Given a square $ABCD$ with two consecutive vertices, say $A$ and $B$ on the positive $x$-axis and positive $y$-axis respectively. Suppose the other vertice $C$ lying in the first quadrant has coordinates $(u , v)$. Then find the area of the square $ABCD$ in terms of $u$ and $v$.
1. Let's set the coordinates of the square as \( A(x, 0) \), \( B(0, y) \), \( C(u, v) \), and \( D \) where \( x, y, u, v > 0 \). The fourth coordinate \( D \) can be written as \( D(u + x, v - y) \).

2. We shift the coordinates such that the new coordinates of the square are: \( A(0, 0) \), \( B(-x, y) \), \( C(u - x, v) \), and \( D(u, v - y) \).

3. Since \( AD \) and \( AB \) are perpendicular, the product of their slopes must be \(-1\). Therefore, we have:
   \[
   \text{slope of } AD \times \text{slope of } AB = -1
   \]
   \[
   \frac{v - y}{u} \times \frac{y}{-x} = -1
   \]
   Simplifying, we get:
   \[
   (v - y)y = ux
   \]

4. Also, since \( AD^2 = AB^2 \), we have:
   \[
   u^2 + (v - y)^2 = x^2 + y^2
   \]

5. Using the two equations, we get:
   \[
   (v - y)y = ux
   \]
   \[
   u^2 + (v - y)^2 = x^2 + y^2
   \]

6. Substituting \( x = v - y \) into the second equation, we get:
   \[
   u^2 + (v - y)^2 = (v - y)^2 + y^2
   \]
   Simplifying, we get:
   \[
   u^2 = y^2
   \]
   Since \( u, y > 0 \), we have \( u = y \).

7. Substituting \( u = y \) into the first equation, we get:
   \[
   (v - y)y = uy
   \]
   \[
   (v - y)y = y^2
   \]
   \[
   v - y = y
   \]
   \[
   v = 2y
   \]

8. Therefore, the side length of the square is \( y \), and the area of the square is:
   \[
   \text{Area} = y^2 = \left(\frac{v}{2}\right)^2 = \frac{v^2}{4}
   \]

The final answer is \(\boxed{\frac{v^2}{4}}\).