Question

How to find the surface area of a regular pyramid?

Answer 1

Answer:

$S=\dfrac{ns^2}{4}\left(\cot{\left(\dfrac{180^{\circ}}{n}\right)}+\sqrt{3}\right)$

Step-by-step 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)

Answer 2

Step-by-step explanation:

$l \times w + l \sqrt{( \frac{w}{2}) ^{2} + {h}^{2} } + w \times \sqrt{( \frac{l}{2} )^{2} + {h}^{2} }$

L = length base

w = width base

h = height