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Answer :
To find the total surface area of a cylindrical water container (bidón), we use the formula for the total surface area of a cylinder:
[tex]$$
\text{Total Surface Area} = \text{Lateral Surface Area} + \text{Area of the Two Circular Bases}
$$[/tex]
Given that the cylinder has a diameter of 20 cm and a height of 45 cm, we first determine the radius:
[tex]$$
r = \frac{\text{diameter}}{2} = \frac{20\,\text{cm}}{2} = 10\,\text{cm}
$$[/tex]
The lateral surface area is calculated using:
[tex]$$
\text{Lateral Surface Area} = 2\pi r h
$$[/tex]
Substituting the values:
[tex]$$
2\pi r h = 2 \times 3.14 \times 10\,\text{cm} \times 45\,\text{cm} = 2826\,\text{cm}^2
$$[/tex]
Next, the area of one circular base is:
[tex]$$
\text{Base Area} = \pi r^2 = 3.14 \times (10\,\text{cm})^2 = 314\,\text{cm}^2
$$[/tex]
Since there are two bases, the total area of the bases is:
[tex]$$
2 \times 314\,\text{cm}^2 = 628\,\text{cm}^2
$$[/tex]
Now, add the lateral surface area and the areas of the two bases to get the total surface area:
[tex]$$
\text{Total Surface Area} = 2826\,\text{cm}^2 + 628\,\text{cm}^2 = 3454\,\text{cm}^2
$$[/tex]
Thus, the area of the total surface of the water container is:
[tex]$$
\boxed{3454\,\text{cm}^2}
$$[/tex]
This corresponds to option D.
[tex]$$
\text{Total Surface Area} = \text{Lateral Surface Area} + \text{Area of the Two Circular Bases}
$$[/tex]
Given that the cylinder has a diameter of 20 cm and a height of 45 cm, we first determine the radius:
[tex]$$
r = \frac{\text{diameter}}{2} = \frac{20\,\text{cm}}{2} = 10\,\text{cm}
$$[/tex]
The lateral surface area is calculated using:
[tex]$$
\text{Lateral Surface Area} = 2\pi r h
$$[/tex]
Substituting the values:
[tex]$$
2\pi r h = 2 \times 3.14 \times 10\,\text{cm} \times 45\,\text{cm} = 2826\,\text{cm}^2
$$[/tex]
Next, the area of one circular base is:
[tex]$$
\text{Base Area} = \pi r^2 = 3.14 \times (10\,\text{cm})^2 = 314\,\text{cm}^2
$$[/tex]
Since there are two bases, the total area of the bases is:
[tex]$$
2 \times 314\,\text{cm}^2 = 628\,\text{cm}^2
$$[/tex]
Now, add the lateral surface area and the areas of the two bases to get the total surface area:
[tex]$$
\text{Total Surface Area} = 2826\,\text{cm}^2 + 628\,\text{cm}^2 = 3454\,\text{cm}^2
$$[/tex]
Thus, the area of the total surface of the water container is:
[tex]$$
\boxed{3454\,\text{cm}^2}
$$[/tex]
This corresponds to option D.
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