1. **State the problem:** We need to find the area of the minor sector OAB of a circle with radius $40$ cm and central angle $36^\circ$. 2. **Formula for the area of a sector:** The area $A$ of a sector with radius $r$ and central angle $\theta$ (in degrees) is given by: $$ A = \frac{\theta}{360} \times \pi r^2 $$ 3. **Substitute the given values:** Here, $r = 40$ cm and $\theta = 36^\circ$. $$ A = \frac{36}{360} \times \pi \times 40^2 $$ 4. **Simplify the fraction:** $$ \frac{36}{360} = \frac{\cancel{36}}{\cancel{360}} = \frac{1}{10} $$ 5. **Calculate the area:** $$ A = \frac{1}{10} \times \pi \times 1600 = 160 \pi $$ 6. **Final answer:** The area of the minor sector OAB is $$ \boxed{160 \pi \text{ cm}^2} $$