A6. On side $C D$ of a square $A B C D$ lies a point $E$. Line segment $A E$ is divided into four equal parts by the points $P, R$ and $T$. Line segment $B E$ is divided into four equal parts by the points $Q, S$ and $U$. Given that $|P Q|=3$.

What is the area of quadrilateral $P Q U T$?
A) $\frac{15}{4}$
B) 4
C) $\frac{17}{4}$
D) $\frac{9}{2}$
E) 5

![](https://cdn.mathpix.com/cropped/2024_04_17_dd45dfb811a24abc1595g-2.jpg?height=368&width=371&top_left_y=507&top_left_x=1522)
A6. B) 4 Triangle $E A B$, triangle $E P Q$, and triangle $E T U$ are similar because $|A E|:|P E|:|T E|=4: 3: 1=|B E|:|Q E|:|U E|$ and $\angle A E B=\angle P E Q=\angle T E U$ (SAS). This implies that $|A B|:|P Q|:|T U|=$ $4: 3: 1$. From this, it follows that $|A B|=4$, so the area of triangle $A B E$ is $\frac{1}{2} \times 4 \times 4=8$.

Due to the similarity, triangles $P Q E$ and $T U E$ therefore have areas $\left(\frac{3}{4}\right)^{2} \times 8=\frac{9}{2}$ and $\left(\frac{1}{4}\right)^{2} \times 8=\frac{1}{2}$, respectively. The area of the

![](https://cdn.mathpix.com/cropped/2024_04_17_ecd98442595314acd6b1g-1.jpg?height=369&width=371&top_left_y=2214&top_left_x=1525)
quadrilateral is thus $\frac{9}{2}-\frac{1}{2}=4$.