Question

a cone has a height of 20 centimeters (cm) and a circular base with a radius of 4 cm. the top 15 cm are cut off leaving a \truncated cone\ that has a circular top. the \truncated cone\ has a volume of \\( \frac{p}{3}\pi \\) cubic centimeters \\( (\mathrm{cm}^{3}) \\), where \\( p \\) is a constant. what is the value of \\( p \\)?

Explanation:

Step1: Find the radius of the smaller cone

Use the property of similar triangles. For two similar cones, the ratio of their radii is equal to the ratio of their heights.
Let the radius of the smaller cone (the part that is cut off) be \(r\). The height of the original cone \(H = 20\) cm and its radius \(R=4\) cm. The height of the smaller cone \(h = 15\) cm.
Since \(\frac{r}{R}=\frac{h}{H}\), substituting the values \(\frac{r}{4}=\frac{15}{20}\), then \(r = 3\) cm.

Step2: Calculate the volume of the original cone and the smaller cone

The volume formula of a cone is \(V=\frac{1}{3}\pi R^{2}H\).
For the original cone with \(R = 4\) cm and \(H = 20\) cm, \(V_{1}=\frac{1}{3}\pi\times4^{2}\times20=\frac{320}{3}\pi\).
For the smaller cone with \(r = 3\) cm and \(h = 15\) cm, \(V_{2}=\frac{1}{3}\pi\times3^{2}\times15 = 45\pi=\frac{135}{3}\pi\).

Step3: Calculate the volume of the truncated cone

The volume of the truncated cone \(V = V_{1}-V_{2}\).
\(V=\frac{320}{3}\pi-\frac{135}{3}\pi=\frac{320 - 135}{3}\pi=\frac{185}{3}\pi\).

Answer:

\(185\)