The midpoints of sides $\mathrm{AB}$, $\mathrm{BC}$, and $\mathrm{AC}$ in $\triangle \: \mathrm{ABC}$ are $\mathrm{M}, \mathrm{N}$, and $\mathrm{P}$, respectively. The medians drawn from $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$ intersect the line segments $\mathrm{MP}$, $\mathrm{MN}$ and $\mathrm{NP}$ at $\mathrm{X}$, $\mathrm{Y}$, and $\mathrm{Z}$, respectively. If the area of $\triangle \: \mathrm{ABC}$ is $1440 \: \mathrm{sq \:cm}$ , then the area, in $\mathrm{sq \:cm}$, of $\triangle \: \mathrm{XYZ}$ is