Solution:

Each half has $3$ red, $5$ blue, and $8$ white triangles.

The $2$ red-red pairs use $2$ red triangles from each half, so the remaining red triangle from each half is used in the $2$ red-white pairs. Thus all red triangles are accounted for.

The $3$ blue-blue pairs use $3$ blue triangles from each half, leaving $2$ blue triangles from each half. Since no more blue-blue pairs occur, those $4$ blue triangles must pair with white triangles.

So on each half, $1$ white is used with red and $2$ whites are used with blue, leaving $8-3=5$ white triangles to pair with white triangles.

Thus, B is the correct answer.