2010年2月17日 · If I understand your question correctly then you have a six sided pyramid (with a regular hexagonal base) with side lengths 2cm and corner to top-point length 6cm? If so, a regular hexagon is made up of six regular (e quilateral) triangles, so you know the distance from a corner to the centre of the hexagon is also 2cm. If you take a cross ...

2009年2月25日 · Can anyone please help me with this problem. I don't even have an idea how a hexagonal spherical pyramid looks like. Find the volume of a hexagonal spherical pyramid whose base angles are 135°, 105°, 122°, 110°, 131°, and 142° on a sphere of radius 20in.

2014年5月29日 · The base of a regular hexagonal pyramid is 2 cm. The angle between the base and a side is 45°. I have to find: 1) The lenght of the longest diagonal of the base 2)The lenght of the shortest diagonal of the base 3)The side lenght of the pyramid 4) Altitude lenght 5) Volume Please help me...

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2018年7月26日 · The Question: By using Riemann’s sum, synthesise a mathematical model for finding the exact volume of any ‘tepee’ tent of side s and height h.

2015年12月4日 · How to find the surface area of a pyramid which has a regular hexagonal base of edge 6 cm and a height of ...

2013年8月22日 · Find the total area of the frustum of a regular square pyramid which is inscribed in the frustum of a cone whose upper and lower base diameters are 4ft. and 6ft.,respectively, and whose altitude is 12ft. The answer is 196sq ft. I don't know where to …

2010年11月30日 · In a beehive, each cell is a regular hexagonal prism, open at one end with a trihedral angle at the other end. It is believed that bees form their cells in such a way as to minimize the surface area for a given volume, thus using the least amount of …

2009年7月15日 · So we have a tetrahedral pyramid with side length a. What's the volume of the inscribed sphere that is tangent to the sides of the pyramid? I just need some feedback on a solution, as I have no idea if it is correct. And I should mention that …

2005年11月16日 · Google for "circle packing".A simple solution (most likely suboptimal) consists in placing the right triangle in the first quadrant of the plane (with the right angle at the origin), imagining a square grid with squares of side length =6, and counting how many complete squares are inside the triangle, plus any partial squares that can still contain a circle of radius 3.