My guess for the smallest value of $X$ is

$65$ cm

where the lengths of the pieces are

$8$, $12$, $18$ and $27$ cm

Reasoning

Let the original triangle have side lengths $x,y,z$. Then, the new triangle will have side lengths $ax, ay, az$ for some rational number $a$. Also, two of the side lengths must shared between the two triangles.

Without loss of generality, suppose $x < y < z$ and $a>1$.

Then we must have $ax = y$ and $ay = z$. (This is due to the fact that $x$ cannot feature in the new triangle and $az$ does not feature in the original).

Hence, the four pieces have lengths $x, ax, a^2x, a^3x$ for integer $x$ and rational $a$. Since each length must be an integer, we find that $x$ must be divisible by the cube of the denominator of $a$, when written in its simplest form. A denominator of $1$ doesn't work for the sides to be non-degenerate triangles so the next simplest case is a denominator of $2$ and a numerator of $3$, i.e, $a = \frac{3}{2}$. The smallest $x$ can be is $8$.

This gives the side lengths as being $8, 12, 18, 27$ and $X = 65$

I think the smallest value of $X$ is

$65$ cm

where the lengths of the pieces are

$8$, $12$, $18$ and $27$ cm

The side lengths of the similar triangles are $(8,12,18)$ and $(12,18,27)$.

Reasoning

Let the original triangle have side lengths $x,y,z$. Then, the new triangle will have side lengths $ax, ay, az$ for some rational number $a$. Also, two of the side lengths must shared between the two triangles.

Without loss of generality, suppose $x < y < z$ and $a>1$.

Then we must have $ax = y$ and $ay = z$. (This is due to the fact that $x$ cannot feature in the new triangle and $az$ does not feature in the original).

Hence, the four pieces have lengths $x, ax, a^2x, a^3x$ for integer $x$ and rational $a$. Since each length must be an integer, we find that $x$ must be divisible by the cube of the denominator of $a$, when written in its simplest form. A denominator of $1$ doesn't work for the sides to be non-degenerate triangles so the next simplest case is a denominator of $2$ and a numerator of $3$, i.e, $a = \frac{3}{2}$. The smallest $x$ can be, in this case, is $8$.

This gives the side lengths as being $8, 12, 18, 27$ and $X = 65$

It's not too difficult to see that raising either the numerator or denominator of $a$, in its simplest form, will necessarily increase the value of $X$ so this is the best we can do.