We need to find the volume of a cone with a diameter of 8 cm and a slant height of 12 cm. The final answer needs to be rounded to the nearest tenth.

1. Problem Description

We need to find the volume of a cone with a diameter of 8 cm and a slant height of 12 cm. The final answer needs to be rounded to the nearest tenth.

2. Solution Steps

First, we need to find the radius, $r$, from the diameter. The radius is half of the diameter.

$r = \frac{diameter}{2} = \frac{8}{2} = 4$ cm

Next, we need to find the height, $h$, of the cone. We can use the Pythagorean theorem, as the radius, height, and slant height form a right triangle. The slant height is the hypotenuse.

$r^2 + h^2 = l^2$

where $l$ is the slant height.

$4^2 + h^2 = 12^2$

$16 + h^2 = 144$

$h^2 = 144 - 16$

$h^2 = 128$

$h = \sqrt{128} = 8\sqrt{2}$ cm

Now, we can calculate the volume of the cone. The formula for the volume of a cone is

$V = \frac{1}{3} \pi r^2 h$

Substitute the values of $r$ and $h$ that we found:

$V = \frac{1}{3} \pi (4^2) (\sqrt{128})$

$V = \frac{1}{3} \pi (16) (8\sqrt{2})$

$V = \frac{128\sqrt{2}}{3} \pi$

$V \approx \frac{128(1.414)}{3} \pi \approx \frac{180.992}{3} \pi \approx 60.330666... \pi$

$V \approx 189.584...$ cm$^3$

Rounding to the nearest tenth:

$V \approx 189.6$ cm$^3$

3. Final Answer

189.6 cm$^3$