Answer :
To determine the maximum dimensions for the pyramid with a given surface area of 250 square centimeters, we need to account for both the area of the base and the lateral surface area. The height [tex]\( h \)[/tex] of the toy is double the side length [tex]\( a \)[/tex]. Let's solve the dimensions step-by-step for both a square base and a hexagonal base.
### 1. Square Base:
#### Step 1: Define Variables and Information
- Let [tex]\( a \)[/tex] be the side length of the square base.
- The height of the pyramid [tex]\( h \)[/tex] is given by [tex]\( h = 2a \)[/tex].
#### Step 2: Calculate the Surface Area
1. Base Area:
[tex]\[ A_{\text{base}} = a^2 \][/tex]
2. Lateral Surface Area:
- Each triangular face: The height of the triangular face from the base to the apex is calculated using the Pythagorean theorem.
[tex]\[ \text{Height of triangular face} = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{(2a)^2 + \left(\frac{a}{2}\right)^2} = \sqrt{4a^2 + \frac{a^2}{4}} = \sqrt{\frac{16a^2 + a^2}{4}} = \sqrt{\frac{17a^2}{4}} = \frac{\sqrt{17}a}{2} \][/tex]
- Area of one triangular face:
[tex]\[ A_{\text{triangle}} = \frac{1}{2} \times a \times \frac{\sqrt{17}a}{2} = \frac{\sqrt{17}a^2}{4} \][/tex]
- Total lateral surface area:
[tex]\[ A_{\text{lateral}} = 4 \times \frac{\sqrt{17}a^2}{4} = \sqrt{17}a^2 \][/tex]
3. Total Surface Area:
[tex]\[ A_{\text{total}} = A_{\text{base}} + A_{\text{lateral}} = a^2 + \sqrt{17}a^2 = a^2(1 + \sqrt{17}) \][/tex]
#### Step 3: Set Up the Equation and Solve for [tex]\( a \)[/tex]
The total surface area is given to be 250 cm²:
[tex]\[ a^2(1 + \sqrt{17}) = 250 \][/tex]
[tex]\[ a^2 = \frac{250}{1 + \sqrt{17}} \][/tex]
Calculating the above:
[tex]\[ a^2 \approx \frac{250}{1 + 4.123} \approx \frac{250}{5.123} \approx 48.8 \implies a \approx \sqrt{48.8} \approx 7 \text{ cm} \quad (\text{rounded}) \][/tex]
[tex]\[ h = 2a \approx 2 \times 7 = 14 \text{ cm} \][/tex]
### 2. Regular Hexagonal Base
#### Step 1: Define Variables and Information
- Let [tex]\( a \)[/tex] be the side length of each hexagon side.
- The height of the pyramid [tex]\( h \)[/tex] is [tex]\( h = 2a \)[/tex].
#### Step 2: Calculate the Surface Area
1. Base Area:
The area of the hexagonal base:
[tex]\[ A_{\text{base}} = \frac{3\sqrt{3}}{2} a^2 \][/tex]
2. Lateral Surface Area:
- Each triangular face:
The height of each triangle is the same as derived before:
[tex]\[ \text{Height of triangular face} = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{\frac{17a^2}{4}} = \frac{\sqrt{17}a}{2} \][/tex]
- Area of one triangular face:
[tex]\[ A_{\text{triangle}} = \frac{1}{2} \times a \times \frac{\sqrt{17}a}{2} = \frac{\sqrt{17}a^2}{4} \][/tex]
- Total lateral surface area:
[tex]\[ A_{\text{lateral}} = 6 \left(\frac{\sqrt{17}a^2}{4}\right) = \frac{6\sqrt{17}a^2}{4} = \frac{3\sqrt{17}a^2}{2} \][/tex]
3. Total Surface Area:
[tex]\[ A_{\text{total}} = A_{\text{base}} + A_{\text{lateral}} = \frac{3\sqrt{3}}{2}a^2 + \frac{3\sqrt{17}}{2}a^2 = \frac{3}{2}a^2(\sqrt{3} + \sqrt{17}) \][/tex]
#### Step 3: Set Up the Equation and Solve for [tex]\( a \)[/tex]
The total surface area is given to be 250 cm²:
[tex]\[ \frac{3}{2}a^2(\sqrt{3} + \sqrt{17}) = 250 \][/tex]
[tex]\[ a^2 = \frac{250 \times 2}{3(\sqrt{3} + \sqrt{17})} = \frac{500}{3(\sqrt{3} + \sqrt{17})} \][/tex]
Calculating the above:
[tex]\[ a^2 \approx \frac{500}{3(1.732 + 4.123)} = \frac{500}{3 \times 5.855} \approx \frac{500}{17.565} \approx 28.5 \implies a \approx \sqrt{28.5} \approx 5 \text{ cm} \quad (\text{rounded}) \][/tex]
[tex]\[ h = 2a \approx 2 \times 5 = 10 \text{ cm} \][/tex]
### Final Answer:
\begin{tabular}{|c|c|c|}
\hline
Shape of Base & Side Length & Height \\
\hline
square & [tex]\(7\)[/tex] & [tex]\(14\)[/tex] \\
\hline
regular hexagon & [tex]\(5\)[/tex] & [tex]\(10\)[/tex] \\
\hline
\end{tabular}
### 1. Square Base:
#### Step 1: Define Variables and Information
- Let [tex]\( a \)[/tex] be the side length of the square base.
