1. **State the problem:** We need to find the area of triangle $\triangle UVW$ where side $UV = 41$ cm, side $VW = 48$ cm, and the angle at vertex $V$ is $64^\circ$. 2. **Formula used:** The area of a triangle when two sides and the included angle are known is given by: $$\text{Area} = \frac{1}{2}ab\sin(C)$$ where $a$ and $b$ are the sides enclosing angle $C$. 3. **Identify sides and angle:** Here, $a = 41$ cm, $b = 48$ cm, and $C = 64^\circ$. 4. **Calculate the area:** $$\text{Area} = \frac{1}{2} \times 41 \times 48 \times \sin(64^\circ)$$ 5. **Calculate intermediate multiplication:** $$\frac{1}{2} \times 41 \times 48 = \frac{1}{2} \times 1968 = \cancel{\frac{1}{2}} \times 1968 = 984$$ 6. **Calculate $\sin(64^\circ)$:** $$\sin(64^\circ) \approx 0.8988$$ 7. **Multiply to find area:** $$\text{Area} = 984 \times 0.8988 \approx 884.3$$ 8. **Final answer:** The area of $\triangle UVW$ is approximately $884.3$ cm$^2$ rounded to the nearest tenth.