The two diagonals of a rhombus have lengths with ratio $3 : 4$ and sum $56$. What is the perimeter of the rhombus?
1. **Determine the lengths of the diagonals:**
   - Let the lengths of the diagonals be \(d_1\) and \(d_2\).
   - Given the ratio of the diagonals is \(3:4\), we can write \(d_1 = 3x\) and \(d_2 = 4x\) for some positive number \(x\).
   - The sum of the diagonals is given as \(56\). Therefore, we have:
     \[
     d_1 + d_2 = 56 \implies 3x + 4x = 56 \implies 7x = 56 \implies x = 8
     \]
   - Substituting \(x = 8\) back, we get:
     \[
     d_1 = 3 \cdot 8 = 24 \quad \text{and} \quad d_2 = 4 \cdot 8 = 32
     \]

2. **Calculate the side length of the rhombus:**
   - In a rhombus, the diagonals bisect each other at right angles.
   - Each half-diagonal forms a right triangle with the side of the rhombus.
   - The half-diagonals are:
     \[
     \frac{d_1}{2} = \frac{24}{2} = 12 \quad \text{and} \quad \frac{d_2}{2} = \frac{32}{2} = 16
     \]
   - Using the Pythagorean theorem in one of these right triangles, where the side of the rhombus is the hypotenuse:
     \[
     s = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20
     \]

3. **Calculate the perimeter of the rhombus:**
   - The perimeter \(P\) of a rhombus is four times the length of one side:
     \[
     P = 4s = 4 \cdot 20 = 80
     \]

The final answer is \(\boxed{80}\).