In the diagram, rectangle $P Q R S$ is made up of six identical squares. Points $U, V, W, X, Y$, and $Z$ are midpoints of sides of the squares, as shown. Which of the following triangles has the greatest area?
(A) $P V U$
(B) $P X Z$
(C) $P V X$
(D) $P Y S$
(E) $P Q W$

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-038.jpg?height=304&width=436&top_left_y=1840&top_left_x=1300)
We can determine which triangle has the greatest area by using a fixed side length of 4 for each of the identical squares and using this to calculate the unknown areas. We begin by constructing $\triangle P V U$ and noticing that it is contained within square $Q A B P$, as shown. The area of $\triangle P V U$ is determined by subtracting the areas of triangles $P Q V, V A U$ and $P B U$ from the area of square $Q A B P$.

Since $Q A=8$ and $A B=8$, then the area of square $Q A B P$ is $8 \times 8=64$.

Since $P Q=8$ and $Q V=2$, then the area of $\triangle P Q V$ is $\frac{1}{2} \times 8 \times 2=8$.

Since $V A=6$ and $A U=6$, then the area of $\triangle V A U$ is $\frac{1}{2} \times 6 \times 6=18$.

Since $P B=8$ and $U B=2$, then the area of $\triangle P B U$ is $\frac{1}{2} \times 8 \times 2=8$.

Therefore, the area of $\triangle P V U$ is $64-8-18-8=30$.

Next, we construct $\triangle P X Z$ and then construct rectangle $C D S P$ by drawing $C D$ parallel to $P S$ through $X$. Further, $X$ is the midpoint of the side of a square and so $C$ and $D$ are also midpoints of the sides of their respective squares.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-173.jpg?height=320&width=440&top_left_y=1350&top_left_x=1520)

Construct $\triangle P V X$ and notice that it is contained within square $Q A B P$, as shown. The area of $\triangle P V X$ is determined by subtracting the areas of triangles $P Q V, V A X$ and $P B X$ from the area of square $Q A B P$.

As we previously determined, the area of square $Q A B P$ is 64 and the area of $\triangle P Q V$ is 8 .

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-174.jpg?height=330&width=440&top_left_y=136&top_left_x=1531)

Since $V A=6$ and $A X=2$, then the area of $\triangle V A X$ is $\frac{1}{2} \times 6 \times 2=6$.

Since $P B=8$ and $X B=6$, then the area of $\triangle P B X$ is $\frac{1}{2} \times 8 \times 6=24$.

Therefore, the area of $\triangle P V X$ is $64-8-6-24=26$.

Construct $\triangle P Y S$ and the perpendicular from $Y$ to $E$ on $P S$, as shown. Since $P S=12$ and $Y E=4$ ( $Y E$ is parallel to $R S$ and thus equal in length to the side of the square), then the area of $\triangle P Y S$ is $\frac{1}{2} \times 12 \times 4=24$.

Construct $\triangle P Q W$, as shown. Since $P Q=8$ and $Q W=6$, then the area of $\triangle P Q W$ is $\frac{1}{2} \times 8 \times 6=24$.

The areas of the 5 triangles are $30,28,26,24$, and 24 . The triangle with greatest area, 30 , is $\triangle P V U$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-174.jpg?height=586&width=423&top_left_y=642&top_left_x=1534)

ANSWER: (A)