Question:

Step 1: Identify the given information and the geometry of the quadrilateral $APBQ$. We are given two points $P$ and $Q$ that are $3$ units apart, meaning $PQ = 3$. A circle centered at $P$ has a radius of $3$ units. This means any point on this circle, such as $A$ or $B$, is $3$ units away from $P$. So, $PA = PB = 3$. A circle centered at $Q$ has a radius of $\sqrt{3}$ units. This means any point on this circle, such as $A$ or $B$, is $\sqrt{3}$ units away from $Q$. So, $QA = QB = \sqrt{3}$. The quadrilateral $APBQ$ is formed by connecting these points. Step 2: Recognize that the quadrilateral $APBQ$ is composed of two congruent triangles, $\triangle APQ$ and $\triangle BPQ$. The quadrilateral $APBQ$ can be viewed as two triangles sharing a common side $PQ$. Consider $\triangle APQ$: Its sides are $PA = 3$, $QA = \sqrt{3}$, and $PQ = 3$. Consider $\triangle BPQ$: Its sides are $PB = 3$, $QB = \sqrt{3}$, and $PQ = 3$. Since $PA = PB = 3$, $QA = QB = \sqrt{3}$, and $PQ$ is common to both, $\triangle APQ$ and $\triangle BPQ$ are congruent by the SSS (Side-Side-Side) criterion. Therefore, the area of quadrilateral $APBQ$ is twice the area of $\triangle APQ$. $$ \text{Area}(APBQ) = 2 \times \text{Area}(\triangle APQ) $$ Step 3: Calculate the area of $\triangle APQ$. The sides of $\triangle APQ$ are $a=3$, $b=\sqrt{3}$, and $c=3$. Since two sides are equal, it is an isosceles triangle. We can use Heron's formula to find its area. First, calculate the semi-perimeter $s$: $$ s = \frac{PA + QA + PQ}{2} = \frac{3 + \sqrt{3} + 3}{2} = \frac{6 + \sqrt{3}}{2} $$ Next, calculate the terms $(s-a)$, $(s-b)$, and $(s-c)$: $$ s - PA = \frac{6 + \sqrt{3}}{2} - 3 = \frac{6 + \sqrt{3} - 6}{2} = \frac{\sqrt{3}}{2} $$ $$ s - QA = \frac{6 + \sqrt{3}}{2} - \sqrt{3} = \frac{6 + \sqrt{3} - 2\sqrt{3}}{2} = \frac{6 - \sqrt{3}}{2} $$ $$ s - PQ = \frac{6 + \sqrt{3}}{2} - 3 = \frac{6 + \sqrt{3} - 6}{2} = \frac{\sqrt{3}}{2} $$ Now, apply Heron's formula for the area of $\triangle APQ$: $$ \text{Area}(\triangle APQ) = \sqrt{s(s-PA)(s-QA)(s-PQ)} $$ $$ \text{Area}(\triangle APQ) = \sqrt{\left(\frac{6 + \sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) \left(\frac{6 - \sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right)} $$ $$ \text{Area}(\triangle APQ) = \sqrt{\frac{(6 + \sqrt{3})(6 - \sqrt{3})(\sqrt{3})(\sqrt{3})}{16}} $$ Using the difference of squares formula $(x+y)(x-y)=x^2-y^2$, we have $(6 + \sqrt{3})(6 - \sqrt{3}) = 6^2 - (\sqrt{3})^2 = 36 - 3 = 33$. Also, $(\sqrt{3})(\sqrt{3}) = 3$. Substitute these values: $$ \text{Area}(\triangle APQ) = \sqrt{\frac{33 \times 3}{16}} = \sqrt{\frac{99}{16}} = \frac{\sqrt{99}}{4} $$ Step 4: Calculate the total area of the quadrilateral $APBQ$. From Step 2, the area of quadrilateral $APBQ$ is twice the area of $\triangle APQ$. $$ \text{Area}(APBQ) = 2 \times \text{Area}(\triangle APQ) $$ $$ \text{Area}(APBQ) = 2 \times \frac{\sqrt{99}}{4} $$ $$ \text{Area}(APBQ) = \frac{\sqrt{99}}{2} $$ Step 5: Conclude the final answer. The area of the quadrilateral $APBQ$ is $\frac{\sqrt{99}}{2}$. The final answer is $\boxed{\frac{\sqrt{99}}{2}}$.