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 ( \triangle a b c ) and ( \triangle x y z ) are congruent.

1. ( a b c ) : side lengths of ( 6,8,7 x-12 )

( x y z ) : side lengths of ( 6,4 y, 3 x )

1. ( a b c ) : side lengths of ( 4 x+3,5,8 y-1 )

( x y z ) : side lengths of ( 6 x-1,5, y+6 )

Explanation:

1.

Step1: Identify the sides

In the first figure, we have \(AB = DC\), \(AD = BC\), and \(AC\) is common.

Step2: Apply SSS congruence

Since all three sides of \(\triangle ABC\) and \(\triangle ADC\) are equal (\(AB = DC\), \(AD = BC\), \(AC=AC\)), by the Side - Side - Side (SSS) congruence theorem, \(\triangle ABC\cong\triangle ADC\).

2.

Step1: Identify the angles and sides

We have \(FJ = FG\) (given by the arrow marks), \(\angle JFH=\angle GFH\) (common angle), and \(FH = FH\) (common side).

Step2: Apply SAS congruence

By the Side - Angle - Side (SAS) congruence theorem, \(\triangle FJH\cong\triangle FGH\) because two sides (\(FJ = FG\), \(FH = FH\)) and the included angle (\(\angle JFH=\angle GFH\)) are equal.

3.

Step1: Identify the sides and parallel lines

Since \(QT\parallel RS\), \(\angle TQS=\angle RSQ\) (alternate interior angles). Also, \(QT = RS\) (given by the tick marks) and \(QS = QS\) (common side).

Step2: Apply SAS congruence

By the Side - Angle - Side (SAS) congruence theorem, \(\triangle TQS\cong\triangle RSQ\) as two sides (\(QT = RS\), \(QS = QS\)) and the included angle (\(\angle TQS=\angle RSQ\)) are equal.

4.

Step1: Identify the angles and sides

We have \(\angle B=\angle C\) (given), \(BE = CE\) (given by the tick marks), and \(\angle AEB=\angle DEC\) (vertically opposite angles).

Step2: Apply AAS congruence

By the Angle - Angle - Side (AAS) congruence theorem, \(\triangle ABE\cong\triangle DCE\) since two angles (\(\angle B=\angle C\), \(\angle AEB=\angle DEC\)) and a non - included side (\(BE = CE\)) are equal.

5.

Step1: Set up equations for congruent sides

Since \(\triangle ABC\cong\triangle XYZ\), we can set up the following equations:
Case 1: If \(7x - 12=4y\) and \(8 = 3x\)
From \(8 = 3x\), we solve for \(x\): \(x=\frac{8}{3}\)
Substitute \(x = \frac{8}{3}\) into \(7x-12 = 4y\): \(7\times\frac{8}{3}-12 = 4y\), \(\frac{56}{3}-12 = 4y\), \(\frac{56 - 36}{3}=4y\), \(\frac{20}{3}=4y\), \(y=\frac{5}{3}\)
Case 2: If \(7x - 12=3x\) and \(8 = 4y\)
From \(7x-12 = 3x\), \(7x-3x=12\), \(4x = 12\), \(x = 3\)
From \(8 = 4y\), \(y = 2\)
We check for the first case:
For \(\triangle ABC\) sides: \(6\), \(8\), \(7\times\frac{8}{3}-12=\frac{56 - 36}{3}=\frac{20}{3}\)
For \(\triangle XYZ\) sides: \(6\), \(4\times\frac{5}{3}=\frac{20}{3}\), \(3\times\frac{8}{3}=8\)
For the second case:
For \(\triangle ABC\) sides: \(6\), \(8\), \(7\times3-12=9\)
For \(\triangle XYZ\) sides: \(6\), \(4\times2 = 8\), \(3\times3=9\)
So \(x = 3\) and \(y = 2\)

6.

Step1: Set up equations for congruent sides

Since \(\triangle ABC\cong\triangle XYZ\)
We have \(4x + 3=6x-1\) (equating two non - equal sides) and \(8y-1=y + 6\)
For \(4x + 3=6x-1\):
\(6x-4x=3 + 1\), \(2x=4\), \(x = 2\)
For \(8y-1=y + 6\):
\(8y-y=6 + 1\), \(7y=7\), \(y = 1\)
Check:
For \(\triangle ABC\) sides: \(4\times2+3=11\), \(5\), \(8\times1-1 = 7\)
For \(\triangle XYZ\) sides: \(6\times2-1=11\), \(5\), \(1 + 6=7\)

Answer:

1. SSS; \(AB = DC\), \(AD = BC\), \(AC = AC\)
2. SAS; \(FJ = FG\), \(\angle JFH=\angle GFH\), \(FH = FH\)
3. SAS; \(QT = RS\), \(\angle TQS=\angle RSQ\), \(QS = QS\)
4. AAS; \(\angle B=\angle C\), \(\angle AEB=\angle DEC\), \(BE = CE\)
5. \(x = 3\), \(y = 2\)
6. \(x = 2\), \(y = 1\)