1. **Stating the problem:** We need to calculate the ratio $\frac{a}{b}$ where $a$ is the height of the semicircle (which is its radius) and $b$ is the radius of the smaller circle tangent to both the rectangle and the semicircle. 2. **Understanding the figure:** The semicircle has diameter equal to the rectangle's width, so its radius is $a$. The smaller circle is tangent to the rectangle's bottom and left sides and also tangent to the semicircle above. 3. **Set up variables:** Let the rectangle's width be $2a$ (since the semicircle diameter equals the rectangle width). The semicircle center is at $(a,0)$ with radius $a$. 4. **Position of the smaller circle:** Let the smaller circle have radius $b$ and be tangent to the left and bottom sides of the rectangle, so its center is at $(b,b)$. 5. **Tangency condition with semicircle:** The distance between the centers of the semicircle and the smaller circle equals the sum of their radii: $$\sqrt{(a - b)^2 + (0 - b)^2} = a + b$$ 6. **Simplify the equation:** $$\sqrt{(a - b)^2 + b^2} = a + b$$ $$\sqrt{a^2 - 2ab + b^2 + b^2} = a + b$$ $$\sqrt{a^2 - 2ab + 2b^2} = a + b$$ 7. **Square both sides:** $$a^2 - 2ab + 2b^2 = (a + b)^2 = a^2 + 2ab + b^2$$ 8. **Bring all terms to one side:** $$a^2 - 2ab + 2b^2 - a^2 - 2ab - b^2 = 0$$ $$-4ab + b^2 = 0$$ 9. **Solve for $b$ in terms of $a$:** $$b^2 = 4ab$$ $$b = 4a$$ or $$b = 0$$ (discard zero radius) 10. **Check the ratio:** Since $b = 4a$, the ratio is $$\frac{a}{b} = \frac{a}{4a} = \frac{1}{4}$$ **Final answer:** $$\boxed{\frac{a}{b} = \frac{1}{4}}$$