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].
[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].
