Question

using the sss congruence theorem
which of the following pairs of values for ( x ) and ( y ) would justify the claim that the two triangles are congruent?
( x = 9, y = 7 )
( x = 3, y = 11 )
( x = 5, y = 5 )
( x = 7, y = 9 )

Explanation:

Step1: Apply SSS Congruence Theorem

For two triangles to be congruent by SSS (Side - Side - Side), their corresponding sides must be equal.
So, \(2x + 3=7\) and \(y - 4=9\).

Step2: Solve for \(x\)

Solve \(2x+3 = 7\).
Subtract 3 from both sides: \(2x=7 - 3\), so \(2x = 4\).
Divide both sides by 2: \(x=\frac{4}{2}=2\) (This is wrong approach, actually we should match the sides as per the figure's side - side correspondence. The correct way is: If \(2x + 3\) corresponds to 7 and \(y-4\) corresponds to 9. Wait, no, actually looking at the options, if we assume the correspondence of sides:
Let's check each option:

• For \(x = 9,y = 7\): \(2x+3=2\times9 + 3=21

eq7\), \(y - 4=7 - 4 = 3
eq9\)

• For \(x = 3,y = 11\): \(2x+3=2\times3+3 = 9\), \(y - 4=11 - 4=7\). But the sides 9 and 7 are in the wrong correspondence (if we assume the upper triangle has sides 7,9 and common side. The lower triangle should have sides \(2x + 3\), \(y - 4\) and common side. By SSS, if \(2x+3 = 7\) and \(y - 4=9\) (wrong). Wait, no, actually, if we consider the two triangles share a common side. Let's assume the sides of the first triangle are \(a = 7\), \(b = 9\) and common side \(c\). The second triangle has sides \(a'=2x + 3\), \(b'=y - 4\) and \(c\) (common). By SSS \(a=a'\) and \(b = b'\). So \(2x+3=9\) (solving \(2x=6\), \(x = 3\)) and \(y-4=7\) (solving \(y=11\)) is wrong. Wait, no, actually looking at the options:

If we use the fact that for SSS, the three sides of one triangle equal three sides of another.
Let’s check \(x = 7,y = 9\):
\(2x+3=2\times7+3=17
eq7\), \(y - 4=9 - 4 = 5
eq9\)
For \(x = 3,y = 11\): \(2x + 3=9\), \(y-4 = 7\) (wrong correspondence)
For \(x=5,y = 5\): \(2x+3=13
eq7\), \(y - 4=1
eq9\)
For \(x = 9,y = 7\): \(2x+3=21
eq7\), \(y - 4=3
eq9\)
Wait, no, actually, we made a mistake. The correct approach:
Since the two triangles are congruent by SSS, the sides must match.
If we assume the sides of the first triangle are \(7\), \(9\) and the common side. The second triangle has \(2x + 3\), \(y - 4\) and common side.
If \(2x+3 = 7\) (solving \(x = 2\)) and \(y - 4=9\) (solving \(y = 13\)) (not an option). But looking at the options, if we assume the correspondence is \(2x+3 = 9\) (so \(x=(9 - 3)/2=3\)) and \(y-4 = 7\) (so \(y=11\)) is wrong. Wait, no, actually, looking at the figure (assuming the two triangles are divided by a common side). The upper triangle has sides \(7\), \(9\) and common side. The lower triangle has \(2x + 3\), \(y - 4\) and common side. By SSS, \(2x+3 = 7\) and \(y - 4=9\) (wrong as per options). But wait, no, actually, if we check the options:
Let’s check \(x = 3,y = 11\):
\(2x+3=2\times3+3=9\), \(y - 4=11 - 4 = 7\). But the sides 9 and 7 are in the wrong order. Wait, no, SSS just requires that all three sides are equal. The common side is equal. If one triangle has sides \(7\), \(9\), \(c\) and the other has \(9\), \(7\), \(c\). By SSS, they are congruent.

Answer:

\(x = 3,y = 11\)