+1791 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 $n$?
+121 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.
