Calculate the sum of the ratio parts.

Step 1: Calculate the sum of the ratio parts. The ratio of the sides of the triangle is $2:3:4$. The sum of the ratio parts is: $2 + 3 + 4 = 9$

Step 2: Determine the length of each side. Let the common multiplier for the ratio be $x$. The perimeter of the triangle is $120 \text{ mm}$. The sum of the sides is equal to the perimeter: $2x + 3x + 4x = 120 \text{ mm}$ $9x = 120 \text{ mm}$ $x = \frac{120}{9} mm = \frac{40}{3} \text{ mm}$ Now, calculate the length of each side: Side 1: $2x = 2 \times \frac{40}{3} mm = \frac{80}{3} mm \approx 26.67 \text{ mm}$ Side 2: $3x = 3 \times \frac{40}{3} mm = 40 \text{ mm}$ Side 3: $4x = 4 \times \frac{40}{3} mm = \frac{160}{3} mm \approx 53.33 \text{ mm}$

Step 3: Construct the triangle using the calculated side lengths. To construct the triangle with sides $\frac{80}{3} \text{ mm}$, $40 \text{ mm}$, and $\frac{160}{3} \text{ mm}$:

1. Draw a line segment, say $AB$, with a length of $40 \text{ mm}$. This will be the base of your triangle.
2. Place the compass point at $A$ and open it to a radius of $\frac{80}{3} \text{ mm}$ (approximately $26.67 \text{ mm}$). Draw an arc above the line segment $AB$.
3. Place the compass point at $B$ and open it to a radius of $\frac{160}{3} \text{ mm}$ (approximately $53.33 \text{ mm}$). Draw another arc above the line segment $AB$, intersecting the first arc.
4. Label the point where the two arcs intersect as $C$.
5. Draw line segments $AC$ and $BC$. Triangle $ABC$ is the required triangle.

The lengths of the sides of the triangle are $\boxed{Side 1 = \frac{80}{3} mm, Side 2 = 40 mm, Side 3 = \frac{160}{3} mm}$.

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