Find the s side of the pyramid. How to calculate the area of a pyramid: base, side and total? The connection between the pyramid and the sphere

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Typical geometric problems on 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 lateral surface area of a regular quadrangular pyramid.

What is a pyramid?

Let us give a strict geometric definition of a pyramid. Suppose we have a polygon with n sides and n angles. Let's 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 with a certain volume, which is called an n-gonal pyramid. For example, let's show in the figure below what a pentagonal pyramid looks like.

The two important elements of any pyramid are its base (n-gon) and its apex. These elements are connected to each other by n triangles, which in general are not equal to each other. The perpendicular descending from the top to the base is called the height of the figure. If it intersects the base at 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 picture below shows what regular pyramids look like with triangular, quadrangular, pentagonal and hexagonal bases.

Surface of the pyramid

Before moving on to the question of the lateral surface area of a regular quadrangular pyramid, we should dwell in more detail 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 entire figure is the sum of the areas of each of its sides.

It is convenient to study a surface using the example of the development of a figure. The development 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 triangles that form the sides of a figure is usually called the lateral surface area. Next we will show how to calculate it for a regular quadrangular pyramid.

Lateral surface area of a quadrangular regular pyramid

To calculate the lateral surface area of the indicated figure, we again turn to the above development. Let's assume that we know the side of the square base. Let's denote it by the 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. From the geometry course we know that the area S t of a triangle is equal to the product of the base and the height, which should be divided in half. That is:

Where h b is the height of an isosceles triangle drawn to the base a. For a pyramid, this height is an apothem. Now it remains to multiply the resulting expression by 4 to obtain the area S b of the lateral surface for the pyramid in question:

S b = 4*S t = 2*h b *a.

This formula contains two parameters: the apothem and the side of the base. If the latter is known in most problem conditions, then the former has to be calculated knowing other quantities. Here are the formulas for calculating the apothem 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 (side of an isosceles triangle) by the symbol L, then the apothem h b is determined by the formula:

h b = √(L 2 - a 2 /4).

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

If the height h of the pyramid is known, then the apothem h b can be calculated as follows:

h b = √(h 2 + a 2 /4).
It is also not difficult to obtain this expression if we consider a right triangle inside the pyramid, 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 area of the lateral surface of the quadrangular is 108 cm 2. It is necessary to calculate the length of its apothem h b if the height of the pyramid is 7 cm.
Let us write the formula for the area S b of the lateral surface in terms of height. We have:

S b = 2*√(h 2 + a 2 /4) *a.

Here we simply substituted the appropriate apothem formula into the expression for S b. Let's square both sides of the equation:

S b 2 = 4*a 2 *h 2 + a 4.

To find the value of a, we make a change of variables:

t 2 + 4*h 2 *t - S b 2 = 0.

Now we substitute the known values and solve the quadratic equation:

t 2 + 196*t - 11664 = 0.

We have written down only the positive root of this equation. Then the sides of the base of the pyramid will be equal to:

a = √t = √47.8355 ≈ 6.916 cm.

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

h b = √(h 2 + a 2 /4) = √(7 2 + 6.916 2 /4) ≈ 7.808 cm.

Side surface of the Cheops pyramid

Let us determine the value of the side 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 initially 146.5 meters. Substitute these numbers into the corresponding formula for S b, we get:

S b = 2*√(h 2 + a 2 /4) *a = 2*√(146.5 2 +230.363 2 /4)*230.363 ≈ 85860 m 2.

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

When preparing for the Unified State Exam in mathematics, students have to systematize their knowledge of algebra and geometry. I would like to combine all known information, for example, on how to calculate the area of a pyramid. Moreover, starting from the base and side edges to the entire surface area. If the situation with the side faces is clear, since they are triangles, then the base is always different.

How to find the area of the base of the pyramid?

It can be absolutely any figure: from an arbitrary triangle to an n-gon. And this base, in addition to the difference in the number of angles, can be a regular figure or an irregular one. In the Unified State Exam tasks that interest schoolchildren, there are only tasks with correct figures at the base. Therefore, we will talk only about them.

Regular triangle

That is, equilateral. The one in which all sides are equal and are designated by 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:

Arbitrary regular n-gon

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

S = (n * a 2) / (4 * tg (180º/n)).

What to do when calculating the lateral and total surface area?

Since the base is a regular figure, all faces of the pyramid are equal. Moreover, each of them is an isosceles triangle, since the side edges are equal. Then, in order to calculate the lateral area of the pyramid, you will need a formula consisting of the 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 the product of the base is multiplied by the height. This height in the pyramid is called apothem. Its designation is “A”. The general formula for lateral surface area is:

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

There are situations when the sides of the base are not known, but the side edges (c) and the flat angle at its apex (α) are given. Then you need to use the following formula to calculate the lateral area of the pyramid:

S = n/2 * in 2 sin α .

