Question

How to find the surface area of a regular pyramid?

Answer 1

Explanation:

`l * w + l \sqrt{( (w)/(2)) ^(2) + {h}^(2) } + w * \sqrt{( (l)/(2) )^(2) + {h}^(2) }`

L = length base

w = width base

h = height

Answer 2

`S=(ns^2)/(4)\left(\cot{\left((180^(\circ))/(n)\right)}+√(3)\right)`

Explanation:

A "regular pyramid" is a pyramid whose base is a regular polygon and whose edges are all the same length. Thus each face is an equilateral triangle.

A hexagonal regular pyramid will look like a hexagonal pancake, as the vertical height of it will be zero. A regular pyramid with 7 or more faces cannot exist, because the apexes of the triangular faces cannot meet at a point.

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For an edge length of "s", the area of each triangular face is (√3)/4×s². There are n of those faces, so the lateral area (LA) will be ...

LA = ns²(√3)/4

The area of the regular polygon base will be the product of half its perimeter and the length of its apothem (a). The apothem is ...

a = (s/2)cot(180°/n)

So, the area of the base (BA) is ...

BA = (1/2)(ns)(s/2)cot(180°/n) = ns²cot(180°/n)/4

The total surface area of the regular pyramid is then ...

S = BA + LA

S = (ns²/4)(cot(180°/n) +√3) . . . . for edge length s and n faces (3≤n≤5)