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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]
This is a list of volume formulas of basic shapes: [4]: 405–406 Cone – 1 3 π r 2 h {\textstyle {\frac {1}{3}}\pi r^{2}h} , where r {\textstyle r} is the base 's radius Cube – a 3 {\textstyle a^{3}} , where a {\textstyle a} is the side's length;
The generation of a bicylinder Calculating the volume of a bicylinder. A bicylinder generated by two cylinders with radius r has the volume =, and the surface area [1] [6] =.. The upper half of a bicylinder is the square case of a domical vault, a dome-shaped solid based on any convex polygon whose cross-sections are similar copies of the polygon, and analogous formulas calculating the volume ...
Green line has two intersections. Yellow line lies tangent to the cylinder, so has infinitely many points of intersection. Line-cylinder intersection is the calculation of any points of intersection, given an analytic geometry description of a line and a cylinder in 3d space. An arbitrary line and cylinder may have no intersection at all.
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 ...
A rock mass with a high RMR before the adjustment factors are applied has a high reinforcement potential, and can be reinforced by, for example, rock bolts, whatever the MRMR value might be after excavation. Contrariwise, rock bolts are not a suitable reinforcement for a rock mass with a low RMR (i.e. has a low reinforcement potential).
Reuse the volume formula and unit information given in 41 to calculate the volume of a cylindrical grain silo with a diameter of 10 cubits and a height of 10 cubits. Give the answer in terms of cubic cubits, khar, and hundreds of quadruple heqats, where 400 heqats = 100 quadruple heqats = 1 hundred-quadruple heqat, all as Egyptian fractions.
The problem includes a diagram indicating the dimensions of the truncated pyramid. Several problems compute the volume of cylindrical granaries (41, 42, and 43 of the RMP), while problem 60 RMP seems to concern a pillar or a cone instead of a pyramid. It is rather small and steep, with a seked (slope) of four palms (per cubit). [10]