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Find the volume of a pyramid with a square base, where the perimeter of the base is 
13.1ft and the height of the pyramid is 
6.7ft. Round your answer to the nearest tenth of a cubic foot.
Answer: 
ft^(3)

Find the volume of a pyramid with a square base, where the perimeter of the base is 13.1ft 13.1 \mathrm{ft} and the height of the pyramid is 6.7ft 6.7 \mathrm{ft} . Round your answer to the nearest tenth of a cubic foot.\newlineAnswer: ft3 \mathrm{ft}^{3}

Full solution

Q. Find the volume of a pyramid with a square base, where the perimeter of the base is 13.1ft 13.1 \mathrm{ft} and the height of the pyramid is 6.7ft 6.7 \mathrm{ft} . Round your answer to the nearest tenth of a cubic foot.\newlineAnswer: ft3 \mathrm{ft}^{3}
  1. Find Side Length: First, we need to find the length of one side of the square base. Since the perimeter of a square is the sum of all its sides, and a square has four equal sides, we divide the perimeter by 44. \newlinePerimeter of the base = 13.1ft13.1\,\text{ft} \newlineLength of one side of the base = Perimeter ÷4\div 4 \newlineLength of one side of the base = 13.1ft÷413.1\,\text{ft} \div 4
  2. Calculate Area: Now, let's calculate the length of one side of the base.\newlineLength of one side of the base = 13.1ft÷413.1 \, \text{ft} \div 4\newlineLength of one side of the base = 3.275ft3.275 \, \text{ft}\newlineWe will round this to the nearest tenth later if necessary.
  3. Calculate Volume: Next, we calculate the area of the square base using the length of one side.\newlineArea of the base = (Length of one side)\newlineArea of the base = (3.275ft)2(3.275 \, \text{ft})^2
  4. Round Volume: Let's perform the calculation for the area of the base.\newlineArea of the base = (3.275ft)2(3.275 \, \text{ft})^2\newlineArea of the base = 10.726625ft210.726625 \, \text{ft}^2\newlineWe will round this to the nearest tenth later if necessary.
  5. Round Volume: Let's perform the calculation for the area of the base.\newlineArea of the base = (3.275ft)2(3.275 \, \text{ft})^2\newlineArea of the base = 10.726625ft210.726625 \, \text{ft}^2\newlineWe will round this to the nearest tenth later if necessary.Now we can calculate the volume of the pyramid. The formula for the volume of a pyramid with a square base is:\newlineVolume = (Area of the base×Height)÷3(\text{Area of the base} \times \text{Height}) \div 3\newlineVolume = (10.726625ft2×6.7ft)÷3(10.726625 \, \text{ft}^2 \times 6.7 \, \text{ft}) \div 3
  6. Round Volume: Let's perform the calculation for the area of the base.\newlineArea of the base = (3.275ft)2(3.275 \, \text{ft})^2\newlineArea of the base = 10.726625ft210.726625 \, \text{ft}^2\newlineWe will round this to the nearest tenth later if necessary.Now we can calculate the volume of the pyramid. The formula for the volume of a pyramid with a square base is:\newlineVolume = (Area of the base×Height)÷3(\text{Area of the base} \times \text{Height}) \div 3\newlineVolume = (10.726625ft2×6.7ft)÷3(10.726625 \, \text{ft}^2 \times 6.7 \, \text{ft}) \div 3Let's perform the calculation for the volume of the pyramid.\newlineVolume = (10.726625ft2×6.7ft)÷3(10.726625 \, \text{ft}^2 \times 6.7 \, \text{ft}) \div 3\newlineVolume = 71.8683875ft3÷371.8683875 \, \text{ft}^3 \div 3\newlineVolume = 23.95612917ft323.95612917 \, \text{ft}^3
  7. Round Volume: Let's perform the calculation for the area of the base.\newlineArea of the base = (3.275ft)2(3.275 \, \text{ft})^2\newlineArea of the base = 10.726625ft210.726625 \, \text{ft}^2\newlineWe will round this to the nearest tenth later if necessary.Now we can calculate the volume of the pyramid. The formula for the volume of a pyramid with a square base is:\newlineVolume = Area of the base×Height3\frac{\text{Area of the base} \times \text{Height}}{3}\newlineVolume = 10.726625ft2×6.7ft3\frac{10.726625 \, \text{ft}^2 \times 6.7 \, \text{ft}}{3}Let's perform the calculation for the volume of the pyramid.\newlineVolume = 10.726625ft2×6.7ft3\frac{10.726625 \, \text{ft}^2 \times 6.7 \, \text{ft}}{3}\newlineVolume = 71.8683875ft33\frac{71.8683875 \, \text{ft}^3}{3}\newlineVolume = 23.95612917ft323.95612917 \, \text{ft}^3Finally, we round the volume to the nearest tenth of a cubic foot as requested.\newlineVolume \approx 24.0ft324.0 \, \text{ft}^3

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