Exercise 7. Each side of a unit square is divided into 3 equal segments. The figure below is drawn from this division. What is the area of the shaded polygon? Justify your answer.

![](https://cdn.mathpix.com/cropped/2024_06_04_d69c62b301e4ba44b0f0g-4.jpg?height=323&width=328&top_left_y=2153&top_left_x=904)

![](https://cdn.mathpix.com/cropped/2024_06_04_d69c62b301e4ba44b0f0g-5.jpg?height=358&width=363&top_left_y=301&top_left_x=599)
## Solution to Exercise 7

Since $I$ and $J$ are the midpoints of $[A K]$ and $[A L]$, $(I J)$ is parallel to $(K L)$ and $I J / K L = 1 / 2$. Therefore, $X J / X K = I J / K L$ so $X J = K J / 3$. This shows that the height from $X$ to $X L J$ is one-third of the height from $K$ to $K J A$, thus it is 2/9. From this, we deduce that the area of $X L J$ and the area of $X J A$ are both $1 / 27$.

A similar reasoning shows that $L N / L I = 1 / 4$, so the area of $J N L$ is 1/72, and the area of $J X N$ is $\frac{1}{27} - \frac{1}{72} = \frac{5}{216}$.

The area of $A K X N J A$ is therefore equal to $S(A K J) + S(J X N) = \frac{1}{9} + \frac{5}{216} = \frac{29}{216}$. The complement within the square of the shaded part being the union of four parts of area $\frac{29}{216}$, its area is $\frac{29}{54}$, and finally the area of the shaded part is equal to $1 - \frac{29}{54} = \frac{25}{54}$.