Problem 4. Two overlapping rectangles $ABCD$ and $AMLP$ have side lengths $\overline{AB}=\overline{ML}=24 \text{ cm}$ and $\overline{BC}=\overline{LP}=6 \text{ cm}$ as shown in the diagram on the right. The intersection point of sides $\overline{AP}$ and $\overline{CD}$ is point $Q$, and the intersection points of side $\overline{ML}$ with sides $\overline{AB}$ and $\overline{CD}$ are points $N$ and $K$, respectively. Calculate the area of quadrilateral $ANKQ$ and the length of segment $NB$, given that $\measuredangle ANL=150^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_c3c23cf15e32fbc44a06g-08.jpg?height=406&width=637&top_left_y=1619&top_left_x=845)
Solution. We have,

$$
\overline{A B}=\overline{C D}=\overline{M L}=\overline{P A}=24 \mathrm{~cm} \text { and } \overline{B C}=\overline{L P}=\overline{D A}=\overline{A M}=6 \mathrm{~cm}
$$

From the fact that quadrilateral $A B C D$ is a rectangle, $N \in A B$ and $K, Q \in C D$, for $A N$ and $K Q$ it holds that $A N \| K Q$, and from the fact that quadrilateral $A M L P$ is a rectangle, $Q \in A P$ and $N, K \in M L$ for $A Q$ and $N K$ it holds that $A Q \| N K$. Therefore, quadrilateral $A N K Q$ is a parallelogram. From $\measuredangle A N L=150^{\circ}$ it follows that $\measuredangle B N L=180^{\circ}-\measuredangle A N L=180^{\circ}-150^{\circ}=30^{\circ}$, i.e., $\measuredangle N A Q=30^{\circ}$. According to this, it holds that $\measuredangle Q A D=90^{\circ}-\measuredangle N A Q=90^{\circ}-30^{\circ}=60^{\circ}$. Triangle $Q D A$ is a right triangle, so $\measuredangle D Q A=90^{\circ}-\measuredangle Q A D=90^{\circ}-60^{\circ}=30^{\circ}$. Therefore, $\overline{A Q}=2 \overline{A D}=$ $12 \mathrm{~cm}$. From the fact that quadrilateral $A N K Q$ is a parallelogram and $A M$ and $A D$ are its heights, we get $P_{A N K Q}=\overline{A N} \cdot \overline{A D}$ and $P_{A N K Q}=\overline{N K} \cdot \overline{A M}$. From this, we have $\overline{A N} \cdot \overline{A D}=\overline{N K} \cdot \overline{A M}$. However, $\overline{A M}=\overline{A D}$ and $\overline{N K}=\overline{A Q}$, so we get $\overline{N K}=\overline{A N}$, i.e., $\overline{A N}=\overline{N K}=\overline{A Q}=12 \mathrm{~cm}$.

According to this, the area of quadrilateral $A N K Q$ is equal to $P_{A N K Q}=72 \mathrm{~cm}^{2}$. From $\overline{N B}=\overline{A B}-\overline{A N}$, we get $\overline{N B}=12 \mathrm{~cm}$.

## IX Department