Question

what values of s and t make \\( \triangle xyz \cong \triangle edc \\)?

Explanation:

Step1: Set up equation for corresponding sides

Since \(\triangle XYZ\cong\triangle EDC\), then \(YZ = EC\) and \(XY=ED\).
For \(YZ = EC\): \(10s + 6=3s+34\).
Subtract \(3s\) from both sides: \(10s-3s + 6=3s-3s+34\), \(7s+6 = 34\).
Subtract \(6\) from both sides: \(7s+6 - 6=34 - 6\), \(7s=28\).
Divide both sides by \(7\): \(s=\frac{28}{7}=4\).

Step2: Find the value of \(t\)

Substitute \(s = 4\) into the equation for \(XY = ED\).
\(XY = 2t\), \(ED=t + 9\).
Since \(XY=ED\), then \(2t=t + 9\).
Subtract \(t\) from both sides: \(2t-t=t - t+9\), \(t=9\).

Answer:

\(s = 4\), \(t = 9\)