The area of the lateral surface of a regular quadrangular pyramid: formulas and examples of problems

2024 Author: Angel Austin | [email protected]. Last modified: 2024-01-02 17:39

Typical geometric problems in the plane and in three-dimensional space are the problems of determining the surface areas of different figures. In this article, we present the formula for the area of the lateral surface of a regular quadrangular pyramid.

What is a pyramid?

Let's give a strict geometric definition of a pyramid. Suppose there is some polygon with n sides and n corners. We choose an arbitrary point in space that will not be in the plane of the specified n-gon, and connect it to each vertex of the polygon. We will get a figure that has some volume, which is called an n-gonal pyramid. For example, let's show in the figure below what a pentagonal pyramid looks like.

Two important elements of any pyramid are its base (n-gon) and top. These elements are connected to each other by n triangles, which in general are not equal to each other. Perpendicular dropped fromtop to bottom is called the figure's height. If it intersects the base in the geometric center (coincides with the center of mass of the polygon), then such a pyramid is called a straight line. If, in addition to this condition, the base is a regular polygon, then the entire pyramid is called regular. The figure below shows what regular pyramids look like with triangular, quadrangular, pentagonal and hexagonal bases.

Pyramid surface

Before turning to the question of the area of the lateral surface of a regular quadrangular pyramid, we should dwell on the concept of the surface itself.

As mentioned above and shown in the figures, any pyramid is formed by a set of faces or sides. One side is the base and n sides are triangles. The surface of the whole figure is the sum of the areas of each of its sides.

It is convenient to study the surface on the example of a figure unfolding. A scan for a regular quadrangular pyramid is shown in the figures below.

We see that its surface area is equal to the sum of four areas of identical isosceles triangles and the area of a square.

The total area of all the triangles that form the sides of the figure is called the area of the side surface. Next, we will show how to calculate it for a regular quadrangular pyramid.

The area of the lateral surface of a quadrangular regular pyramid

To calculate the area of the lateralsurface of the specified figure, we again turn to the above scan. Suppose we know the side of the square base. Let's denote it by symbol a. It can be seen that each of the four identical triangles has a base of length a. To calculate their total area, you need to know this value for one triangle. It is known from the geometry course that the area of a triangle S_{t is equal to the product of the base and the height, which should be divided in half. That is:}

S_t=1/2h_ba.

Where h_b is the height of an isosceles triangle drawn to the base a. For a pyramid, this height is the apothem. Now it remains to multiply the resulting expression by 4 to get the area S_{bof the lateral surface for the pyramid in question:}

S_b=4S_t=2h_ba.

This formula contains two parameters: the apothem and the side of the base. If the latter is known in most conditions of the problems, then the former has to be calculated knowing other quantities. Here are the formulas for calculating apotema h_{b for two cases:}

• when the length of the side rib is known;
• when the height of the pyramid is known.

If we denote the length of the lateral edge (the side of an isosceles triangle) with the symbol L, then the apotema h_{b is determined by the formula:}

h_b=√(L² - a^2/4).

This expression is the result of applying the Pythagorean theorem for the lateral surface triangle.

If knownthe height h of the pyramid, then the apotema h_{b can be calculated as follows:}

h_b=√(h² + a^2/4).

Getting this expression is also not difficult if we consider inside the pyramid a right-angled triangle formed by the legs h and a/2 and the hypotenuse h_b.

Let's show how to apply these formulas by solving two interesting problems.

Problem with known surface area

It is known that the lateral surface area of a regular quadrangular pyramid is 108 cm². It is necessary to calculate the value of the length of its apothem h_{b, if the height of the pyramid is 7 cm.}

Let's write the formula for the area S_{bof the lateral surface through the height. We have:}

S_b=2√(h² + a^{2/4) a.}

Here we just substituted the corresponding apotema formula into the expression for S_{b. Let's square both sides of the equation:}

S_b²=4a²h² + a^4.

To find the value of a, let's make a change of variables:

a2=t;

t²+ 4h²t - S_b ^2=0.

We now substitute the known values and solve the quadratic equation:

t2+ 196t - 11664=0.

t ≈ 47, 8355.

We wrote out only the positive root of this equation. Then the sides of the base of the pyramid will be:

a=√t=√47.8355 ≈ 6.916 cm.

To get the length of the apotema,just use the formula:

h_b=√(h² + a²/4)=√(7 ²+ 6, 916^{2/4) ≈ 7, 808 see}

Side surface of Cheops pyramid

Determine the value of the lateral surface area for the largest Egyptian pyramid. It is known that at its base lies a square with a side length of 230.363 meters. The height of the structure was originally 146.5 meters. Substitute these numbers into the corresponding formula for S_{b, we get:}

S_b=2√(h² + a²/4) a=2√(146, 5²+230, 363²/4)230, 363 ≈ 85860 m^2.

The found value is slightly larger than the area of 17 football fields.