Question

what values of w and x make \\( \triangle cde \cong \triangle pqr \\)?

Explanation:

Step1: Set equal the corresponding sides

Since \(\triangle CDE\cong\triangle PQR\), then \(CE = PR\). So, \(12w - 4=11w\).
Solve for \(w\):
\(12w-11w=4\)
\(w = 4\)

Step2: Substitute \(w = 4\) into the other side - equation

Also, \(CD=PQ\). Substitute \(w = 4\) into \(CD=x + 10w-29\) and \(PQ=2x+w + 4\).
\(CD=x+10\times4 - 29=x + 11\), \(PQ=2x+4 + 4=2x+8\)
Since \(CD = PQ\), then \(x + 11=2x+8\)
Solve for \(x\):
\(x-2x=8 - 11\)
\(-x=-3\)
\(x = 3\)

Answer:

\(w = 4\), \(x = 3\)