯¯¯¯≅YX¯¯¯¯¯¯¯¯
definition of a perpendicular bisector
3. ∠WXU
and ∠YXU
are right angles. definition of a perpendicular bisector
4. ∠WXU≅∠YXU
Right Angle Congruence Theorem
5. XU¯¯¯¯¯¯¯¯≅XU¯¯¯¯¯¯¯¯
Reflexive Property of Congruence
6. UW¯¯¯¯¯¯¯¯¯≅UY¯¯¯¯¯¯¯¯
CPCTC Theorem
What is the missing step in her proof?
Option 1: There is no missing step in Harriet's proof.
Option 2: Harriet forgot the step where she proves that ∠XUW≅∠XUY
because of the CPCTC Theorem.
Option 3: Harriet forgot the step where she proves that △WXU≅△YXU
because of the SAS Congruence Theorem.
(1 point)
The missing statement in Harriet's proof is ?
The missing statement in Harriet's proof is:
Option 3: Harriet forgot the step where she proves that △WXU≅△YXU because of the SAS Congruence Theorem.
Use the image to answer the question.
The outline of a kite is in the shape of a diamond. The vertices of this diamond are labeled upper F, upper G, upper K, and upper J, respectively. Each of these points are connected via lines that intersect at a central, interior point: upper H.
In quadrilateral FGKJ
, FK¯¯¯¯¯¯¯¯
is a perpendicular bisector of GJ¯¯¯¯¯¯¯
. The length of segment GF¯¯¯¯¯¯¯¯
is 6+x
units, and the length of segment JF¯¯¯¯¯¯¯
is 3x−5
units. What is the value of x?
(1 point)
The value of x is
.
Remaining Attempts : 3
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To find the value of x, we can set the lengths of segment GF¯¯¯¯¯¯¯¯ and segment JF¯¯¯¯¯¯¯ equal to each other and solve for x.
GF¯¯¯¯¯¯¯¯ = JF¯¯¯¯¯¯¯
6 + x = 3x - 5
Combine like terms:
5 + x = 3x - 6
Add 6 to both sides:
11 + x = 3x
Subtract x from both sides:
11 = 2x
Divide both sides by 2 to solve for x:
x = 5.5
Therefore, the value of x is 5.5.