Let the diagonals be 3x and 4x this gives x= 8 So the length of diagonals 24 and 32 respectively Now , Since the diagonals of rhombus bisect each other at $90^{\circ}$ Side length of rhombus = $\sqrt{16^2+12^2}=20$ Thus perimeter of the rhombus is $4 \cdot 20 =80$

Since the two diagonals have length ratios of $3:4:$, their sum can be divided into 7 parts. Hence, each section is 8, with the diagonal lengths: $24:32$. Since all triangles are congruent by transitive properties and HL Theorem, each triangle has length $12$ and $16$. We see that this is a 3-4-5 triangle. Hence, each of the side length of the perimeter is $20$. The answer is $\boxed{80}$