A square is inscribed in a right triangle. Find the area of the square.

To determine the area of a square inscribed in a right triangle with legs of length 1 and 3, we can use a geometric approach.

Let's denote the right triangle $ABC$ where $A = (0, 0)$, $B = (1, 0)$, and $C = (0, 3)$. The right angle is at $A$, the x-axis extends along $AB$, and the y-axis along $AC$.

Let the side length of the inscribed square be $s$. When a square is inscribed in the triangle, the square's lower left corner will coincide with point $A$ (the right angle), and the top right corner will lie somewhere along the hypotenuse $BC$.

We need to find the relationship between $s$ and the coordinates of the square touching the hypotenuse. The coordinates where the square touches the hypotenuse will be at the point $(s, s)$ since the square's sides are parallel to the axes.

Now, we need to find the equation of line $BC$. The slope $m$ of line segment $BC$ can be calculated as follows:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{0 - 1} = -3.$$

The equation of line $BC$, which has a y-intercept at point $C$ $(0, 3)$, can be expressed as:

$$y - 3 = -3(x - 0) \quad \Rightarrow \quad y = -3x + 3.$$

Now, we want the point $(s, s)$ on the line $BC$. Therefore, substituting $x = s$ into the line equation gives:

$$s = -3s + 3.$$

Solving this equation for $s$:

$$s + 3s = 3 \quad \Rightarrow \quad 4s = 3 \quad \Rightarrow \quad s = \frac{3}{4}.$$

The area $A$ of the square is given by $s^2$:

$$A = s^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16}.$$

Thus, the area of the square inscribed in the right triangle is

$$\boxed{\frac{9}{16}}.$$