Let \( F \) be the foot of the perpendicular from \( B \) to \( AC \).

In \( \triangle BCF \), since \( \cos \hat{C} = \frac{3}{4} \), we have:

\[ \cos \hat{C} = \frac{\text{CF}}{BC} = \frac{\text{CF}}{x} = \frac{3}{4} \]

\[ \text{CF} = \frac{3}{4} x \]

Apply Pythagoras’ theorem in \( \triangle BCF \):

\[ \text{BF}^2 = x^2 – \left( \frac{3}{4} x \right)^2 = x^2 – \frac{9}{16} x^2 = \frac{16}{16} x^2 – \frac{9}{16} x^2 = \frac{7}{16} x^2 \]

\[ \text{BF} = \sqrt{\frac{7}{16} x^2} = \frac{\sqrt{7}}{4} x \]

Since \( \text{AC} = 2x \), and \( \text{AC} = \text{AF} + \text{CF} \):

\[ \text{AF} = \text{AC} – \text{CF} = 2x – \frac{3}{4} x = \frac{8}{4} x – \frac{3}{4} x = \frac{5}{4} x \]

In \( \triangle ABF \), apply Pythagoras’ theorem:

\[ AB^2 = \text{AF}^2 + \text{BF}^2 \]

\[ 10^2 = \left( \frac{5}{4} x \right)^2 + \left( \frac{\sqrt{7}}{4} x \right)^2 \]

\[ 100 = \frac{25}{16} x^2 + \frac{7}{16} x^2 = \frac{25 + 7}{16} x^2 = \frac{32}{16} x^2 = 2 x^2 \]

\[ x^2 = 50 \]

Calculate the area of \( \triangle ABC \):

Area = \( \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot \text{AC} \cdot \text{BF} \).

\[ \text{Area} = \frac{1}{2} \cdot 2x \cdot \frac{\sqrt{7}}{4} x = \frac{1}{2} \cdot \frac{2 \sqrt{7}}{4} x^2 = \frac{\sqrt{7}}{4} x^2 \]

Substitute \( x^2 = 50 \):

\[ \text{Area} = \frac{\sqrt{7}}{4} \cdot 50 = \frac{50 \sqrt{7}}{4} = \frac{25 \sqrt{7}}{2} \]

Answer: The area of triangle \( ABC \) is \( \frac{25 \sqrt{7}}{2} \).