3. In a convex quadrilateral $ABCD$, points $\mathrm{M}, \mathrm{N}, \mathrm{P}$, and $Q$ are the midpoints of sides $AB, BC$, CD, and DA, respectively. If $\overline{\mathrm{AC}}=30 \mathrm{~cm}$, $\overline{\mathrm{MP}}=26 \mathrm{~cm}$, and $\overline{\mathrm{NQ}}=28 \mathrm{~cm}$, calculate the area of quadrilateral MNPQ.
Solution: We will show that MNPQ is a parallelogram. In $\triangle ABC$, the segment MN is a midline. Therefore, $\overline{\mathrm{MN}}=\frac{\overline{\mathrm{AC}}}{2}$

![](https://cdn.mathpix.com/cropped/2024_06_05_e7266f2e82216ac5d245g-2.jpg?height=481&width=416&top_left_y=536&top_left_x=1032)
and MN\|AC. Similarly, in $\triangle ACD$, the segment PQ is a midline. Therefore, $\overline{\mathrm{PQ}}=\frac{\overline{\mathrm{AC}}}{2}$ and $\mathrm{PQ} \| \mathrm{AC}$. Hence, $\overline{\mathrm{MN}}=\overline{\mathrm{PQ}}$ and $\mathrm{MN} \| \mathrm{PQ}$, i.e., quadrilateral MNPQ is a parallelogram. Let O be the intersection of the diagonals of parallelogram ABCD. Since the diagonals divide the parallelogram into four triangles of equal area, we have: $\overline{\mathrm{ON}}=\frac{1}{2} \overline{\mathrm{QN}}=14 \mathrm{~cm} ; \overline{\mathrm{OM}}=\frac{1}{2} \overline{\mathrm{PM}}=13 \mathrm{~cm} ; \overline{\mathrm{MN}}=\overline{\mathrm{QP}}=\frac{1}{2} \overline{\mathrm{AC}}=15$ cm. $\mathrm{P}_{\text {MNPQ }}=4 \cdot \mathrm{P}_{\text {MON }}=$

$$
=4 \cdot \sqrt{21 \cdot(21-13) \cdot(21-14) \cdot(21-15)}=4 \cdot \sqrt{21 \cdot 8 \cdot 7 \cdot 6}=4 \cdot 84=336 .
$$

The area of quadrilateral MNPQ is $336 \mathrm{~cm}^{2}$.