24. if two lines are perpendicular to the same line then the lines are
a. parallel b. perpendicular c. skew d. oblique
25. if parallel lines are intersected by a transversal then which angles are always supplementary?
a. alternate interior angles b. same side exterior angles
c. corresponding angles d. vertical angles
26. find the value of x in the diagram at right. lines are parallel
a. 74 b. 11.5
c. 46.5 d. 90
27. which is not a shortcut to congruent triangles?
a. aas b. sas c. cpctc d. sss
28. which shortcut can be used to show the triangles are congruent?
a. aas b. sas
c. asa d. sss
29. which shortcut can be used to show the triangles are congruent?
a. aas b. hl
c. asa d. sas
30. what else is needed to show δtwo ≅ δhip by hl?
a. $\overline{wo} \cong \overline{hi}$ b. $\overline{to} \cong \overline{hp}$
c. $\angle o \cong \angle i$ d. $\angle t \cong \angle p$

Question

24. if two lines are perpendicular to the same line then the lines are
a. parallel  b. perpendicular  c. skew  d. oblique
25. if parallel lines are intersected by a transversal then which angles are always supplementary?
a. alternate interior angles  b. same side exterior angles
c. corresponding angles  d. vertical angles
26. find the value of x in the diagram at right. lines are parallel
a. 74  b. 11.5
c. 46.5  d. 90
27. which is not a shortcut to congruent triangles?
a. aas  b. sas  c. cpctc  d. sss
28. which shortcut can be used to show the triangles are congruent?
a. aas  b. sas
c. asa  d. sss
29. which shortcut can be used to show the triangles are congruent?
a. aas  b. hl
c. asa  d. sas
30. what else is needed to show δtwo ≅ δhip by hl?
a. $\overline{wo} \cong \overline{hi}$  b. $\overline{to} \cong \overline{hp}$
c. $\angle o \cong \angle i$  d. $\angle t \cong \angle p$

Accepted Answer

### Question 24
# Explanation:
## Step1: Recall line perpendicularity rules
In a plane, if two lines are perpendicular to the same line, they are parallel. (In 3D, they could be skew, but the question likely refers to a plane context as it's a basic geometry question.)
## Step2: Evaluate options
- A. parallel: Matches the plane - case rule.
- B. perpendicular: Two lines perpendicular to the same line aren't necessarily perpendicular to each other.
- C. skew: Applies to 3D, less likely here.
- D. oblique: Not related to this rule.
# Answer: A. parallel

### Question 25
# Explanation:
## Step1: Recall angle relationships with transversal
- Alternate interior angles: Congruent.
- Same - side exterior angles: Supplementary (sum to $180^{\circ}$).
- Corresponding angles: Congruent.
- Vertical angles: Congruent.
## Step2: Match with the property of being supplementary
Same - side exterior angles are supplementary when parallel lines are cut by a transversal.
# Answer: B. same side exterior angles

### Question 26
# Explanation:
## Step1: Identify angle relationship
The angles $2x + 32$ and $125$ are same - side interior angles (or we can use the fact that if two parallel lines are cut by a transversal, same - side interior angles are supplementary, or we can also consider the exterior - interior relationship. Wait, actually, the angle $2x + 32$ and $125$ should be supplementary? Wait, no, let's look at the diagram. If the lines are parallel, and the angle given is $125$, and the other angle is $2x+32$, maybe they are same - side interior angles. So $2x + 32+125 = 180$? Wait, no, maybe the angle $2x + 32$ and $125$ are related as alternate exterior or something. Wait, no, let's solve $2x+32 = 180 - 125$? Wait, no, maybe the angle $2x + 32$ and $125$ are supplementary? Wait, $180-125 = 55$? No, that's not right. Wait, maybe the angle $2x + 32$ and $125$ are equal? No, that would not make sense. Wait, maybe I made a mistake. Wait, the correct approach: If two parallel lines are cut by a transversal, the sum of same - side interior angles is $180^{\circ}$. So $2x + 32+125=180$? No, $2x+32 = 180 - 125$? $180 - 125=55$, $2x+32 = 55$? No, that gives $x = 11.5$, which is option B. Wait, let's check: $2\times11.5+32=23 + 32 = 55$, and $180 - 125 = 55$. So the two angles are supplementary (same - side interior angles), so $2x + 32+125 = 180$ → $2x=180 - 125 - 32=23$ → $x = 11.5$
## Step1: Set up the equation
Since the lines are parallel, the angles $2x + 32$ and $125$ are same - side interior angles, so they are supplementary. So $2x+32 + 125=180$
## Step2: Solve for x
$2x+157 = 180$
$2x=180 - 157$
$2x = 23$
$x=\frac{23}{2}=11.5$
# Answer: B. 11.5

### Question 27
# Explanation:
## Step1: Recall triangle congruence shortcuts
- AAS: Angle - Angle - Side, a congruence shortcut.
- SAS: Side - Angle - Side, a congruence shortcut.
- CPCTC: Corresponding Parts of Congruent Triangles are Congruent. It is used to prove parts of triangles congruent after proving the triangles congruent, not a shortcut to prove triangles congruent.
- SSS: Side - Side - Side, a congruence shortcut.
## Step2: Identify the non - shortcut
CPCTC is not a shortcut to prove triangles congruent.
# Answer: C. CPCTC

