1. Parabola and its focus

   Given parabola: $y^2=4x$ Comparing with the standard form $y^2=4ax$, we get $a=1$ So the focus is $S(1,0)$ and the vertex is $O(0,0)$.

2. Parametric points of a focal chord

   A general point on the parabola $y^2=4ax$ is $P(at^2,2at)$ For $a=1$, this becomes $P(t^2,2t)$

   If $PQ$ is a focal chord, then the corresponding parameters are $t$ and $-\frac{1}{t}$.

   Hence, $P(t^2,2t), \qquad Q\left(\frac{1}{t^2},-\frac{2}{t}\right)$

3. Use the length of focal chord

   Length of focal chord of parabola $y^2=4ax$ with endpoints corresponding to $t$ and $-\frac{1}{t}$ is $PQ=a\left(t+\frac{1}{t}\right)^2$

   Here $a=1$ and $PQ=6.25=\frac{25}{4}$, so $\left(t+\frac{1}{t}\right)^2=\frac{25}{4}$

4. Area of triangle $POQ$

   Since $O=(0,0)$, area of $\triangle POQ$ is $\text{Area} = \frac12 \left|x_1y_2-x_2y_1\right|$

   Substituting $P(t^2,2t), \quad Q\left(\frac{1}{t^2},-\frac{2}{t}\right)$

   we get $\text{Area} = \frac12 \left| t^2\left(-\frac{2}{t}\right)-\frac{1}{t^2}(2t)\right|$ $=\frac12 \left|-2t-\frac{2}{t}\right|$ $=\left|t+\frac{1}{t}\right|$

5. Compute the area

   From step 3, $\left(t+\frac{1}{t}\right)^2=\frac{25}{4}$ Therefore, $\left|t+\frac{1}{t}\right|=\frac{5}{2}$

   So the area is $\frac{5}{2}$

6. Find 10 times the area

   $10 \times \text{Area} = 10 \times \frac{5}{2} = 25$

7. Final answer

   $\boxed{25}$

The derived answer matches the stored correct answer.