4. Let $\triangle A B K$ be a triangle with acute angles at vertices $A$ and $B$. Points $D$ and $N$ are chosen on side $\overline{K A}$, and points $C$ and $M$ on side $\overline{K B}$ such that the lines $C D, M N$, and $A B$ are parallel. The lines $A B$ and $C D$ are $4 \mathrm{~cm}$ away from the line $N M$. The midpoint of segment $\overline{N M}$ is $4 \mathrm{~cm}$ away from line $A K$ and $3 \mathrm{~cm}$ away from line $B K$. If the area of quadrilateral $A B C D$ is $80 \mathrm{~cm}^{2}$, determine the area of triangle $\triangle A B K$.
4. Let $\triangle A B K$ be a triangle with acute angles at vertices $A$ and $B$. Points $D$ and $N$ are chosen on side $\overline{K A}$, and points $C$ and $M$ on side $\overline{K B}$ such that lines $D C, N M$, and $A B$ are parallel. Lines $A B$ and $D C$ are $4 \mathrm{~cm}$ away from line $N M$. The midpoint of segment $\overline{N M}$ is $4 \mathrm{~cm}$ away from line $A K$ and $3 \mathrm{~cm}$ away from line $B K$. If the area of quadrilateral $A B C D$ is $80 \mathrm{~cm}^{2}$, determine the area of triangle $\triangle A B K$.

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_ba00de53bdad84e80dffg-22.jpg?height=950&width=1470&top_left_y=1284&top_left_x=290)

Let $|S P|=3 \mathrm{~cm}$, and $|S Q|=4 \mathrm{~cm}$. Quadrilateral $A B C D$ is a trapezoid with $\overline{A B}$ and $\overline{C D}$ as its bases because $A B \| C D$.

The area of the trapezoid is given by $P=s v$, where $s=\frac{a+b}{2}$, and $a$ and $b$ are the lengths of the bases of the trapezoid.

Thus, $80=s \cdot 8$, from which $s=\frac{a+b}{2}=10$, so $a+b=20$.

The right triangles $\triangle S M P$ and $\triangle S Q N$ have hypotenuses of length $\frac{s}{2}=5 \mathrm{~cm}$.

Using the Pythagorean theorem, we calculate $|M P|=4 \mathrm{~cm}$ and $|N Q|=3 \mathrm{~cm}$.

Let $E$ and $F$ be the feet of the perpendiculars from vertices $D$ and $C$ to the bases, and $Z$ be the foot of the perpendicular from vertex $K$ to the base of triangle $\triangle A B K$.

Denote the lengths of segments $\overline{E Z}$ and $\overline{Z F}$ as $l$ and $m$ respectively, and the lengths of segments $\overline{A E}$ and $\overline{F B}$ as $y$ and $x$ respectively.

By the AA similarity criterion, right triangles $\triangle S M P$ and $\triangle B C F$ are similar because $|\angle P M S|=|\angle C B F|$ (angles between parallel lines).

From this similarity, we have $|S P|:|M P|=|C F|:|B F|$, or $3: 4=8: x$, so $x=\frac{32}{3}$.

Similarly, from the similarity of $\triangle N Q S$ and $\triangle A E D$, we have $|S Q|:|Q N|=|D E|:|A E|$, or $4: 3=8: y$, so $y=6$.

Since quadrilateral $C D E F$ is a rectangle (adjacent sides are perpendicular), $|E F|=|C D|=b$, so

$x+y=a-b$.

Solving the system $a-b=\frac{32}{3}+6, a+b=20$, we get $a=\frac{55}{3}, b=\frac{5}{3}$.

Triangles $\triangle A Z K$ and $\triangle A E D$ are similar by the AA similarity criterion, so the ratio $v: 8=(6+l): 6$ gives $v=8+\frac{4}{3} l$.

Triangles $\triangle Z B K$ and $\triangle F B C$ are similar by the AA similarity criterion, so the ratio $v: 8=\left(m+\frac{32}{3}\right): \frac{32}{3}$ gives $v=\frac{3}{4} m+8$.

Since $m+l=a-x-y$, we have $m+l=\frac{55}{3}-6-\frac{32}{3}=\frac{5}{3}$.

Solving the system $\frac{3}{4} m+8=8+\frac{4}{3} l, m+l=\frac{5}{3}$, we get

$l=\frac{3}{5}, \quad v=\frac{44}{5}$.

The area of triangle $\triangle A B K$ is

$P=\frac{a \cdot v}{2}=\frac{\frac{55}{3} \cdot \frac{44}{5}}{2}=\frac{242}{3} \mathrm{~cm}^{2}$.