Consider the square $ABCD$ with side length 1, as shown in the following diagram. A horizontal segment $EW$ and a vertical segment $NS$, both of length $1/2$, are entirely within the square and intersect at point $X$ forming a right angle of $90^{\circ}$.

a) What is the area of the quadrilateral NESW?

b) What is the sum of the areas of the triangles $EDN, AES, BSW$, and $CNW$?

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-25.jpg?height=555&width=534&top_left_y=254&top_left_x=843)
Solution

a) The area of the quadrilateral $N E S W$ is given by

$$
\begin{aligned}
A_{N E S W} & =A_{E N W}+A_{E S W} \\
& =\frac{E W \cdot N X}{2}+\frac{E W \cdot X S}{2} \\
& =\frac{E W \cdot(N X+X S)}{2} \\
& =\frac{1 / 2 \cdot 1 / 2}{2} \\
& =\frac{1}{8}
\end{aligned}
$$

b) The sum of the areas of triangles $C D N$ and $A B S$ is $1 / 4$, as each has a base of length 1 and their heights sum to $1 / 2$. Similarly, the sum of the areas of triangles $A E D$ and $B C W$ is also $1 / 4$. The sum of the areas of the four triangles mentioned in the problem is the complement of the areas of triangles $C D N, A B S, A D E, B C W$ and the quadrilateral $N E S M$, therefore, the area sought is

$$
1-\frac{1}{8}-\frac{1}{4}-\frac{1}{4}=\frac{3}{8}
$$