The length DE, which is the base of the triangle, fits 4 times along the side DC. So 4 copies of the triangle fit with their bases along DC:

The white unshaded triangles are congruent to the grey shaded triangles, so they have the same area. There are 4 shaded triangles and 4 unshaded triangles, so there are 8 triangles altogether.

So the area of each triangle is $\frac18$ of the area of the square.

Suppose the square has side length $1$, so that its area is $1$. We can do this because we don't need to know what the area of the triangle is as a number, we just want to know it as a fraction of the area of the square. And finding numbers as fractions of $1$ is easy - for example, $\frac12$ of $1$ is just $\frac12$.

Then the height of the triangle is $1$, since it is the side length of the square.

$4\times=1$. So the length DE, which is the base of the triangle, is $\frac14$

So the area of the triangle is $\frac12\times\frac14\times1=\frac12\times\frac14=\frac18$

And $\frac18$ as a fraction of $1$ is just $\frac18$. So the area of the triangle is $\frac18$ of the area of the square.

You can find more short problems, arranged by curriculum topic, in our short problems collection.