2.60. Determine the lateral surface area and volume of a truncated cone with slant height $l$, circumscribed about a sphere of radius $r$.
2.60. To find the lateral surface area of a truncated cone, we use the formula \( S_{\text {bok }}=\pi\left(r_{1}+r_{2}\right) l \), where \( r_{1} \) and \( r_{2} \) are the radii of the bases of the truncated cone. We draw a plane through the height of the cone. In the cross-section, we get an isosceles trapezoid circumscribed around a circle of radius \( r \), and \( A D=2 r_{2}, B C=2 r_{1} \) (Fig. 2.49). For a circumscribed quadrilateral, we have \( B C+A D=A B+C D \), i.e., \( 2 r_{1}+2 r_{2}=2 l \), from which \( S_{\text {bok }}=\pi l^{2} \).

Volume of the truncated cone

\[
V=\frac{1}{3} \pi H\left(\left(r_{1+}+r_{2}\right)^{2}-r_{1} r_{2}\right), \text { where } H=2 r, r_{1}+r_{2}=l, r_{1} r_{2}=r^{2} \text {. }
\]

The last relation is obtained from the right triangle \( O C D \), in which \( O K^{2}=C K \cdot K D \) ( \(\angle C O D=90^{\circ}\), since \( O C \) and \( O D \) are the angle bisectors of the trapezoid and, therefore, \(\angle O C D+\angle C D O=90^{\circ}\) ). Thus,

\[
V=\frac{2}{3} \pi r\left(l^{2}-r^{2}\right)
\]

\[
\text { Answer: } \pi l^{2} ; \frac{2}{3} \pi r\left(l^{2}-r^{2}\right)
\]

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-487.jpg?height=296&width=467&top_left_y=1319&top_left_x=57)

Fig. 2.49