Find the midpoint of \( [AC] \), with \( A(6, 0) \) and \( C(3, 3\sqrt{3}) \).

Midpoint formula: \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

\[ M = \left( \frac{6 + 3}{2}, \frac{0 + 3\sqrt{3}}{2} \right) = \left( \frac{9}{2}, \frac{3\sqrt{3}}{2} \right) \]

Answer: The coordinates of the midpoint of \( [AC] \) are \( \left( \frac{9}{2}, \frac{3\sqrt{3}}{2} \right) \).

Find the equation of the line passing through points \( O(0, 0) \) and \( B \), given \( OABC \) is symmetrical about \( [OB] \).

Since \( OABC \) is symmetrical about \( [OB] \), the line \( OB \) passes through the midpoint \( M \left( \frac{9}{2}, \frac{3\sqrt{3}}{2} \right) \) of \( [AC] \).

Calculate the slope of line \( OB \):

\[ m = \frac{\frac{3\sqrt{3}}{2} – 0}{\frac{9}{2} – 0} = \frac{\frac{3\sqrt{3}}{2}}{\frac{9}{2}} = \frac{3\sqrt{3}}{9} = \frac{\sqrt{3}}{3} \]

The line passes through the origin \( O(0, 0) \), so the equation is:

\[ y = \frac{\sqrt{3}}{3} x \]

Answer: The equation of the line \( OB \) is \( y = \frac{\sqrt{3}}{3} x \).

Given \( [OA] \) is perpendicular to \( [AB] \), find the area of quadrilateral \( OABC \).

Since \( OABC \) is symmetrical about \( [OB] \), the area of quadrilateral \( OABC \) is twice the area of triangle \( OAB \).

Find the coordinates of \( B \). Since \( [OA] \) is perpendicular to \( [AB] \), and \( A(6, 0) \), the slope of \( [OA] \) is:

\[ m_{OA} = \frac{0 – 0}{6 – 0} = 0 \]

The slope of \( [AB] \) must be undefined (vertical line) to be perpendicular to \( [OA] \). Thus, \( B \) has the same x-coordinate as \( A \), so \( x_B = 6 \).

Since \( B \) lies on the line \( OB \), \( y = \frac{\sqrt{3}}{3} x \), substitute \( x = 6 \):

\[ y_B = \frac{\sqrt{3}}{3} \cdot 6 = 2\sqrt{3} \]

So, \( B \) has coordinates \( (6, 2\sqrt{3}) \).

Calculate the area of triangle \( OAB \), with vertices \( O(0, 0) \), \( A(6, 0) \), \( B(6, 2\sqrt{3}) \):

Use the area formula: \( \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} \).

Base \( [OA] \) lies along the x-axis from \( (0, 0) \) to \( (6, 0) \), so base = 6.

Height is the y-coordinate of \( B \), which is \( 2\sqrt{3} \).

\[ \text{Area of } \triangle OAB = \frac{1}{2} \cdot 6 \cdot 2\sqrt{3} = 6\sqrt{3} \]

Area of quadrilateral \( OABC = 2 \cdot \text{Area of } \triangle OAB \):

\[ 2 \cdot 6\sqrt{3} = 12\sqrt{3} \]

Answer: The area of quadrilateral \( OABC \) is \( 12\sqrt{3} \).