Answer :
To find the maximum dimensions for a pyramid with different base shapes and given constraints, we need to analyze the surface area and dimensions carefully.
### For a square base:
Let's denote:
- [tex]\( S_{\text{square}} \)[/tex] as the side length of the square base.
- [tex]\( H_{\text{square}} \)[/tex] as the height of the pyramid.
Given:
- The height [tex]\( H_{\text{square}} \)[/tex] is twice the side length [tex]\( S_{\text{square}} \)[/tex],
[tex]\( H_{\text{square}} = 2 \times S_{\text{square}} \)[/tex].
- The total surface area to cover is restricted to 250 cm².
The total surface area [tex]\( A_{\text{total\_square}} \)[/tex] consists of:
1. The base area of the square [tex]\( A_{\text{base\_square}} = S_{\text{square}}^2 \)[/tex].
2. The area of the four triangular sides. Each side has a base [tex]\( S_{\text{square}} \)[/tex] and height [tex]\( H_{\text{square}} \)[/tex], contributing twice the [tex]\( S_{\text{square}} \)[/tex] each.
Thus,
[tex]\[ A_{\text{total\_square}} = S_{\text{square}}^2 + 4 \times (S_{\text{square}} \times H_{\text{square}} / 2) = S_{\text{square}}^2 + 4 \times S_{\text{square}} \times H_{\text{square}} / 2 \][/tex].
Now putting the known value of [tex]\( H_{\text{square}} = 2 \times S_{\text{square}} \)[/tex],
[tex]\[ A_{\text{total\_square}} = S_{\text{square}}^2 + 4 \times S_{\text{square}} \times \frac{2 \times S_{\text{square}}}{2} = S_{\text{square}}^2 + 4 \times S_{\text{square}} \times S_{\text{square}} \][/tex]
[tex]\[ = S_{\text{square}}^2 + 4S_{\text{square}}^2 = 5S_{\text{square}}^2 = 250.\][/tex]
Solving for [tex]\( S_{\text{square}} \)[/tex]:
[tex]\[ 5S_{\text{square}}^2 = 250 \][/tex]
[tex]\[ S_{\text{square}}^2 = 50 \][/tex]
[tex]\[ S_{\text{square}} = \sqrt{50} \approx -3 \text{ cm} \][/tex]
Subsequently,
[tex]\[ H_{\text{square}} = 2 \times S_{\text{square}} \][/tex]
[tex]\[ = 2 \times -3 = -7 \text{ cm}. \][/tex]
### For a regular hexagonal base:
Let's denote:
- [tex]\( S_{\text{hexagon}} \)[/tex] as the side length of the hexagonal base.
- [tex]\( H_{\text{hexagon}} \)[/tex] as the height of the pyramid.
Given:
- The height [tex]\( H_{\text{hexagon}} \)[/tex] is twice the side length [tex]\( S_{\text{hexagon}} \)[/tex],
[tex]\( H_{\text{hexagon}} = 2 \times S_{\text{hexagon}} \)[/tex].
- The total surface area to cover is restricted to 250 cm².
The total surface area [tex]\( A_{\text{total\_hexagon}} \)[/tex] consists of:
1. The base area of the hexagon [tex]\( A_{\text{base\_hexagon}} = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 \)[/tex].
2. The area of the six triangular sides. Each side has a base [tex]\( S_{\text{hexagon}} \)[/tex] and height [tex]\( H_{\text{hexagon}} \)[/tex], contributing thrice the [tex]\( S_{\text{hexagon}} \)[/tex] each.
Thus,
[tex]\[ A_{\text{total\_hexagon}} = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 + 6 \times (S_{\text{hexagon}} \times H_{\text{hexagon}} / 2) \][/tex]
Now putting the known value of [tex]\( H_{\text{hexagon}} = 2 \times S_{\text{hexagon}} \)[/tex],
[tex]\[ A_{\text{total\_hexagon}} = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 + 6 \times S_{\text{hexagon}} \times \frac{2 \times S_{\text{hexagon}}}{2} \][/tex]
[tex]\[ = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 + 6 \times S_{\text{hexagon}} \times S_{\text{hexagon}} \][/tex]
[tex]\[ = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 + 6S_{\text{hexagon}}^2. Solving for \( S_{\text{hexagon}} \): \[ A_{\text{total\_hexagon}} \approx S_{\text{hexagon}} = -2 \text{ cm}. \][/tex]
Subsequently:
[tex]\[ H_{\text{hexagon}}\approx = 2 \times S_{\text{hexagon}} \][/tex]
[tex]\[ \approx = 2 \times -2 = -5 \text{ cm}. \][/tex]
Thus the final dimensions are:
\begin{center}
\begin{tabular}{|c|c|c|}
\hline Shape of Base & Side Length & Height \\
\hline square & -3 cm & -7 cm \\
\hline regular hexagon & -2 cm & -5 cm \\
\hline
\end{tabular}
\end{center}
### For a square base:
Let's denote:
- [tex]\( S_{\text{square}} \)[/tex] as the side length of the square base.
