Let ABCD be a trapezium whose vertices lie on the parabola $$y^{2}=4x$$. Let the sides AD and BC of the trapezium be parallel to y -axis. If the diagonal AC is of length $$\frac{25}{4}$$ and it passes through the point (1,0), then the area of ABCD is

The parabola is $$y^2 = 4x$$, so $$a = 1$$ and the focus is at $$F = (1, 0)$$.

Any point on the parabola can be written as $$(t^2, 2t)$$.

Since AD and BC are parallel to the y-axis, the vertices of the trapezium lie on two vertical lines (latus rectum chords). Let the two vertical lines be $$x = t_1^2$$ and $$x = t_2^2$$.

For a vertical chord at $$x = t^2$$, the parabola gives $$y^2 = 4t^2$$, so $$y = \pm 2t$$. The chord length is $$4t$$.

Let $$A = (a_1^2, 2a_1)$$, $$D = (a_1^2, -2a_1)$$ be on the line $$x = a_1^2$$, and $$B = (a_2^2, 2a_2)$$, $$C = (a_2^2, -2a_2)$$ be on the line $$x = a_2^2$$.

The diagonal AC goes from $$(a_1^2, 2a_1)$$ to $$(a_2^2, -2a_2)$$. We are told this diagonal passes through the focus $$(1, 0)$$.

The line through $$(a_1^2, 2a_1)$$ and $$(a_2^2, -2a_2)$$ has the parametric form. The condition that $$(1, 0)$$ lies on line AC gives us:

$$\frac{0 - 2a_1}{1 - a_1^2} = \frac{-2a_2 - 2a_1}{a_2^2 - a_1^2}$$

$$\frac{-2a_1}{1 - a_1^2} = \frac{-2(a_1 + a_2)}{(a_2 - a_1)(a_2 + a_1)}$$

$$\frac{-2a_1}{1 - a_1^2} = \frac{-2}{a_2 - a_1}$$

Cross-multiplying: $$-2a_1(a_2 - a_1) = -2(1 - a_1^2)$$

$$a_1 a_2 - a_1^2 = 1 - a_1^2$$

$$a_1 a_2 = 1$$ $$-(1)$$

This is the well-known focal chord property: for a chord of the parabola $$y^2 = 4x$$ passing through the focus, $$t_1 t_2 = -1$$. Here, since A and C are on opposite sides (A has $$y = 2a_1$$ and C has $$y = -2a_2$$), the effective parameters are $$a_1$$ and $$-a_2$$, giving $$a_1 \cdot (-a_2) = -1$$, i.e., $$a_1 a_2 = 1$$.

Now we compute $$|AC|$$. The distance from $$A = (a_1^2, 2a_1)$$ to $$C = (a_2^2, -2a_2)$$ is:

$$|AC|^2 = (a_1^2 - a_2^2)^2 + (2a_1 + 2a_2)^2$$

$$= (a_1 - a_2)^2(a_1 + a_2)^2 + 4(a_1 + a_2)^2$$

$$= (a_1 + a_2)^2[(a_1 - a_2)^2 + 4]$$

$$= (a_1 + a_2)^2[(a_1 + a_2)^2 - 4a_1 a_2 + 4]$$

Using $$a_1 a_2 = 1$$:

$$= (a_1 + a_2)^2[(a_1 + a_2)^2 - 4 + 4] = (a_1 + a_2)^2 \cdot (a_1 + a_2)^2 = (a_1 + a_2)^4$$

So $$|AC| = (a_1 + a_2)^2$$.

Given $$|AC| = \frac{25}{4}$$, we get $$(a_1 + a_2)^2 = \frac{25}{4}$$, so $$a_1 + a_2 = \frac{5}{2}$$ (taking positive value).

The area of the trapezium ABCD with parallel sides AD and BC along the y-axis is:

$$\text{Area} = \frac{1}{2}(|AD| + |BC|) \times h$$

where $$|AD| = 4a_1$$, $$|BC| = 4a_2$$, and the height $$h = |a_1^2 - a_2^2| = |a_1 - a_2|(a_1 + a_2)$$.

We need $$a_1 - a_2$$. From $$a_1 + a_2 = \frac{5}{2}$$ and $$a_1 a_2 = 1$$:

$$(a_1 - a_2)^2 = (a_1 + a_2)^2 - 4a_1 a_2 = \frac{25}{4} - 4 = \frac{9}{4}$$

So $$|a_1 - a_2| = \frac{3}{2}$$.

Therefore: $$\text{Area} = \frac{1}{2}(4a_1 + 4a_2) \times |a_1 - a_2|(a_1 + a_2)$$

$$= \frac{1}{2} \times 4(a_1 + a_2) \times |a_1 - a_2| \times (a_1 + a_2)$$

$$= 2(a_1 + a_2)^2 |a_1 - a_2|$$

$$= 2 \times \frac{25}{4} \times \frac{3}{2} = \frac{75}{4}$$

The area of trapezium ABCD is $$\frac{75}{4}$$, which matches Option A.