Let's go through each of the problems in this image.

First Problem

For the first triangle pair, we have:

1. Triangle $\triangle ABC$ with sides $AC = x$, $AB = 25$, and the other triangle $\triangle XYZ$ with sides $XZ = 6$ and $XY = 10$.

Since the triangles are similar, we can set up a proportion using corresponding sides: $\frac{AC}{XZ} = \frac{AB}{XY}$ Substitute the known values: $\frac{x}{6} = \frac{25}{10}$ Now, cross-multiply and solve for $x$: $10x = 6 \times 25$ $10x = 150$ $x = 15$

So, $AC = 15$.

Second Problem

In the second triangle pair, we have:

1. Triangle $\triangle ABC$ with sides $AB = 50$, $BC = 40$, and the other triangle $\triangle YZX$ with sides $YZ = 20$ and $ZX = x$.

Using the similarity property again: $\frac{BC}{YZ} = \frac{AB}{YX}$ Substitute the known values: \