Solution: As shown in Figure 1, let the side length of the square $ABCD$ be $a$, and $AH = x, DG = y, CF = z, BE = w$, then $HD = a - x, GC = a - y, FB = a - z, EA = a - w$.
Thus, from the given conditions, we have
$$\left\{\begin{array}{l} x(a - w) = 4, \\ y(a - x) = 14, \\ z(a - y) = 2t, \\ w(a - z) = 56. \end{array}\right.$$

Let $EG, FH, BD$ intersect at point $I$. Since
$\triangle IDH \sim \triangle IBF, \triangle IGD \sim \triangle IEB$,
we have $\frac{DH}{BF} = \frac{DI}{BI} = \frac{DG}{BE}$,
i.e., $(a - x)w = (a - z)y$.
Substituting (2) and (4) into (5) gives $\frac{14}{y} \cdot w = \frac{56}{w} \cdot y$, i.e., $w = 2y$.
Substituting (6) into (1) gives
$x(a - 2y) = 4$.
Substituting (6) into (4) gives
$y(a - z) = 28$.
Eliminating $z$ from (8) and (3), and simplifying, we get
$$ay^2 + (-a^2 - 28 + 2t)y + 28a = 0.$$

Eliminating $x$ from (7) and (2), and simplifying, we get
$$2ay^2 + (-a^2 - 24)y + 14a = 0.$$

Solving (9) and (11) gives
$$y = \frac{a^2 + 20 + 2t}{3a}.$$

Substituting (12) into (11) and simplifying, we get
$$a^4 + (-74 - 2t)a^2 + (640 - 16t - 8t^2) = 0.$$

By the quadratic formula, we get
$$a^2 = 37 + t \pm 3\sqrt{t^2 + 10t + 81}.$$

Noting that the sum of the areas of the four triangles at the corners of the square is $37 + t$, from (12) (taking the positive sign), we know that the area of quadrilateral $EFGH$ is $3\sqrt{t^2 + 10t + 81}$ (where $t$ is any positive real number).
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University, 510405; Zuo Huaiqing, No. 6 Middle School of Guangzhou, Guangdong Province, 510000)