Pyramid (polygon in circle base) - Volume

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Jul 24, 2020, 6:28:07 PM

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Jul 15, 2014, 4:09:37 PM

{Pyramid}_{(volume)} = \frac{1}{6} \cdot h \cdot n \cdot r^{2} \cdot \sin (\frac{2 π}{n})

$"Pyramid"_"(volume)" = 1/6 * "h" * "n" * "r" ^2 * sin((2pi)/ "n" )$

(n) number of sides

$(n) "number of sides"$

(r) radius

$(r)"radius"$

(h) height

$(h)"height"$

Notes

This formula is based on the standard volume of a pyramid which is $\frac{1}{3}$ $1/3$ * baseArea * height. It is also based on the area formula for a polygon that is inscribed within a circle $\frac{1}{2} \cdot n \cdot r \cdot \sin (2 \frac{π}{n})$ $1/2 * n * r * sin(2 pi/n)$ . The fractions of $\frac{1}{3}$ $1/3$ and $\frac{1}{2}$ $1/2$ have been combined into $\frac{1}{6}$ $1/6$ .

This equation, Pyramid (polygon in circle base) - Volume, references 1 page

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• Polygon - area2 by KurtHeckman

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• Pyramid Calc by KurtHeckman

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Pyramid (polygon in circle base) - Volume

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