- The height of the pyramid [tex]\( h \)[/tex] is given by [tex]\( h = 2a \)[/tex].
#### Step 2: Calculate the Surface Area
1. Base Area:
[tex]\[ A_{\text{base}} = a^2 \][/tex]
2. Lateral Surface Area:
- Each triangular face: The height of the triangular face from the base to the apex is calculated using the Pythagorean theorem.
[tex]\[ \text{Height of triangular face} = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{(2a)^2 + \left(\frac{a}{2}\right)^2} = \sqrt{4a^2 + \frac{a^2}{4}} = \sqrt{\frac{16a^2 + a^2}{4}} = \sqrt{\frac{17a^2}{4}} = \frac{\sqrt{17}a}{2} \][/tex]
- Area of one triangular face:
[tex]\[ A_{\text{triangle}} = \frac{1}{2} \times a \times \frac{\sqrt{17}a}{2} = \frac{\sqrt{17}a^2}{4} \][/tex]
- Total lateral surface area:
[tex]\[ A_{\text{lateral}} = 4 \times \frac{\sqrt{17}a^2}{4} = \sqrt{17}a^2 \][/tex]
3. Total Surface Area:
[tex]\[ A_{\text{total}} = A_{\text{base}} + A_{\text{lateral}} = a^2 + \sqrt{17}a^2 = a^2(1 + \sqrt{17}) \][/tex]
#### Step 3: Set Up the Equation and Solve for [tex]\( a \)[/tex]
The total surface area is given to be 250 cm²:
[tex]\[ a^2(1 + \sqrt{17}) = 250 \][/tex]
[tex]\[ a^2 = \frac{250}{1 + \sqrt{17}} \][/tex]
Calculating the above:
[tex]\[ a^2 \approx \frac{250}{1 + 4.123} \approx \frac{250}{5.123} \approx 48.8 \implies a \approx \sqrt{48.8} \approx 7 \text{ cm} \quad (\text{rounded}) \][/tex]
[tex]\[ h = 2a \approx 2 \times 7 = 14 \text{ cm} \][/tex]
### 2. Regular Hexagonal Base
#### Step 1: Define Variables and Information
- Let [tex]\( a \)[/tex] be the side length of each hexagon side.
- The height of the pyramid [tex]\( h \)[/tex] is [tex]\( h = 2a \)[/tex].
#### Step 2: Calculate the Surface Area
1. Base Area:
The area of the hexagonal base:
[tex]\[ A_{\text{base}} = \frac{3\sqrt{3}}{2} a^2 \][/tex]
2. Lateral Surface Area:
- Each triangular face:
The height of each triangle is the same as derived before:
[tex]\[ \text{Height of triangular face} = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{\frac{17a^2}{4}} = \frac{\sqrt{17}a}{2} \][/tex]
- Area of one triangular face:
[tex]\[ A_{\text{triangle}} = \frac{1}{2} \times a \times \frac{\sqrt{17}a}{2} = \frac{\sqrt{17}a^2}{4} \][/tex]
- Total lateral surface area:
[tex]\[ A_{\text{lateral}} = 6 \left(\frac{\sqrt{17}a^2}{4}\right) = \frac{6\sqrt{17}a^2}{4} = \frac{3\sqrt{17}a^2}{2} \][/tex]
3. Total Surface Area:
[tex]\[ A_{\text{total}} = A_{\text{base}} + A_{\text{lateral}} = \frac{3\sqrt{3}}{2}a^2 + \frac{3\sqrt{17}}{2}a^2 = \frac{3}{2}a^2(\sqrt{3} + \sqrt{17}) \][/tex]
#### Step 3: Set Up the Equation and Solve for [tex]\( a \)[/tex]
The total surface area is given to be 250 cm²:
[tex]\[ \frac{3}{2}a^2(\sqrt{3} + \sqrt{17}) = 250 \][/tex]
[tex]\[ a^2 = \frac{250 \times 2}{3(\sqrt{3} + \sqrt{17})} = \frac{500}{3(\sqrt{3} + \sqrt{17})} \][/tex]
Calculating the above:
[tex]\[ a^2 \approx \frac{500}{3(1.732 + 4.123)} = \frac{500}{3 \times 5.855} \approx \frac{500}{17.565} \approx 28.5 \implies a \approx \sqrt{28.5} \approx 5 \text{ cm} \quad (\text{rounded}) \][/tex]
[tex]\[ h = 2a \approx 2 \times 5 = 10 \text{ cm} \][/tex]
### Final Answer:
\begin{tabular}{|c|c|c|}
\hline
Shape of Base & Side Length & Height \\
\hline
square & [tex]\(7\)[/tex] & [tex]\(14\)[/tex] \\
\hline
regular hexagon & [tex]\(5\)[/tex] & [tex]\(10\)[/tex] \\
\hline
\end{tabular}
