Area of triangle formed by intersection of medians and midpoint segments

In $\triangle ABC$, $M, N, P$ are midpoints of $AB, BC, AC$. Thus $\triangle MNP$ is the medial triangle of $\triangle ABC$, and its area is $\frac{1}{4} \text{Area}(\triangle ABC) = 360$ sq cm.

Median from vertex $A$ passes through midpoint $N$ of $BC$. Since line segment $MP$ joins the midpoints of $AB$ and $AC$, $MP \parallel BC$ and $MP$ bisects $AN$. Thus, the intersection $X$ of median $AN$ with $MP$ is the midpoint of $MP$.

Similarly, $Y$ is the midpoint of $MN$ and $Z$ is the midpoint of $NP$.

Hence, $\triangle XYZ$ is the medial triangle of $\triangle MNP$. $\text{Area}(\triangle XYZ) = \frac{1}{4} \text{Area}(\triangle MNP) = \frac{1}{16} \text{Area}(\triangle ABC) = \frac{1440}{16} = 90\text{ sq cm}$