Task No. 1

Condition. Find the total area of the pyramid if its base has a side of 4 cm and the apothem has a value of √3 cm.

Solution. You need to start by calculating the perimeter of the base. Since this is a regular triangle, then P = 3*4 = 12 cm. Since the apothem is known, we can immediately calculate the area of the entire lateral surface: ½*12*√3 = 6√3 cm 2.

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

To determine the entire area, you will need to add the two resulting values: 6√3 + 4√3 = 10√3 cm 2.

Answer. 10√3 cm 2.

Problem No. 2

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

Solution. Since the polyhedron is quadrangular and regular, its base is a square. Once you know the area of the base and side faces, you will be able to calculate the area of the pyramid. The formula for the square is given above. And for the side faces, all sides of the triangle are known. Therefore, you can use Heron's formula to calculate their areas.

The first calculations are simple and lead to the following number: 49 mm 2. For the second value, you will need to calculate the semi-perimeter: (7 + 16*2): 2 = 19.5 mm. Now you 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 will need to multiply it by 4.

It turns out: 49 + 4 * 54.644 = 267.576 mm 2.

Answer. The desired value is 267.576 mm 2.

Problem No. 3

Condition. For a regular quadrangular pyramid, you need to calculate the area. The side of the square is known to be 6 cm and the height is 4 cm.

Solution. The easiest way is to use the formula with the product of perimeter and apothem. The first value is easy to find. The second one is a little more complicated.

We will have to remember the Pythagorean theorem and consider It is formed by the height of the pyramid and the apothem, which is the hypotenuse. The second leg is equal to half the side of the square, since the height of the polyhedron falls at its middle.

The required apothem (hypotenuse of a right triangle) is equal to √(3 2 + 4 2) = 5 (cm).

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

Answer. 96 cm 2.

Problem No. 4

Condition. The correct side is given: the sides of its base are 22 mm, the side edges are 61 mm. What is the lateral surface area of this polyhedron?

Solution. The reasoning in it is the same as that described in task No. 2. Only there was given a pyramid with a square at the base, and now it is a hexagon.

First of all, the base area is calculated using the above formula: (6*22 2) / (4*tg (180º/6)) = 726/(tg30º) = 726√3 cm 2.

Now you need to find out the semi-perimeter of an isosceles triangle, which is the side face. (22+61*2):2 = 72 cm. All that remains is to use Heron’s formula to calculate the area of each such triangle, and then multiply it by six and add it to the one obtained for the base.

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

Answer. The base is 726√3 cm 2, the side surface is 3960 cm 2, the entire area is 5217 cm 2.

The area of the lateral surface of an arbitrary pyramid is equal to the sum of the areas of its lateral faces. It makes sense to give a special formula for expressing this area in the case of a regular pyramid. So, let us be given a regular pyramid, at the base of which lies a regular n-gon with side equal to a. Let h be the height of the side face, also called apothem pyramids. The area of one side face is equal to 1/2ah, and the entire side surface of the pyramid has an area equal to n/2ha. Since na is the perimeter of the base of the pyramid, we can write the found formula in the form:

Lateral surface area of a regular pyramid is equal to the product of its apothem and half the perimeter of the base.

Regarding total surface area, then we simply add the area of the base to the side one.

Inscribed and circumscribed sphere and sphere. It should be noted that the center of the sphere inscribed in the pyramid lies at the intersection of the bisector planes of the internal dihedral angles of the pyramid. The center of the sphere described near the pyramid lies at the intersection of planes passing through the midpoints of the edges of the pyramid and perpendicular to them.

Truncated pyramid. If a pyramid is cut by a plane parallel to its base, then the part enclosed between the cutting plane and the base is called truncated pyramid. The figure shows a pyramid; discarding its part lying above the cutting plane, we get a truncated pyramid. It is clear that the small discarded pyramid is homothetic to the large pyramid with the center of homothety at the apex. The similarity coefficient is equal to the ratio of heights: k=h 2 /h 1, or side edges, or other corresponding linear dimensions of both pyramids. We know that the areas of similar figures are related like squares of linear dimensions; so the areas of the bases of both pyramids (i.e. the area of the bases of the truncated pyramid) are related as

Here S 1 is the area of the lower base, and S 2 is the area of the upper base of the truncated pyramid. The lateral surfaces of the pyramids are in the same relation. A similar rule exists for volumes.

Volumes of similar bodies are related like cubes of their linear dimensions; for example, the volumes of pyramids are related as the product of their heights and the area of the bases, from which our rule is immediately obtained. It is of a completely general nature and directly follows from the fact that volume always has a dimension of the third power of length. Using this rule, we derive a formula expressing the volume of a truncated pyramid through the height and area of the bases.

