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

The correct answer is 90.

Let us analyze the relationship between the triangles step-by-step.

Step 1: Understand the Medial Triangle of △ABC
The points $M$, $N$, and $P$ are the midpoints of the sides $\mathrm{AB}$, $\mathrm{BC}$, and $\mathrm{AC}$ of $△\mathrm{ABC}$, respectively.
By the Midpoint Theorem, the segments connecting these midpoints are parallel to the opposite sides of the triangle and have half of their lengths:
$\mathrm{MP}\parallel \mathrm{BC}$ and $\mathrm{MP}=\frac{1}{2}\mathrm{BC}$
$\mathrm{MN}\parallel \mathrm{AC}$ and $\mathrm{MN}=\frac{1}{2}\mathrm{AC}$
$\mathrm{NP}\parallel \mathrm{AB}$ and $\mathrm{NP}=\frac{1}{2}\mathrm{AB}$
Thus, $△\mathrm{MNP}$ is the medial triangle of $△\mathrm{ABC}$.
The area of a medial triangle is always exactly one-fourth of the area of the main triangle:
$\text{Area}(△\mathrm{MNP})=\frac{1}{4}\times \text{Area}(△\mathrm{ABC})$

Step 2: Find the Positions of Points X, Y, and Z
Let us consider the median $\mathrm{AN}$ drawn from vertex $A$ to the midpoint $N$ of side $\mathrm{BC}$. This median intersects the segment $\mathrm{MP}$ at point $X$.
Since the line segment $\mathrm{MP}$ is parallel to $\mathrm{BC}$, the triangles $△\mathrm{AMX}$ and $△\mathrm{ABN}$ are similar, and $△\mathrm{AXP}$ and $△\mathrm{ANC}$ are similar.
Using similarity ratios, we get:
$\frac{\mathrm{MX}}{\mathrm{BN}}=\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{1}{2}\Rightarrow \mathrm{MX}=\frac{1}{2}\mathrm{BN}$
$\frac{\mathrm{XP}}{\mathrm{NC}}=\frac{\mathrm{AP}}{\mathrm{AC}}=\frac{1}{2}\Rightarrow \mathrm{XP}=\frac{1}{2}\mathrm{NC}$
Since $N$ is the midpoint of $\mathrm{BC}$, we have $\mathrm{BN}=\mathrm{NC}$.
Therefore, it follows that $\mathrm{MX}=\mathrm{XP}$, which means $X$ is the midpoint of the segment $\mathrm{MP}$.
By applying the same reasoning to the other medians:
- The median $\mathrm{BP}$ intersects $\mathrm{MN}$ at $Y$, making $Y$ the midpoint of $\mathrm{MN}$.
- The median $\mathrm{CM}$ intersects $\mathrm{NP}$ at $Z$, making $Z$ the midpoint of $\mathrm{NP}$.

Step 3: Relate the Area of △XYZ to △MNP
Since $X$, $Y$, and $Z$ are the midpoints of the sides of $△\mathrm{MNP}$, the triangle $△\mathrm{XYZ}$ is the medial triangle of $△\mathrm{MNP}$.
Therefore, the area of $△\mathrm{XYZ}$ is one-fourth of the area of $△\mathrm{MNP}$:
$\text{Area}(△\mathrm{XYZ})=\frac{1}{4}\times \text{Area}(△\mathrm{MNP})$

Step 4: Compute the Final Area
Expressing the area of $△\mathrm{XYZ}$ in terms of the area of $△\mathrm{ABC}$ yields:
$\text{Area}(△\mathrm{XYZ})=\frac{1}{4}\times \left(\frac{1}{4}\times \text{Area}(△\mathrm{ABC})\right)=\frac{1}{16}\times \text{Area}(△\mathrm{ABC})$
Given that the area of $△\mathrm{ABC}$ is $1440{\text{cm}}^2$, we substitute this value into the equation:
$\text{Area}(△\mathrm{XYZ})=\frac{1440}{16}=90{\text{cm}}^2$