3. Given is a rectangle $A B C D$ whose side lengths differ by $4 \mathrm{~cm}$, and its perimeter is 16 $\mathrm{dm}$, with $|A B|>|B C|$. Point $M$ is the midpoint of side $\overline{B C}$ of the rectangle. Point $N$ lies on side $\overline{C D}$ and it is given that $|N C|=3|D N|$. Segments $\overline{A M}$ and $\overline{B N}$ intersect at point $P$. Is the area of triangle $A B P$ greater or the area of quadrilateral $P M C N$? By how much?
3. 

![](https://cdn.mathpix.com/cropped/2024_05_30_3b064f6bc407e279e59eg-05.jpg?height=331&width=574&top_left_y=1282&top_left_x=747)

Let $a=|A B|=|C D|$ and $b=|B C|=|A D|$. Then $a=b+4$.

The perimeter of the rectangle is $16 \mathrm{dm}=160 \mathrm{~cm}$, so we have:

$160=2 \cdot a+2 \cdot b$,

$160=2 \cdot(b+4)+2 \cdot b$,

$160=2 \cdot b+8+2 \cdot b$,

$160=4 \cdot b+8$,

$4 \cdot b=152$

$b=38 \mathrm{~cm}$.

Then $a=38+4=42 \mathrm{~cm}$.

Since $|N C|=3|D N|$, we have $|N C|=\frac{3}{4}|D C|=\frac{3}{4} \cdot 42=31.5 \mathrm{~cm}$.

Point $M$ is the midpoint of side $\overline{B C}$, so $|B M|=|M C|=\frac{1}{2}|B C|=\frac{1}{2} \cdot 38=19 \mathrm{~cm}$.

$P_{\triangle A B P}=P_{\triangle A B M}-P_{\triangle B M P}=\frac{42 \cdot 19}{2}-P_{\triangle B M P}=399-P_{\triangle B M P}$,

$P_{P M C N}=P_{\triangle B C N}-P_{\triangle B M P}=\frac{38 \cdot 31.5}{2}-P_{\triangle B M P}=598.5-P_{\triangle B M P}$,

$P_{P M C N}-P_{\triangle A B P}=598.5-P_{\triangle B M P}-\left(399-P_{\triangle B M P}\right)=598.5-P_{\triangle B M P}-399+P_{\triangle B M P}=199.5 \mathrm{~cm}^{2}$.

The area of quadrilateral $P M C N$ is $199.5 \mathrm{~cm}^{2}$ larger than the area of triangle $A B P$.