Let a truncated pyramid with height h and base areas S 1 and S 2 be given. If we imagine that it is extended to a full pyramid, then the coefficient of similarity between the full pyramid and the small pyramid is easy to find as the root of the ratio S 2 /S 1 . The height of a truncated pyramid is expressed as h = h 1 - h 2 = h 1 (1 - k). Now we have for the volume of a truncated pyramid (V 1 and V 2 denote the volumes of the full and small pyramids)

formula for the volume of a truncated pyramid

Let us derive the formula for the area S of the lateral surface of a regular truncated pyramid through the perimeters P 1 and P 2 of the bases and the length of the apothem a. We reason in exactly the same way as when deriving the formula for volume. We supplement the pyramid with the upper part, we have P 2 = kP 1, S 2 =k 2 S 1, where k is the similarity coefficient, P 1 and P 2 are the perimeters of the bases, and S 1 and S 2 are the areas of the lateral surfaces of the entire resulting pyramid and its the top part accordingly. For the lateral surface we find (a 1 and a 2 are apothems of the pyramids, a = a 1 - a 2 = a 1 (1-k))

formula for the lateral surface area of a regular truncated pyramid

In this lesson:

Problem 1. Find the total surface area of the pyramid
Problem 2. Find the lateral surface area of a regular triangular pyramid

See also related materials:
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Note . If you need to solve a geometry problem that is not here, write about it in the forum. In problems, instead of the "square root" symbol, the sqrt() function is used, in which sqrt is the square root symbol, and the radical expression is indicated in parentheses. For simple radical expressions, the sign "√" can be used.

Problem 1. Find the total surface area of a regular pyramid

The height of the base of a regular triangular pyramid is 3 cm, and the angle between the side face and the base of the pyramid is 45 degrees.
Find the total surface area of the pyramid

Solution.

At the base of a regular triangular pyramid lies an equilateral triangle.
Therefore, to solve the problem, we will use the properties of a regular triangle:

We know the height of the triangle, from where we can find its area.
h = √3/2a
a = h / (√3/2)
a = 3 / (√3/2)
a = 6 / √3

Whence the area of the base will be equal to:
S = √3/4 a 2
S = √3/4 (6 / √3) 2
S = 3√3

In order to find the area of the side face, we calculate the height KM. According to the problem, the angle OKM is 45 degrees.
Thus:
OK / MK = cos 45
Let's use the table of values of trigonometric functions and substitute the known values.

OK / MK = √2/2

Let's take into account that OK is equal to the radius of the inscribed circle. Then
OK = √3/6a
OK = √3/6 * 6/√3 = 1

Then
OK / MK = √2/2
1/MK = √2/2
MK = 2/√2

The area of the side face is then equal to half the product of the height and the base of the triangle.
Sside = 1/2 (6 / √3) (2/√2) = 6/√6

Thus, the total surface area of the pyramid will be equal to
S = 3√3 + 3 * 6/√6
S = 3√3 + 18/√6

Answer: 3√3 + 18/√6

Problem 2. Find the lateral surface area of a regular pyramid

In a regular triangular pyramid, the height is 10 cm and the side of the base is 16 cm . Find the lateral surface area .

Solution.

Since the base of a regular triangular pyramid is an equilateral triangle, AO is the radius of the circle circumscribed around the base.
(This follows from)

The radius of a circle circumscribed around an equilateral triangle can be found from its properties

Whence the length of the edges of a regular triangular pyramid will be equal to:
AM 2 = MO 2 + AO 2
the height of the pyramid is known by condition (10 cm), AO = 16√3/3
AM 2 = 100 + 256/3
AM = √(556/3)

Each side of the pyramid is an isosceles triangle. Find the area of an isosceles triangle from the first formula presented below

S = 1/2 * 16 sqrt((√(556/3) + 8) (√(556/3) - 8))
S = 8 sqrt((556/3) - 64)
S = 8 sqrt(364/3)
S = 16 sqrt(91/3)

Since all three faces of a regular pyramid are equal, the lateral surface area will be equal to
3S = 48 √(91/3)

Answer: 48 √(91/3)

Problem 3. Find the total surface area of a regular pyramid

The side of a regular triangular pyramid is 3 cm and the angle between the side face and the base of the pyramid is 45 degrees. Find the total surface area of the pyramid.

Solution.
Since the pyramid is regular, there is an equilateral triangle at its base. Therefore the area of the base is

So = 9 * √3/4

In order to find the area of the side face, we calculate the height KM. According to the problem, the angle OKM is 45 degrees.
Thus:
OK / MK = cos 45
Let's take advantage