1. **State the problem:** We need to find the ratio of the area of a sector with central angle $\alpha$ radians to the area of the circle inscribed in that sector. 2. **Recall formulas:** - Area of a sector of a circle with radius $r$ and central angle $\alpha$ (in radians) is given by: $$\text{Area}_{\text{sector}} = \frac{1}{2} r^2 \alpha$$ - The inscribed circle in the sector is the largest circle tangent to both radii and the arc. Its radius $r_{\text{inscribed}}$ is given by: $$r_{\text{inscribed}} = r \frac{\sin(\frac{\alpha}{2})}{1 + \sin(\frac{\alpha}{2})}$$ 3. **Area of the inscribed circle:** $$\text{Area}_{\text{inscribed}} = \pi r_{\text{inscribed}}^2 = \pi \left(r \frac{\sin(\frac{\alpha}{2})}{1 + \sin(\frac{\alpha}{2})}\right)^2 = \pi r^2 \left(\frac{\sin(\frac{\alpha}{2})}{1 + \sin(\frac{\alpha}{2})}\right)^2$$ 4. **Find the ratio:** $$\text{Ratio} = \frac{\text{Area}_{\text{sector}}}{\text{Area}_{\text{inscribed}}} = \frac{\frac{1}{2} r^2 \alpha}{\pi r^2 \left(\frac{\sin(\frac{\alpha}{2})}{1 + \sin(\frac{\alpha}{2})}\right)^2}$$ 5. **Simplify the ratio:** $$= \frac{\frac{1}{2} \alpha}{\pi \left(\frac{\sin(\frac{\alpha}{2})}{1 + \sin(\frac{\alpha}{2})}\right)^2} = \frac{\frac{1}{2} \alpha}{\pi \frac{\sin^2(\frac{\alpha}{2})}{(1 + \sin(\frac{\alpha}{2}))^2}} = \frac{\frac{1}{2} \alpha (1 + \sin(\frac{\alpha}{2}))^2}{\pi \sin^2(\frac{\alpha}{2})}$$ 6. **Final answer:** $$\boxed{\text{Ratio} = \frac{\alpha (1 + \sin(\frac{\alpha}{2}))^2}{2 \pi \sin^2(\frac{\alpha}{2})}}$$ This ratio expresses how many times larger the sector's area is compared to the inscribed circle's area, depending only on the central angle $\alpha$.