1. **Problem statement:** A rectangular piece of paper has sides of length $a$ cm and $b$ cm. It is cut in half vertically, creating two smaller rectangles each with dimensions $\frac{a}{2}$ by $b$. The ratio of the longer side to the shorter side in the original rectangle is the same as in each smaller rectangle. We need to find $a$ in terms of $b$. 2. **Set up the ratio for the original rectangle:** The longer side is $a$ (assuming $a > b$), and the shorter side is $b$. So the ratio is: $$\frac{a}{b}$$ 3. **Set up the ratio for the smaller rectangle:** After cutting, each smaller rectangle has sides $b$ and $\frac{a}{2}$. The longer side here is $b$ (since the problem states the ratio is the same, and the ratio compares longer to shorter side, so $b > \frac{a}{2}$). The shorter side is $\frac{a}{2}$. So the ratio is: $$\frac{b}{\frac{a}{2}}$$ 4. **Equate the two ratios:** $$\frac{a}{b} = \frac{b}{\frac{a}{2}}$$ 5. **Simplify the right side:** $$\frac{b}{\frac{a}{2}} = b \times \frac{2}{a} = \frac{2b}{a}$$ 6. **So the equation becomes:** $$\frac{a}{b} = \frac{2b}{a}$$ 7. **Cross multiply:** $$a \times a = b \times 2b$$ $$a^2 = 2b^2$$ 8. **Solve for $a$:** $$a = \sqrt{2b^2} = b\sqrt{2}$$ **Final answer:** $$a = b\sqrt{2}$$