How to calculate the area of the pyramid: the base, side and full?

When preparing for the USE in mathematics, students have to systematize knowledge of algebra and geometry. I want to combine all known information, for example, on how to calculate the area of the pyramid. And starting from the base and the side faces to the area of the entire surface. If the situation with the side faces is clear, since they are triangles, the base is always different.

How to be when finding the area of the base of the pyramid?

It can be any kind of figure: from an arbitrary triangle to an n-gon. And this foundation, apart from the difference in the number of angles, can be a correct figure or incorrect. In the school assignments of interest to schoolchildren, only jobs with the correct figures at the base are encountered. Therefore, we will only talk about them.

Right Triangle

That is equilateral. The one with all sides equal and marked with the letter "a". In this case, the area of the base of the pyramid is calculated by the formula:

S = (a ² * √3) / 4.

Square

The formula for calculating its area is the simplest, here "a" is again the side:

S = a ² .

An arbitrary regular n-gon

The side of the polygon has the same notation. For the number of angles, use the Latin letter n.

S = (n * a ² ) / (4 * tg (180º / n)).

What should I do when calculating the area of the lateral and full surface?

Since the base has the right figure, all faces of the pyramid are equal. Moreover, each of them is an isosceles triangle, since the lateral edges are equal. Then, in order to calculate the lateral area of the pyramid, we need a formula consisting of a sum of identical monomials. The number of terms is determined by the number of sides of the base.

The area of an isosceles triangle is calculated by the formula in which half of the product of the base is multiplied by the height. This height in the pyramid is called apophema. Its designation is "A". The general formula for the area of the lateral surface is as follows:

S = ½ P * A, where P is the perimeter of the base of the pyramid.

There are situations where the sides of the base are not known, but there are side edges (c) and a flat angle at its vertex (α). Then it is supposed to use such a formula to calculate the lateral area of the pyramid:

S = n / 2 * in ² sin α .

Task number 1

Condition. Find the total area of the pyramid if it has an equilateral triangle with a side of 4 cm, and the apophema has a value of √3 cm.

Decision. It starts with the calculation of the perimeter of the base. Since this is a regular triangle, then P = 3 * 4 = 12 cm. Since the apopheme is known, we can immediately calculate the area of the entire lateral surface: ½ * 12 * √3 = 6√3 cm ² .

For a triangle at the bottom, we get the following area value: (4 ² * √3) / 4 = 4√3 cm ² .

To determine the total area, it is necessary to add two resulting values: 6√3 + 4√3 = 10√3 cm ² .

Answer. 10√3 cm ² .

Task number 2

Condition . There is a regular quadrangular pyramid. The length of the side of the base is 7 mm, the lateral edge is 16 mm. It is necessary to know the area of its surface.

Decision. Since the polyhedron is quadrangular and regular, there is a square at its base. Having learned the area of the base and side faces, it will be possible to count the area of the pyramid. The formula for the square is given above. And the side faces are known to all sides of the triangle. Therefore, you can use Geron's formula to calculate their areas.

The first calculations are simple and lead to such a number: 49 mm ² . For the second value, you will need to calculate the semiperimeter: (7 + 16 * 2): 2 = 19.5 mm. Now we can calculate the area of an isosceles triangle: √ (19.5 * (19.5-7) * (19.5-16) ² ) = √2985.9375 = 54.644 mm ² . There are only four such triangles, so when calculating the final number, you need to multiply it by 4.

It turns out: 49 + 4 * 54,644 = 267,576 mm ² .

Answer . The sought value is 267.576 mm ² .

Task number 3

Condition . In a regular quadrangular pyramid, it is necessary to calculate the area. It knows the side of the square - 6 cm and the height - 4 cm.

Decision. The simplest way is to use the formula with the product of the perimeter and apophema. The first value is easy to find. The second is a little more complicated.

I'll have to remember the Pythagorean theorem and consider a rectangular triangle. It is formed by the height of the pyramid and apophema, which is the hypotenuse. The second leg is equal to half the side of the square, since the height of the polyhedron falls to its middle.

The desired apopheme (hypotenuse of a right-angled triangle) is √ (3 ² + 4 ² ) = 5 (cm).

Now we can calculate the required quantity: ½ * (4 * 6) * 5 + 6 ² = 96 (cm ² ).

Answer. 96 cm ² .

Task number 4

Condition. Given a regular hexagonal pyramid. The sides of its base are 22 mm, the lateral ribs are 61 mm. What is the area of the lateral surface of this polyhedron?

Decision. The arguments in it are the same as those described in Problem 2. Only there was given a pyramid with a square at the bottom, and now it's a hexagon.

The first step is to calculate the base area according to the above formula: (6 * 22 ² ) / (4 * tg (180º / 6)) = 726 / (tg30º) = 726√3 cm ² .

Now it is necessary to know the half-perimeter of an isosceles triangle, which is a lateral face. (22 + 61 * 2): 2 = 72 cm. It remains according to Heron's formula to calculate the area of each such triangle, and then multiply it by six and add it to the one that turned out for the base.

Calculations using Heron's formula: √ (72 * (72-22) * (72-61) ² ) = √435600 = 660 cm ² . Calculations that will give the area of the lateral surface: 660 * 6 = 3960 cm ² . It remains to add them to find out the entire surface: 5217.47≈5217 cm ² .

Answer. The bases are 726√3 cm ² , the lateral surface is 3960 cm ² , the total area is 5217 cm ² .