### Question 28
# Explanation:
## Step1: Analyze the triangle markings
The two triangles have two sides marked as equal and the included angle marked as equal (from the diagram, the markings show two sides and the included angle).
## Step2: Match with congruence shortcuts
SAS (Side - Angle - Side) is used when two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle.
- AAS: Angle - Angle - Side, not matching the markings.
- B. SAS: Matches the two - sides - included - angle markings.
- C. ASA: Angle - Side - Angle, not matching.
- D. SSS: Side - Side - Side, not matching.
# Answer: B. SAS

### Question 29
# Explanation:
## Step1: Analyze the triangle markings
The triangle has a common side, two sides marked as equal, and the included angle (from the angle markings) or wait, the diagram shows a triangle with a median (or altitude) and two sides marked as equal, and the angle between the equal sides and the common side. Wait, the markings show two sides equal, the included angle (from the angle arcs) and the common side. Wait, no, the HL (Hypotenuse - Leg) is for right triangles. But in this case, the triangle has two sides marked as equal, and the angle between them? Wait, no, the diagram shows a triangle with a line from the vertex to the base, splitting it, and two sides marked as equal, and the angle at the vertex marked as equal. Wait, actually, the triangles have two sides equal and the included angle? No, the correct shortcut here: The triangles have two sides equal (marked) and the included angle? Wait, no, the option D is SAS? Wait, no, the diagram shows a triangle with a common side, two sides equal, and the angle between the equal sides and the common side. Wait, the correct shortcut is SAS? Wait, no, the options are A. AAS, B. HL, C. ASA, D. SAS. Wait, the triangle has two sides marked as equal, and the included angle (from the angle markings) and the common side. So SAS (Side - Angle - Side) is applicable. Wait, but let's re - examine. The triangle has two sides equal (marked with the same number of ticks), and the included angle (marked with the same arc) and the common side. So SAS is the correct shortcut.
## Step1: Identify the markings
The triangles have two sides marked as equal, and the included angle (from the angle arcs) equal, and a common side.
## Step2: Match with congruence shortcuts
SAS (Side - Angle - Side) is used when two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle.
# Answer: D. SAS

### Question 30
# Explanation:
## Step1: Recall HL (Hypotenuse - Leg) congruence
HL congruence for right triangles requires that the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and one leg of another right triangle.
## Step2: Analyze the given triangles
We have two right triangles ($\triangle TWO$ and $\triangle HIP$). We need to find what's needed for HL. Looking at the options:
- A. $\overline{WO}\cong\overline{HI}$: $WO$ and $HI$ are legs? Wait, $W$ and $I$ are right angles. So the hypotenuses would be $TO$ and $HP$? No, wait, the legs: $WO$ and $HI$ are legs, and the hypotenuses would be $TW$ and $HP$? No, wait, the correct approach: For HL, we need the hypotenuse and one leg. So if we have $\overline{TO}\cong\overline{HP}$ (hypotenuses) and we need a leg. Wait, no, the option B is $\overline{TO}\cong\overline{HP}$? Wait, no, the option A is $\overline{WO}\cong\overline{HI}$ (legs), and we need the hypotenuse. Wait, the triangles are right - angled at $W$ and $I$. So the hypotenuses are $TO$ (for $\triangle TWO$) and $HP$ (for $\triangle HIP$)? No, $TW$ and $HI$ are legs? Wait, maybe I got the labels wrong. Let's assume that $\triangle TWO$ has right angle at $W$, so legs are $TW$ and $WO$, hypotenuse $TO$. $\triangle HIP$ has right angle at $I$, legs are $HI$ and $IP$, hypotenuse $HP$. For HL, we need hypotenuse ($TO\cong HP$) and one leg ($WO\cong HI$ or $TW\cong IP$). So the option A is $\overline{WO}\cong\overline{HI}$ (a leg), and we need the hypotenuse? Wait, no, the question is "What else is needed to show $\triangle TWO\cong\triangle HIP$ by HL?". So we already have some information. From the diagram, maybe $TW\cong IP$ (marked with ticks) and we need the hypotenuse or a leg. Wait, the option A is $\overline{WO}\cong\overline{HI}$ (a leg), option B is $\overline{TO}\cong\overline{HP}$ (hypotenuse). For HL, we need hypotenuse and one leg. If we assume that $TW\cong IP$ (a leg) is already given (from the ticks on the legs), then we need the hypotenuse $\overline{TO}\cong\overline{HP}$? No, wait, the option A is $\overline{WO}\cong\overline{HI}$. Let's re - check:
HL requires hypotenuse and one leg. So if we have right triangles, and we need to show HL, we need one leg and the hypotenuse. So if $\overline{WO}\cong\overline{HI}$ (a leg) and we have the hypotenuse? Wait, maybe the correct answer is A. $\overline{WO}\cong\overline{HI}$
## Step1: Recall HL congruence conditions
HL (Hypotenuse - Leg) congruence for right triangles: If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent.
## Step2: Analyze the triangles and options
- $\triangle TWO$ and $\triangle HIP$ are right - angled (at $W$ and $I$ respectively).
- We need to identify which pair of segments (leg or hypotenuse) is needed. $\overline{WO}$ and $\overline{HI}$ are legs of the right triangles. If $\overline{WO}\cong\overline{HI}$ (a leg) and we have the hypotenuse (or vice - versa) congruent, then HL can be applied.
# Answer: A. $\overline{WO}\cong\overline{HI}$

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QUESTION IMAGE

Question

Explanation:

Question 24

Step1: Recall line perpendicularity rules

Step2: Evaluate options

Step1: Recall angle relationships with transversal

Step2: Match with the property of being supplementary

Step1: Identify angle relationship

Step1: Set up the equation

Step2: Solve for x

Answer:

Question 25