The truncated right circular cone below has a large base radius $8$ cm and a small base radius of $4$ cm. The height of the truncated cone is $10$ cm. The volume of this solid is $n \pi$ cubic cm, where $n$ is an integer. What is $n$?

tomtom Aug 23, 2023

#1**0 **

The volume of a truncated cone can be calculated using the formula:

\[V = \frac{h}{3}(A_1 + A_2 + \sqrt{A_1 \cdot A_2}),\]

where:

- \(h\) is the height of the truncated cone,

- \(A_1\) and \(A_2\) are the areas of the two circular bases.

In this case, the height \(h = 10\) cm, and the radii of the large and small bases are \(r_1 = 8\) cm and \(r_2 = 4\) cm respectively.

The formula for the area of a circle is \(A = \pi r^2\).

So, the areas of the two circular bases are:

- \(A_1 = \pi (8^2) = 64 \pi\) square cm,

- \(A_2 = \pi (4^2) = 16 \pi\) square cm.

Now, substitute the values into the formula for the volume:

\[V = \frac{10}{3}(64\pi + 16\pi + \sqrt{64\pi \cdot 16\pi}).\]

Simplify inside the square root:

\[\sqrt{64\pi \cdot 16\pi} = 128\pi.\]

Now, substitute this value back into the formula:

\[V = \frac{10}{3}(64\pi + 16\pi + 128\pi) = \frac{10}{3}(208\pi).\]

Simplify:

\[V = \frac{2080\pi}{3}.\]

So, the value of \(n\) is \(2080\), and the volume of the truncated cone is \(2080\pi\) cubic cm.

SpectraSynth Aug 23, 2023