Площа поверхні піраміди та конуса

Quite simply, the surface of a pyramid or a cone is the area of its base plus its lateral surface area.

For a general cone, there is no way to simplify, and this article would end right about now.

However, when people say "pyramid" or "cone" in standard usage, there are a couple of implicit restrictions that will allow us to have better formulas, and these cases are the subject of this discussion.

If the slant height is uniform, we have an extremely powerful tool to calculate the surface area of a pyramid or cone.

Let us take the first case: a right circular cone. Suppose that the radius is $r$ and that the height is $h$.

Call this slant height $s$. However, using the Pythagorean theorem, it is easy to see that $s=\sqrt{r^2+h^2}$, and that this is true regardless of which slant you take. Thus, the slant height is uniform.

However, this is important to note: since the slant height is uniform, every point on the edge of the base is equidistant from the apex. Thus, unraveling the lateral surface gives us a sector of a circle, with radius $s$ and arc length $2\pi r$, which is the circumference of the base.

Thus, we may calculate the lateral surface area: $$L=\pi s^2\cdot \frac{2\pi r}{2\pi s}=\pi rs$$

The base is just a circle, so the base area is simply $\pi r^2$. Thus, the total surface area is as follows: $$A=\pi r^2+\pi rs$$

Now, let us take a pyramid with uniform slant height. Now, the slant will obviously not be the same everywhere. What we mean by slant height instead is the perpendicular distance from the apex to each edge of the base.

For the slant height to be uniform, the base must be a regular polygon, and the pyramid must be right.

Let $n$ be the number of edges of the base, $s$ be the slant height, and $b$ be the side length of each side of the base.

Now, the base area is the area of an $n$-gon with side length $b$, which is as follows: $$B=\frac{1}{4}nb^2\cot(\pi/n)$$ Now, for the lateral surface area, notice that it is composed of $n$ congruent triangles.

Each triangle has base $b$ and height $s$. Thus, the lateral surface area is: $$L=\frac{n}{2}bs$$ Note that this equals one half the product of the slant height and the perimeter. This actually also applies to the case of the right circular cone, since the perimeter of the base is $2\pi r$.

By the way, if $a$ is the apothem of the polygonal base and $h$ is the height of the pyramid, then $s=\sqrt{a^2+h^2}$.

Thus, we have the total surface area: $$A=B+L$$ $$A=\frac{nb}{4}(b\cot(\pi/n)+2s)$$

A similar idea may be employed even if the polygon is not regular; perhaps in a simple quadrilateral, such as a rectangle, or a triangle, or maybe even a hexagon. But beyond a few very simple cases, the surface area becomes outrageous, and one must abandon all attempts at a general formula.

Consider a pyramid, with a square base of side-length $a$ and height $h$ from its tip to its base (with the apex directly above the center of the square). We wish to find its surface area.

Here's what our pyramid looks like:

We call the height of each triangular face the "slant height" of the pyramid (in the case of regular pyramids like the ones we're working with, the slant height will be the same for each face).

We can use our good friend, the Pythagorean theorem, to find the slant height. We know one leg of our triangle is $h$.

Now, apply the Pythagorean theorem to find the slant height, and thus the area of each triangle.

By the Pythagorean theorem, our slant height $s = \sqrt{\frac{a^2}{4}+h^2}$. Thus, the area of each face is $\frac{as}{2}$.

What is the total surface area of the pyramid, in terms of $a,h$?

The total lateral area of the pyramid is $4 \cdot \frac{as}{2} = 2as = 2a\sqrt{\frac{a^2}{4}+h^2} = a\sqrt{a^2+4h^2}$, so our total area is that plus the base, or $\boxed{a\sqrt{a^2+4h^2}+a^2}$.

This is very much not a formula you should memorize -- if you instead remember how we came up with it, you'll be able to find the surface area of any square pyramid with ease.

Quick pyramid/cone surface area formula reference sheet

(It's hidden so you don't ruin the other explanations for yourself -- you should read those first if you haven't already!)

Cone of radius $r$, height $h$: $A = \pi r (r + h)$

Regular pyramid with base of area $b$, perimeter $p$, and slant height $s$: $A = b + \frac{ps}{2}$

Given apothem $a$ and height $h$, $s^2 = a^2 + h^2$.

(Remember that the apothem is the perpendicular from the center of the regular polygon base to one of its sides).