A total of 2 squares each have side length r . A total of 6 equilateral

Choice C is correct. It’s given that a total of $2$ squares each have side length $r$. Therefore, each of the squares has perimeter $4r$. Since there are a total of $2$ squares, the sum of the perimeters of these squares is $4r+4r$, which is equivalent to $2\left(4r\right)$, or $8r$. It’s also given that a total of $6$ equilateral triangles each have side length $t$. Therefore, each of the equilateral triangles has perimeter $3t$. Since there are a total of $6$ equilateral triangles, the sum of the perimeters of these triangles is $3t+3t+3t+3t+3t+3t$, which is equivalent to $6\left(3t\right)$, or $18t$. Since the sum of the perimeters of the squares is $8r$ and the sum of the perimeters of the triangles is $18t$, the sum of the perimeters of all these squares and triangles is $8r+18t$. It’s given that the sum of the perimeters of all these squares and triangles is $210$. Therefore, the equation $8r+18t=210$ represents this situation.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.