Question

what values of v and w make \\( \triangle xyz \cong \triangle efd \\)?

Explanation:

Step1: Set equal corresponding sides

Since \(\triangle XYZ\cong\triangle EFD\), then \(YZ = FD\). So \(v + 32=3v\).
Subtract \(v\) from both sides: \(32 = 3v - v\), which simplifies to \(32 = 2v\).
Divide both sides by \(2\): \(v=\frac{32}{2}=16\).

Step2: Substitute \(v = 16\) into the other side - equality equation

Also, \(XY=EF\). So \(4v+w - 49=20w+v - 20\).
Substitute \(v = 16\) into the equation: \(4\times16+w - 49=20w+16 - 20\).
Calculate \(4\times16 = 64\), then the left - hand side is \(64+w - 49=w + 15\), and the right - hand side is \(20w-4\).
So \(w + 15=20w-4\).
Subtract \(w\) from both sides: \(15=19w-4\).
Add \(4\) to both sides: \(19 = 19w\).
Divide both sides by \(19\): \(w = 1\).

Answer:

\(v = 16\), \(w = 1\)