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

\[
\boxed{\frac{9}{16}}.
\]