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A grain silo is composed of a cylinder and a hemisphere. The diameter is 4.4 meters, and the height of its cylindrical portion is 6.2 meters.



What is the approximate total volume of the silo? Use 3.14 for [tex]\pi[/tex] and round the answer to the nearest tenth of a cubic meter.



A. [tex]37.1 \, m^3[/tex]

B. [tex]71.9 \, m^3[/tex]

C. [tex]116.5 \, m^3[/tex]

D. [tex]130.8 \, m^3[/tex]

Answer :

We begin by noting that the silo is made up of a cylinder with a hemisphere on top. The given diameter is 4.4 meters, so the radius is

$$
r = \frac{4.4}{2} = 2.2 \text{ m}.
$$

The height of the cylindrical portion is 6.2 meters. With the given value $\pi = 3.14$, we first calculate the volume of the cylinder using

$$
V_{\text{cyl}} = \pi r^2 h.
$$

Substituting the known values,

$$
V_{\text{cyl}} = 3.14 \times (2.2)^2 \times 6.2 \approx 94.225 \text{ m}^3.
$$

Next, we calculate the volume of the hemispherical portion. The formula for the volume of a hemisphere is

$$
V_{\text{hemisphere}} = \frac{2}{3}\pi r^3.
$$

Substituting the value for $r$,

$$
V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.2)^3 \approx 22.289 \text{ m}^3.
$$

Now, the total volume of the silo is the sum of the cylinder and hemisphere volumes:

$$
V_{\text{total}} = V_{\text{cyl}} + V_{\text{hemisphere}} \approx 94.225 + 22.289 \approx 116.514 \text{ m}^3.
$$

Rounding to the nearest tenth, we obtain

$$
V_{\text{total}} \approx 116.5 \text{ m}^3.
$$

Thus, the approximate total volume of the silo is $\boxed{116.5 \text{ m}^3}$.

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