1. **Problem statement:** We need to find the area of a square with side length 5 cm and then find the dimensions of a triangle with the same area and one angle of 40°. 2. **Area of the square:** The formula for the area of a square is $$A = s^2$$ where $s$ is the side length. 3. Calculate the area of the square: $$A = 5^2 = 25 \text{ cm}^2$$ 4. **Triangle area formula:** The area of a triangle can be calculated using two sides and the included angle: $$A = \frac{1}{2}ab\sin(C)$$ where $a$ and $b$ are sides and $C$ is the included angle. 5. We know $A = 25$ and $C = 40^\circ$. We want to find sides $a$ and $b$ such that: $$25 = \frac{1}{2}ab\sin(40^\circ)$$ 6. Rearranging for $ab$: $$ab = \frac{2 \times 25}{\sin(40^\circ)} = \frac{50}{\sin(40^\circ)}$$ 7. Calculate $\sin(40^\circ)$ approximately: $$\sin(40^\circ) \approx 0.6428$$ 8. Substitute: $$ab = \frac{50}{0.6428} \approx 77.8$$ 9. This means the product of the two sides enclosing the 40° angle must be approximately 77.8 cm². 10. For example, if $a = 10$ cm, then $b = \frac{77.8}{10} = 7.78$ cm. **Final answer:** The square has area 25 cm². A triangle with one angle 40° and sides $a$ and $b$ satisfying $ab \approx 77.8$ will have the same area. For instance, $a=10$ cm and $b=7.78$ cm.