11.220 A truncated cone is described around a sphere, the area of the lower base of which is a times larger than the area of its upper base. How many times is the volume of the truncated cone larger than the volume of the sphere?
Solution.

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The axial section of the given truncated cone is an isosceles trapezoid \(ABCD\), in which a circle with center \(O\) and radius \(r\) can be inscribed, where \(r\) is the radius of the inscribed sphere. Let the circle \(O\) touch the lateral side \(AD\) of the trapezoid \(ABCD\) at point \(K\).

Let \(AB = 2x\), then \(DC = 2\sqrt{ax}\) (by the condition), \(AK = x\), \(DK = \sqrt{ax}\).

(Theorem 3). \(\angle AOD = 90^\circ\) (Theorem 9) \(\Rightarrow r^2 = OK^2 = AK \cdot DK = \sqrt{a} x^2\)

(Theorem 5). The volume \(V_k\) of the given cone is \(\frac{2r}{3} \left(\pi x^2 + \pi a x^2 + \pi \sqrt{a} x^2\right)\), the volume \(V_u\) of the inscribed sphere is \(\frac{4}{3} \pi r^3 \Rightarrow \frac{V_k}{V_\phi} = \frac{a + \sqrt{a} + 1}{2 \sqrt{a}}\).

Answer: \(\frac{a + \sqrt{a} + 1}{2 \sqrt{a}}\).