truncated cone

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.