Search results
Results from the WOW.Com Content Network
A polyhedron's surface area is the sum of the areas of its faces. The surface area of a right square pyramid can be expressed as = +, where and are the areas of one of its triangles and its base, respectively. The area of a triangle is half of the product of its base and side, with the area of a square being the length of the side squared.
Such a formula would be needed for building pyramids. In the next problem (Problem 57), the height of a pyramid is calculated from the base length and the seked (Egyptian for slope), while problem 58 gives the length of the base and the height and uses these measurements to compute the seked.
A sphere of radius r has surface area 4πr 2.. The surface area (symbol A) of a solid object is a measure of the total area that the surface of the object occupies. [1] The mathematical definition of surface area in the presence of curved surfaces is considerably more involved than the definition of arc length of one-dimensional curves, or of the surface area for polyhedra (i.e., objects with ...
The base regularity of a pyramid's base may be classified based on the type of polygon: one example is the star pyramid in which its base is the regular star polygon. [24] The truncated pyramid is a pyramid cut off by a plane; if the truncation plane is parallel to the base of a pyramid, it is called a frustum.
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 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]
A similar problem appears in the Lahun Mathematical Papyri in London. [13] [14] Circles: Problem 48 of the RMP compares the area of a circle (approximated by an octagon) and its circumscribing square. This problem's result is used in problem 50, where the scribe finds the area of a round field of diameter 9 khet. [8]
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]