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  2. Pyramid (geometry) - Wikipedia

    en.wikipedia.org/wiki/Pyramid_(geometry)

    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 ...

  3. Frustum - Wikipedia

    en.wikipedia.org/wiki/Frustum

    The formula for the volume of a pyramidal square frustum was introduced by the ancient Egyptian mathematics in what is called the Moscow Mathematical Papyrus, written in the 13th dynasty (c. 1850 BC): = (+ +), where a and b are the base and top side lengths, and h is the height.

  4. Square pyramid - Wikipedia

    en.wikipedia.org/wiki/Square_pyramid

    In general, the volume of a pyramid is equal to one-third of the area of its base multiplied by its height. [8] Expressed in a formula for a square pyramid, this is: [9] =. Many mathematicians have discovered the formula for calculating the volume of a square pyramid in ancient times.

  5. List of formulas in elementary geometry - Wikipedia

    en.wikipedia.org/wiki/List_of_formulas_in...

    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

  6. Ancient Egyptian mathematics - Wikipedia

    en.wikipedia.org/wiki/Ancient_Egyptian_mathematics

    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 the reciprocal of the slope), while problem 58 gives the length of the base and the height and uses these measurements to compute the seqed.

  7. Moscow Mathematical Papyrus - Wikipedia

    en.wikipedia.org/wiki/Moscow_Mathematical_Papyrus

    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]

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  9. Tetrahedron - Wikipedia

    en.wikipedia.org/wiki/Tetrahedron

    The volume of a tetrahedron can be obtained in many ways. It can be given by using the formula of the pyramid's volume: =. where is the base' area and is the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apices to the opposite faces are inversely proportional to the areas of ...