Let's solve this step by step.
1. Know the formula: The volume \(\( V \)\) of a cone with radius \(\( r \)\) and height \(\( h \)\) is given by:
\(\[ V = \frac{1}{3} \pi r^2 h \]\)
2. Insert the known values: The volume \(\( V \)\) is given as \(\( 20\pi \)\) cubic meters and the height \(\( h \)\) is \(\( 10 \)\) meters. Plug these values into the formula:
\(\[ 20\pi = \frac{1}{3} \pi r^2 \cdot 10 \]\)
3. Simplify and solve for \(\( r^2 \)\):
\(\[ 20\pi = \frac{10}{3} \pi r^2 \]\)
\(\[ 20 = \frac{10}{3} r^2 \]\)
\(\[ r^2 = \frac{20}{\left(\frac{10}{3}\right)} \]\)
\(\[ r^2 = \frac{20 \cdot 3}{10} \]\)
\(\[ r^2 = \frac{60}{10} \]\)
\(\[ r^2 = 6 \]\)
4. Find the radius \(\( r \)\) by taking the square root of \(\( r^2 \)\):
\(\[ r = \sqrt{6} \]\)
5. Calculate the diameter \(\( d \)\), which is twice the radius:
\(\[ d = 2r = 2\sqrt{6} \]\)
6. Finding the numerical value for \(\( \sqrt{6} \)\) and doubling it:
\(\[ \sqrt{6} \approx 2.449 \]\)
\(\[ d \approx 2 \times 2.449 \]\)
\(\[ d \approx 4.898 \]\)
7. Round the diameter to the nearest tenth:
\(\[ d \approx 4.9 \]\)
So the diameter of the cone is approximately \(\( 4.9 \)\) meters when rounded to the nearest tenth.
