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The Egyptians knew the correct formula for the volume of such a truncated square pyramid, but no proof of this equation is given in the Moscow papyrus. The volume of a conical or pyramidal frustum is the volume of the solid before slicing its "apex" off, minus the volume of this "apex":
The volume of a pyramid was recorded back in ancient Egypt, where they calculated the volume of a square frustum, suggesting they acquainted the volume of a square pyramid. [26] The formula of volume for a general pyramid was discovered by Indian mathematician Aryabhata, where he quoted in his Aryabhatiya that the volume of a pyramid is ...
Beyond the discovery of the volume of a square pyramid, the problem of finding the slope and height of a square pyramid can be found in the Rhind Mathematical Papyrus. [10] The Babylonian mathematicians also considered the volume of a frustum, but gave an incorrect formula for it. [11]
The fourteenth problem of the Moscow Mathematical calculates the volume of a frustum. Problem 14 states that a pyramid has been truncated in such a way that the top area is a square of length 2 units, the bottom a square of length 4 units, and the height 6 units, as shown. The volume is found to be 56 cubic units, which is correct. [1]
Area#Area formulas – Size of a two-dimensional surface; Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities
The volume is computed as F times the volume of the pyramid whose base is a regular p-gon and whose height is the inradius r. That is, =. The following table lists the various radii of the Platonic solids together with their surface area and volume.
A triangular-pyramid version of the cannonball problem, which is to yield a perfect square from the N th Tetrahedral number, would have N = 48. That means that the (24 × 2 = ) 48th tetrahedral number equals to (70 2 × 2 2 = 140 2 = ) 19600. This is comparable with the 24th square pyramid having a total of 70 2 cannonballs. [5]
The surface-area-to-volume ratio has physical dimension inverse length (L −1) and is therefore expressed in units of inverse metre (m-1) or its prefixed unit multiples and submultiples. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3. The surface to volume ratio for this cube is thus