## 7. Cutting

Draw two concentric semicircles with radii of 2 and 4 cm on a sheet of paper. Cut out the area of paper enclosed between these semicircles and their common diameter, and join the straight-line edges of the resulting paper "half-ring" together. What is the volume of the resulting truncated cone?
7. A truncated cone is bounded by two circles with radii of 1 and 2 cm, respectively (half of 2 cm and 4 cm). The distance between corresponding points of these circles is 2 cm $(4-2=2)$. Using the Pythagorean theorem, we find that the distance between the planes of these circles is $\sqrt{3}$ cm (since $2^{2}-1^{2}=3$).

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-152.jpg?height=703&width=380&top_left_y=1139&top_left_x=835)

The volume of the truncated cone is the difference in the volumes of two complete cones constructed on the given circles as bases and obtained by extending the generators of the truncated cone. Therefore,

$$
\begin{gathered}
V=1 / 3\left(\pi \cdot 2^{2}\right)(2 \sqrt{3})-1 / 3\left(\pi \cdot 1^{2}\right) \sqrt{3}, \\
V=7 \pi / \sqrt{3} \approx 12.697 .
\end{gathered}
$$

${ }^{1}$ Extend the median $A X$ of triangle $A B C$ to a point $X A^{\prime}=A X$ beyond point $X$; note that the quadrilateral $A B A^{\prime} C$ is a parallelogram, the sum of the squares of whose sides is equal to the sum of the squares of its diagonals.-Ed. note.

Thus, the volume of the truncated cone is approximately $12.697 \mathrm{~cm}^{3}$.

(Hopefully, you remember that the volume of a cone is one-third the product of the base area and the height.)