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# truncated cone

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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 $n$?

Aug 23, 2023

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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.

Aug 23, 2023