Question:

Step 1: Identify given geometric and trigonometric relations. The problem provides several relations derived from its geometric setup and coordinate geometry. These include: The slope of $AG$ is given as $-\frac{h}{2}$. The trigonometric relation for $\tan 30^\circ$ is given as: $$ \tan 30^\circ = \frac{a}{1+\frac{a^2}{h^2}} $$ Additionally, a set of equalities relating variables $a, b, h,$ and $k$ is established: $$ a^2 + b^2 = 9 $$ This is further equated to $-\frac{a}{2}$ and $\frac{3ab}{h+k}$: $$ 9 = -\frac{a}{2} = \frac{3ab}{h+k} $$ These relations collectively describe various properties of the geometric configuration. Step 2: Calculate the area of the right triangle. Using the relations established in Step 1, the solution proceeds to determine the area of the right triangle, which is represented by $\frac{1}{2}ab$. The calculation given is: $$ \frac{1}{2}ab = \frac{3\sqrt{3}}{2\sqrt{3}} $$ The solution then states the final value after simplifying this expression: $$ \frac{1}{2}ab = \sqrt{3} $$ Step 3: State the final answer. Based on the calculation in Step 2, the value of $\frac{1}{2}ab$ is $\sqrt{3}$. This matches option (A). The final answer is $\sqrt{3}$. The final answer is $\boxed{\text{(A) } \sqrt{3}}$.