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Find the volume of a right circular cone that has a height of 4.1ft and a base with a circumference of 16.2ft. Round your answer to the nearest tenth of a cubic foot. Answer: ft^(3)

Find the volume of a right circular cone that has a height of 4.1ft 4.1 \mathrm{ft} and a base with a circumference of 16.2ft 16.2 \mathrm{ft} . Round your answer to the nearest tenth of a cubic foot.\newlineAnswer: ft3 \mathrm{ft}^{3}

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Q. Find the volume of a right circular cone that has a height of 4.1ft 4.1 \mathrm{ft} and a base with a circumference of 16.2ft 16.2 \mathrm{ft} . Round your answer to the nearest tenth of a cubic foot.\newlineAnswer: ft3 \mathrm{ft}^{3}
  1. Find Radius from Circumference: To find the volume of a cone, we need to know the radius of the base and the height. The formula for the volume of a cone is V=(13)πr2hV = (\frac{1}{3})\pi r^2 h, where rr is the radius and hh is the height. We are given the height (h=4.1h = 4.1 ft) but not the radius. However, we are given the circumference of the base, which we can use to find the radius. The formula for the circumference of a circle is C=2πrC = 2\pi r.
  2. Calculate Radius: First, we need to find the radius rr of the base of the cone using the circumference. We rearrange the circumference formula to solve for rr: r=C2πr = \frac{C}{2\pi}. The circumference CC is 16.216.2 ft, so we plug that into the formula: r=16.2 ft2πr = \frac{16.2 \text{ ft}}{2\pi}.
  3. Calculate Volume with Radius and Height: Now we calculate the radius: r=16.2 ft2π≈16.2 ft6.2832≈2.578 ftr = \frac{16.2 \, \text{ft}}{2\pi} \approx \frac{16.2 \, \text{ft}}{6.2832} \approx 2.578 \, \text{ft}. We will use this radius to calculate the volume of the cone.
  4. Calculate Final Volume: Next, we use the volume formula for a cone with the radius we just found and the given height: V=13πr2hV = \frac{1}{3}\pi r^2 h. Plugging in the values, we get V=13π(2.578 ft)2(4.1 ft)V = \frac{1}{3}\pi(2.578 \, \text{ft})^2(4.1 \, \text{ft}).
  5. Round to Nearest Tenth: We calculate the volume: V=(13)π(2.578 ft)2(4.1 ft)≈(13)π(6.644 ft2)(4.1 ft)≈(13)(20.84 ft3)(π)≈21.89 ft3V = (\frac{1}{3})\pi(2.578 \text{ ft})^2(4.1 \text{ ft}) \approx (\frac{1}{3})\pi(6.644 \text{ ft}^2)(4.1 \text{ ft}) \approx (\frac{1}{3})(20.84 \text{ ft}^3)(\pi) \approx 21.89 \text{ ft}^3. We round this to the nearest tenth of a cubic foot.
  6. Round to Nearest Tenth: We calculate the volume: V=(13)π(2.578 ft)2(4.1 ft)≈(13)π(6.644 ft2)(4.1 ft)≈(13)(20.84 ft3)(π)≈21.89 ft3V = (\frac{1}{3})\pi(2.578 \text{ ft})^2(4.1 \text{ ft}) \approx (\frac{1}{3})\pi(6.644 \text{ ft}^2)(4.1 \text{ ft}) \approx (\frac{1}{3})(20.84 \text{ ft}^3)(\pi) \approx 21.89 \text{ ft}^3. We round this to the nearest tenth of a cubic foot.Rounding to the nearest tenth, the volume of the cone is approximately 21.9 ft321.9 \text{ ft}^3.

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