Truncated cone and sphere

A sphere is inscribed in a truncated cone with base diameters D1=10 cm and D2=20 cm, touching both bases and the surface. What is its diameter?

Correct answer:

d =  14.1421 cm

Step-by-step explanation:

$$\begin{gathered} D_1=10\text{cm} \\ {D}_2=20\text{cm} \\ ABCD: \\ a=D_1=10\text{cm} \\ c=D_2=20\text{cm} \\ b=d \\ a+c=b+d=2b \\ b=\frac{a+c}{2}=\frac{10+20}{2}=15\text{cm} \\ h=\sqrt{b^2-(a/2-c/2{)}^2}=\sqrt{15^2-(10/2-20/2{)}^2}=10\sqrt{2}\text{cm}\doteq 14.1421\text{cm} \\ h=2r=d \\ d=h=14.1421=10\sqrt{2}=14.1421\text{cm} \end{gathered}$$

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