Let $ABCD$ be a cyclic quadrilateral with $AB = 1$, $BC = 2$, $CD = 3$, $DA = 4$. Find the square of the area of quadrilateral $ABCD$.
1. **Identify the sides and semiperimeter:**
   Given the sides of the cyclic quadrilateral \(ABCD\) are \(AB = 1\), \(BC = 2\), \(CD = 3\), and \(DA = 4\). 
   The semiperimeter \(s\) is calculated as:
   \[
   s = \frac{AB + BC + CD + DA}{2} = \frac{1 + 2 + 3 + 4}{2} = 5
   \]

2. **Apply Brahmagupta's formula:**
   Brahmagupta's formula for the area \(K\) of a cyclic quadrilateral with sides \(a, b, c, d\) and semiperimeter \(s\) is:
   \[
   K = \sqrt{(s-a)(s-b)(s-c)(s-d)}
   \]
   Substituting the given values \(a = 1\), \(b = 2\), \(c = 3\), \(d = 4\), and \(s = 5\):
   \[
   K = \sqrt{(5-1)(5-2)(5-3)(5-4)} = \sqrt{4 \cdot 3 \cdot 2 \cdot 1} = \sqrt{24}
   \]

3. **Calculate the square of the area:**
   The problem asks for the square of the area of the quadrilateral:
   \[
   K^2 = (\sqrt{24})^2 = 24
   \]

The final answer is \(\boxed{24}\).