name sss, sas, asa, aas determine which triangle congruence theorem can be used to show that the triangles are congruent. explain your reasoning. 1. 2. 3. 4. find the values of ( x ) and ( y ) for which ( riangle abc ) and ( riangle xyz ) are congruent. 5. ( abc ): side lengths of ( 6,8,7x - 12 ) ( xyz ): side lengths of ( 6,4y,3x ) 6. ( abc ): side lengths of ( 4x + 3,5,8y - 1 ) ( xyz ): side lengths of ( 6x - 1,5,y + 6 )

Question

name sss, sas, asa, aas determine which triangle congruence theorem can be used to show that the triangles are congruent. explain your reasoning. 1. 2. 3. 4. find the values of ( x ) and ( y ) for which ( 	riangle abc ) and ( 	riangle xyz ) are congruent. 5. ( abc ): side lengths of ( 6,8,7x - 12 ) ( xyz ): side lengths of ( 6,4y,3x ) 6. ( abc ): side lengths of ( 4x + 3,5,8y - 1 ) ( xyz ): side lengths of ( 6x - 1,5,y + 6 )

Accepted Answer

### 1.
# Explanation:
## Step1: Identify the sides
In $\triangle ABC$ and $\triangle ADC$, $AB = CD$, $AD = BC$, and $AC$ is common.
## Step2: Apply SSS congruence
Since three sides of $\triangle ABC$ are equal to three sides of $\triangle ADC$, by SSS (Side - Side - Side) congruence theorem, $\triangle ABC\cong\triangle ADC$.

# Answer:
SSS (Side - Side - Side) congruence theorem. Because $AB = CD$, $AD = BC$, and $AC$ is common.

### 2.
# Explanation:
## Step1: Identify the sides and angles
In $\triangle FJH$ and $\triangle FGH$, $FJ = FG$ (given by the mark), $FH$ is common, and $\angle JFH=\angle GFH$ (given by the mark).
## Step2: Apply SAS congruence
Since two sides and the included angle of $\triangle FJH$ are equal to two sides and the included angle of $\triangle FGH$, by SAS (Side - Angle - Side) congruence theorem, $\triangle FJH\cong\triangle FGH$.

# Answer:
SAS (Side - Angle - Side) congruence theorem. Because $FJ = FG$, $FH$ is common, and $\angle JFH=\angle GFH$.

### 3.
# Explanation:
## Step1: Identify the sides and angles
In $\triangle QTS$ and $\triangle RQS$, $QT = RQ$ (given by the mark), $TS = RS$ (given by the mark), and $QS$ is common.
## Step2: Apply SSS congruence
Since three sides of $\triangle QTS$ are equal to three sides of $\triangle RQS$, by SSS (Side - Side - Side) congruence theorem, $\triangle QTS\cong\triangle RQS$.

# Answer:
SSS (Side - Side - Side) congruence theorem. Because $QT = RQ$, $TS = RS$, and $QS$ is common.

### 4.
# Explanation:
## Step1: Identify the sides and angles
In $\triangle ABE$ and $\triangle CDE$, $AB = CD$ (given by the mark), $\angle B=\angle C$ (given by the mark), and $\angle AEB=\angle DEC$ (vertically opposite angles).
## Step2: Apply AAS congruence
Since two angles and a non - included side of $\triangle ABE$ are equal to two angles and a non - included side of $\triangle CDE$, by AAS (Angle - Angle - Side) congruence theorem, $\triangle ABE\cong\triangle CDE$.

# Answer:
AAS (Angle - Angle - Side) congruence theorem. Because $AB = CD$, $\angle B=\angle C$, and $\angle AEB=\angle DEC$.

### 5.
# Explanation:
## Step1: Set up equations
Since $\triangle ABC\cong\triangle XYZ$, we have two cases for the side - length equalities.
Case 1: $7x - 12=3x$ and $8 = 4y$
## Step2: Solve the equations
For $7x - 12=3x$, subtract $3x$ from both sides: $7x-3x - 12=0$, $4x=12$, $x = 3$.
For $8 = 4y$, divide both sides by $4$: $y = 2$.

# Answer:
$x = 3$, $y = 2$

### 6.
# Explanation:
## Step1: Set up equations
Since $\triangle ABC\cong\triangle XYZ$, we have $4x + 3=6x - 1$ and $8y-1=y + 6$
## Step2: Solve the equations
For $4x + 3=6x - 1$, subtract $4x$ from both sides: $3=2x - 1$, add $1$ to both sides: $4 = 2x$, $x = 2$.
For $8y-1=y + 6$, subtract $y$ from both sides: $7y-1=6$, add $1$ to both sides: $7y=7$, $y = 1$.

# Answer:
$x = 2$, $y = 1$

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QUESTION IMAGE

Question

Explanation:

1.

Step1: Identify the sides

Step2: Apply SSS congruence

Step1: Identify the sides and angles

Step2: Apply SAS congruence

Step1: Identify the sides and angles

Step2: Apply SSS congruence

Answer:

2.