The area of trapezium whose vertices lie on parabola ${\mathrm{y}}^2=4\mathrm{x}$ and its diagonals pass through (1,. 0) and having length $\frac{25}{4}$ units each, is

$$\begin{gathered} Given parabola y^2=4x \\ \Rightarrow a=1 \end{gathered}$$

Diagonals are Intersect at focus    

$\therefore \mathrm{AC}, \mathrm{BD}$ are focal chords.

$$\begin{gathered} \mathrm{A}=\left({\mathrm{at}}^2, 2\mathrm{at}\right)=\left({\mathrm{t}}^2, 2\mathrm{t}\right) \\ \mathrm{B}=\left({\mathrm{t}}^2, -2\mathrm{t}\right) \\ \mathrm{C}=\left(\frac{\mathrm{a}}{{\mathrm{t}}^2}, \frac{-2\mathrm{a}}{\mathrm{t}}\right)=\left(\frac{1}{{\mathrm{t}}^2}, \frac{-2}{\mathrm{t}}\right), \mathrm{D}=\left(\frac{1}{{\mathrm{t}}^2}, \frac{2}{\mathrm{t}}\right) \end{gathered}$$

Length of diagonal $\mathrm{AC}={\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)}^2$

                         $$\begin{gathered} {\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)}^2=\frac{25}{4} \Rightarrow \mathrm{t}+\frac{1}{\mathrm{t}}=\frac{5}{2} \\ \Rightarrow \mathrm{t}= 2 \left(\mathrm{or}\right) \frac{1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{A}=(4, 4), \mathrm{B}(4, -4), \mathrm{C}\left(\frac{1}{4}, -1\right), \mathrm{D}\left(\frac{1}{4}, 1\right) \\ \Rightarrow \mathrm{AB}=8, \mathrm{CD}=2 \\ dis\tan ce between AB and CD=\frac{15}{4} \\ Now \mathrm{Area}=\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right) (dis\tan ce between AB and CD) \\ =\frac{1}{2} \left(10\right) \left(\frac{15}{4}\right)=\frac{75}{4} \end{gathered}$$