The area of triangle $I J S$ is equal to half the area of the rectangle with sides $I J$ and $I A$. Side $I J$ is a side of the square $I J H E$, and side $I A$ is the sum of the side of the square $I J H E$ and the side of the square $E F C A$.

Square $I J H E$ has a side that is twice the side of square $E F C A$, and that measures $1 \mathrm{~cm}$. Therefore, side $I J$ measures $2 \mathrm{~cm}$ and side $I A$ measures $3 \mathrm{~cm}$. The area of triangle $I J S$ is equal to

$$
\frac{1}{2} \cdot 2 \cdot 3=3\left(\mathrm{~cm}^{2}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_8fdde125de0a7cae0b19g-5.jpg?height=788&width=828&top_left_y=634&top_left_x=617)

Note. By more finely dividing the indicated square grid, the same result can be achieved through visual counting of the squares, or triangular segments.
In the figure, squares $ABCD$, $EFCA$, $GHEC$, and $IJHE$ are depicted. Points $S$, $B$, $F$, and $G$ are the centers of these squares, respectively. Segment $AC$ is $1 \, \text{cm}$ long.

Determine the area of triangle $IJS$.

(E. Semerádová)

![](https://cdn.mathpix.com/cropped/2024_04_17_8fdde125de0a7cae0b19g-4.jpg?height=676&width=811&top_left_y=1558&top_left_x=631)

Hint. The area of the triangle is half the area of some rectangle.