To find the volume \( V \) of a cone, we use the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the radius of the base and \( h \) is the height of the cone.
In this problem, the radius \( r \) is given as 12 feet, and the height \( h \) is half the radius:
\[ h = \frac{1}{2} \cdot r = \frac{1}{2} \cdot 12 = 6 \text{ feet} \]
Now we can substitute the values of \( r \) and \( h \) into the volume formula:
\[ V = \frac{1}{3} \cdot 3.14 \cdot (12)^2 \cdot 6 \]
Calculating \( (12)^2 \):
\[ (12)^2 = 144 \]
Now we substitute this back into the volume formula:
\[ V = \frac{1}{3} \cdot 3.14 \cdot 144 \cdot 6 \]
Next, we multiply \( 144 \) by \( 6 \):
\[ 144 \cdot 6 = 864 \]
Now we substitute this product back in:
\[ V = \frac{1}{3} \cdot 3.14 \cdot 864 \]
Now calculating \( 3.14 \cdot 864 \):
\[ 3.14 \cdot 864 = 2715.36 \]
Now we divide by \( 3 \):
\[ V = \frac{2715.36}{3} \approx 905.12 \]
Finally, rounding to the nearest tenth, we have:
\[ \text{Volume} \approx 905.1 \text{ cubic feet} \]
Thus, the volume of Erik’s cone is 905.1 cubic feet.