High School

A grain silo is composed of a cylinder and a hemisphere. The diameter is 4.4 meters. 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. 37.1 m³
B. 71.9 m³
C. 116.5 m³
D. 130.8 m³

Answer :

We begin by determining the dimensions of the silo. The silo has a circular base with a diameter of [tex]$4.4$[/tex] meters, so its radius is

[tex]$$
r = \frac{4.4}{2} = 2.2 \text{ meters}.
$$[/tex]

The silo is made of two parts:

1. A cylinder with a height of [tex]$6.2$[/tex] meters.
2. A cone (topped portion) with a height of [tex]$4.4$[/tex] meters.

We use the formula for the volume of a cylinder:

[tex]$$
V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}},
$$[/tex]

By substituting the values, we have

[tex]$$
V_{\text{cylinder}} = 3.14 \times (2.2)^2 \times 6.2.
$$[/tex]

Next, the volume of a cone is given by:

[tex]$$
V_{\text{cone}} = \frac{1}{3} \pi r^2 h_{\text{cone}},
$$[/tex]

So,

[tex]$$
V_{\text{cone}} = \frac{1}{3} \times 3.14 \times (2.2)^2 \times 4.4.
$$[/tex]

After calculating both volumes, we add them to obtain the total volume:

[tex]$$
V_{\text{total}} = V_{\text{cylinder}} + V_{\text{cone}}.
$$[/tex]

Based on the calculations, the cylindrical portion has a volume of approximately [tex]$94.22512$[/tex] cubic meters and the conical portion has a volume of approximately [tex]$22.28981$[/tex] cubic meters. Their sum is

[tex]$$
V_{\text{total}} \approx 94.22512 + 22.28981 \approx 116.51493 \text{ cubic meters}.
$$[/tex]

Finally, rounding this total volume to the nearest [tex]$0.05$[/tex] cubic meter gives

[tex]$$
\textbf{116.5 } m^3.
$$[/tex]

Thus, the approximate total volume of the silo is [tex]$\boxed{116.5\ m^3}$[/tex].