If $M$ is the mid-point of the side $BC$ of $\Delta ABC$, and the area of $\Delta ABM$ is $19{\text{cm}}^2$, then the area of $\Delta ABC$ is:

A median of a triangle divides it into two triangles of equal area.Since $M$ is the mid-point of $BC$, $AM$ is the median of $\Delta ABC$ corresponding to the side $BC$.Therefore:
$$\text{Area}(\Delta ABM)=\text{Area}(\Delta ACM)=19{\text{cm}}^2$$
The total area of $\Delta ABC$ is the sum of the areas of these two sub-triangles:
$$\text{Area}(\Delta ABC)=\text{Area}(\Delta ABM)+\text{Area}(\Delta ACM)=19+19=38{\text{cm}}^2$$