1. **State the problem:** We need to find the area of triangle ABC where the right angle is at vertex A. 2. **Identify the sides:** Given that AB is vertical with length $\sqrt{3}$ cm, AC is horizontal with length $\sqrt{2}$ cm, and BC is the hypotenuse. 3. **Formula for area of a right triangle:** $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ 4. **Apply the formula:** Here, base = AC = $\sqrt{2}$ and height = AB = $\sqrt{3}$. 5. **Calculate the area:** $$\text{Area} = \frac{1}{2} \times \sqrt{2} \times \sqrt{3}$$ 6. **Simplify the product inside:** $$\sqrt{2} \times \sqrt{3} = \sqrt{2 \times 3} = \sqrt{6}$$ 7. **Final area:** $$\text{Area} = \frac{1}{2} \times \sqrt{6} = \frac{\sqrt{6}}{2}$$ **Answer:** The area of triangle ABC is $\frac{\sqrt{6}}{2}$ cm$^2$ in simplified surd form.