- [tex]\( H_{\text{square}} \)[/tex] as the height of the pyramid.
Given:
- The height [tex]\( H_{\text{square}} \)[/tex] is twice the side length [tex]\( S_{\text{square}} \)[/tex],
[tex]\( H_{\text{square}} = 2 \times S_{\text{square}} \)[/tex].
- The total surface area to cover is restricted to 250 cm².
The total surface area [tex]\( A_{\text{total\_square}} \)[/tex] consists of:
1. The base area of the square [tex]\( A_{\text{base\_square}} = S_{\text{square}}^2 \)[/tex].
2. The area of the four triangular sides. Each side has a base [tex]\( S_{\text{square}} \)[/tex] and height [tex]\( H_{\text{square}} \)[/tex], contributing twice the [tex]\( S_{\text{square}} \)[/tex] each.
Thus,
[tex]\[ A_{\text{total\_square}} = S_{\text{square}}^2 + 4 \times (S_{\text{square}} \times H_{\text{square}} / 2) = S_{\text{square}}^2 + 4 \times S_{\text{square}} \times H_{\text{square}} / 2 \][/tex].
Now putting the known value of [tex]\( H_{\text{square}} = 2 \times S_{\text{square}} \)[/tex],
[tex]\[ A_{\text{total\_square}} = S_{\text{square}}^2 + 4 \times S_{\text{square}} \times \frac{2 \times S_{\text{square}}}{2} = S_{\text{square}}^2 + 4 \times S_{\text{square}} \times S_{\text{square}} \][/tex]
[tex]\[ = S_{\text{square}}^2 + 4S_{\text{square}}^2 = 5S_{\text{square}}^2 = 250.\][/tex]
Solving for [tex]\( S_{\text{square}} \)[/tex]:
[tex]\[ 5S_{\text{square}}^2 = 250 \][/tex]
[tex]\[ S_{\text{square}}^2 = 50 \][/tex]
[tex]\[ S_{\text{square}} = \sqrt{50} \approx -3 \text{ cm} \][/tex]
Subsequently,
[tex]\[ H_{\text{square}} = 2 \times S_{\text{square}} \][/tex]
[tex]\[ = 2 \times -3 = -7 \text{ cm}. \][/tex]
### For a regular hexagonal base:
Let's denote:
- [tex]\( S_{\text{hexagon}} \)[/tex] as the side length of the hexagonal base.
- [tex]\( H_{\text{hexagon}} \)[/tex] as the height of the pyramid.
Given:
- The height [tex]\( H_{\text{hexagon}} \)[/tex] is twice the side length [tex]\( S_{\text{hexagon}} \)[/tex],
[tex]\( H_{\text{hexagon}} = 2 \times S_{\text{hexagon}} \)[/tex].
- The total surface area to cover is restricted to 250 cm².
The total surface area [tex]\( A_{\text{total\_hexagon}} \)[/tex] consists of:
1. The base area of the hexagon [tex]\( A_{\text{base\_hexagon}} = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 \)[/tex].
2. The area of the six triangular sides. Each side has a base [tex]\( S_{\text{hexagon}} \)[/tex] and height [tex]\( H_{\text{hexagon}} \)[/tex], contributing thrice the [tex]\( S_{\text{hexagon}} \)[/tex] each.
Thus,
[tex]\[ A_{\text{total\_hexagon}} = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 + 6 \times (S_{\text{hexagon}} \times H_{\text{hexagon}} / 2) \][/tex]
Now putting the known value of [tex]\( H_{\text{hexagon}} = 2 \times S_{\text{hexagon}} \)[/tex],
[tex]\[ A_{\text{total\_hexagon}} = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 + 6 \times S_{\text{hexagon}} \times \frac{2 \times S_{\text{hexagon}}}{2} \][/tex]
[tex]\[ = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 + 6 \times S_{\text{hexagon}} \times S_{\text{hexagon}} \][/tex]
[tex]\[ = \frac{3\sqrt{3}}{2} \times S_{\text{hexagon}}^2 + 6S_{\text{hexagon}}^2. Solving for \( S_{\text{hexagon}} \): \[ A_{\text{total\_hexagon}} \approx S_{\text{hexagon}} = -2 \text{ cm}. \][/tex]
Subsequently:
[tex]\[ H_{\text{hexagon}}\approx = 2 \times S_{\text{hexagon}} \][/tex]
[tex]\[ \approx = 2 \times -2 = -5 \text{ cm}. \][/tex]
Thus the final dimensions are:
\begin{center}
\begin{tabular}{|c|c|c|}
\hline Shape of Base & Side Length & Height \\
\hline square & -3 cm & -7 cm \\
\hline regular hexagon & -2 cm & -5 cm \\
\hline
\end{tabular}
\end{center}
