To determine the diameter of a hemisphere with a given volume, we will first need to recall the formula used to calculate the volume of a hemisphere:
The volume \(\( V \)\) of a hemisphere is given by the formula:
\(\[ V = \frac{2}{3} \pi r^3 \]\)
where \(\( r \)\) is the radius of the hemisphere.
Given the volume of the hemisphere is \(\( V = 257 \)\) m³, we can use the volume formula to solve for the radius.
Rearrange the formula to solve for radius \(\( r \)\):
\(\[ r^3 = \frac{3V}{2\pi} \]\)
Substitute the given volume into the formula:
\(\[ r^3 = \frac{3 \times 257}{2\pi} \]\)
Calculate the value of \(\( r^3 \)\) first:
\(\[ r^3 = \frac{771}{2\pi} \]\)
Now, we need to compute the cube root of the result to find the value of the radius \(\( r \)\). Without a calculator, we approximate \(\( \pi \approx 3.14159 \)\).
\(\[ r^3 \approx \frac{771}{2 \times 3.14159} \]\)
\(\[ r^3 \approx \frac{771}{6.28318} \]\)
\(\[ r^3 \approx 122.715 \]\)
Once we have a value for \(\( r^3 \)\), we take the cube root:
\(\[ r \approx \sqrt[3]{122.715} \]\)
\(\[ r \approx 4.976 \]\)
Now remember that the diameter \(\( d \)\) is twice the radius:
\(\[ d = 2r \]\)
\(\[ d \approx 2 \times 4.976 \]\)
\(\[ d \approx 9.952 \]\)
To round to the nearest tenth, we will round \(\( 9.952 \)\) to one decimal place:
\(\[ d \approx 10.0 \]\)
Therefore, the diameter of the hemisphere is approximately 10.0 meters to the nearest tenth of a meter.
