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Sagot :
To determine the correct total surface area of the tower, we need to consider the surface areas of both the prism and the pyramid separately, and then combine them appropriately.
### Step-by-Step Solution:
1. Calculate the Surface Area of the Prism:
- The prism has a square base with a length of 20 meters.
- It also has a height of 40 meters.
The surface area of the prism includes:
- The area of the four rectangular sides.
- The area of the bottom square base (the top base is covered by the pyramid).
Surface area of the four rectangular sides:
[tex]\[ 4 \times (\text{base length} \times \text{prism height}) = 4 \times (20 \, \text{m} \times 40 \, \text{m}) = 4 \times 800 \, \text{m}^2 = 3200 \, \text{m}^2 \][/tex]
Area of the bottom square base:
[tex]\[ (\text{base length})^2 = 20 \, \text{m} \times 20 \, \text{m} = 400 \, \text{m}^2 \][/tex]
Total surface area of the prism:
[tex]\[ 3200 \, \text{m}^2 + 400 \, \text{m}^2 = 3600 \, \text{m}^2 \][/tex]
2. Calculate the Surface Area of the Pyramid:
- The pyramid has the same square base size as the prism with a base length of 20 meters.
- Its slant height is [tex]\(10 \sqrt{2}\)[/tex] meters.
The surface area of the pyramid includes:
- The lateral surface area (the area of the four triangular faces).
The lateral area of one triangular face:
[tex]\[ \frac{1}{2} \times (\text{base length}) \times (\text{slant height}) = \frac{1}{2} \times 20 \, \text{m} \times 10 \sqrt{2} \, \text{m} = 100 \sqrt{2} \, \text{m}^2 \][/tex]
Since there are four triangular faces:
[tex]\[ 4 \times 100 \sqrt{2} \, \text{m}^2 = 400 \sqrt{2} \, \text{m}^2 \][/tex]
3. Combine the Surface Areas:
- The total surface area of the tower is the sum of the surface area of the prism and the lateral surface area of the pyramid (excluding the base of the pyramid since it's already counted in the prism's base).
Total surface area:
[tex]\[ 3600 \, \text{m}^2 + 400 \sqrt{2} \, \text{m}^2 \][/tex]
Therefore, the correct expression for the total surface area, including the bottom base, is:
[tex]\[ 400 \sqrt{2} + 3600 \, \text{m}^2 \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{400 \sqrt{2} + 3600 \, \text{m}^2} \][/tex]
### Step-by-Step Solution:
1. Calculate the Surface Area of the Prism:
- The prism has a square base with a length of 20 meters.
- It also has a height of 40 meters.
The surface area of the prism includes:
- The area of the four rectangular sides.
- The area of the bottom square base (the top base is covered by the pyramid).
Surface area of the four rectangular sides:
[tex]\[ 4 \times (\text{base length} \times \text{prism height}) = 4 \times (20 \, \text{m} \times 40 \, \text{m}) = 4 \times 800 \, \text{m}^2 = 3200 \, \text{m}^2 \][/tex]
Area of the bottom square base:
[tex]\[ (\text{base length})^2 = 20 \, \text{m} \times 20 \, \text{m} = 400 \, \text{m}^2 \][/tex]
Total surface area of the prism:
[tex]\[ 3200 \, \text{m}^2 + 400 \, \text{m}^2 = 3600 \, \text{m}^2 \][/tex]
2. Calculate the Surface Area of the Pyramid:
- The pyramid has the same square base size as the prism with a base length of 20 meters.
- Its slant height is [tex]\(10 \sqrt{2}\)[/tex] meters.
The surface area of the pyramid includes:
- The lateral surface area (the area of the four triangular faces).
The lateral area of one triangular face:
[tex]\[ \frac{1}{2} \times (\text{base length}) \times (\text{slant height}) = \frac{1}{2} \times 20 \, \text{m} \times 10 \sqrt{2} \, \text{m} = 100 \sqrt{2} \, \text{m}^2 \][/tex]
Since there are four triangular faces:
[tex]\[ 4 \times 100 \sqrt{2} \, \text{m}^2 = 400 \sqrt{2} \, \text{m}^2 \][/tex]
3. Combine the Surface Areas:
- The total surface area of the tower is the sum of the surface area of the prism and the lateral surface area of the pyramid (excluding the base of the pyramid since it's already counted in the prism's base).
Total surface area:
[tex]\[ 3600 \, \text{m}^2 + 400 \sqrt{2} \, \text{m}^2 \][/tex]
Therefore, the correct expression for the total surface area, including the bottom base, is:
[tex]\[ 400 \sqrt{2} + 3600 \, \text{m}^2 \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{400 \sqrt{2} + 3600 \, \text{m}^2} \][/tex]
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