What is the area of a regular pyramid? Find the surface area of a regular triangular pyramid
- This is a polyhedral figure, at the base of which lies a polygon, and the remaining faces are represented by triangles with a common vertex.
If the base is a square, then a pyramid is called quadrangular, if the triangle is triangular. The height of the pyramid is drawn from its top perpendicular to the base. Also used to calculate the area apothem is the height of the side face lowered from its vertex.
The formula for the area of the lateral surface of a pyramid is the sum of the areas of its lateral faces, which are equal to each other. However, this method of calculation is used very rarely. Basically, the area of \u200b\u200bthe pyramid is calculated through the perimeter of the base and the apothem:
Consider an example of calculating the area of the lateral surface of a pyramid.
Let a pyramid with base ABCDE and apex F be given. AB =BC =CD =DE =EA =3 cm. Apothem a = 5 cm. Find the area of the lateral surface of the pyramid.
Let's find the perimeter. Since all the faces of the base are equal, then the perimeter of the pentagon will be equal to:
Now you can find side area pyramids: 
Area of a regular triangular pyramid
A regular triangular pyramid consists of a base containing right triangle and three side faces that are equal in area.
The formula for the lateral surface area of a regular triangular pyramid can be calculated in many ways. You can apply the usual formula for calculating through the perimeter and apothem, or you can find the area of \u200b\u200bone face and multiply it by three. Since the face of the pyramid is a triangle, we apply the formula for the area of a triangle. It will require an apothem and the length of the base. Consider an example of calculating the lateral surface area of a regular triangular pyramid.
Given a pyramid with an apothem a = 4 cm and a base face b = 2 cm. Find the area of the lateral surface of the pyramid.
First, find the area of one of the side faces. In this case it will be:
Substitute the values in the formula: 
Since in a regular pyramid all sides are the same, the area of the side surface of the pyramid will be equal to the sum of the areas of the three faces. Respectively: 
The area of the truncated pyramid
Truncated A pyramid is a polyhedron formed by a pyramid and its section parallel to the base.
The formula for the lateral surface area of a truncated pyramid is very simple. The area is equal to the product of half the sum of the perimeters of the bases and the apothem: 
Definition. Side face- this is a triangle in which one angle lies at the top of the pyramid, and the opposite side of it coincides with the side of the base (polygon).
Definition. Side ribs- it common sides side edges. A pyramid has as many edges as there are corners in a polygon.
Definition. pyramid height is a perpendicular dropped from the top to the base of the pyramid.
Definition. Apothem- this is the perpendicular of the side face of the pyramid, lowered from the top of the pyramid to the side of the base.
Definition. Diagonal section- this is a section of the pyramid by a plane passing through the top of the pyramid and the diagonal of the base.
Definition. Correct pyramid- This is a pyramid in which the base is a regular polygon, and the height descends to the center of the base.
Volume and surface area of the pyramid
Formula. pyramid volume through base area and height:
pyramid properties
If all side edges are equal, then a circle can be circumscribed around the base of the pyramid, and the center of the base coincides with the center of the circle. Also, the perpendicular dropped from the top passes through the center of the base (circle).
If all side ribs are equal, then they are inclined to the base plane at the same angles.
Side ribs are equal when they form with the plane of the base equal angles or if a circle can be circumscribed around the base of the pyramid.
If the side faces are inclined to the plane of the base at one angle, then a circle can be inscribed in the base of the pyramid, and the top of the pyramid is projected into its center.
If the side faces are inclined to the base plane at one angle, then the apothems of the side faces are equal.
Properties of a regular pyramid
1. The top of the pyramid is equidistant from all corners of the base.
2. All side edges are equal.
3. All side ribs are inclined at the same angles to the base.
4. Apothems of all side faces are equal.
5. The areas of all side faces are equal.
6. All faces have the same dihedral (flat) angles.
7. A sphere can be described around the pyramid. The center of the described sphere will be the intersection point of the perpendiculars that pass through the middle of the edges.
8. A sphere can be inscribed in a pyramid. The center of the inscribed sphere will be the intersection point of the bisectors emanating from the angle between the edge and the base.
9. If the center of the inscribed sphere coincides with the center of the circumscribed sphere, then the sum of the flat angles at the apex is equal to π or vice versa, one angle is equal to π / n, where n is the number of angles at the base of the pyramid.
The connection of the pyramid with the sphere
A sphere can be described around the pyramid when at the base of the pyramid lies a polyhedron around which a circle can be described (a necessary and sufficient condition). The center of the sphere will be the point of intersection of planes passing perpendicularly through the midpoints of the side edges of the pyramid.
A sphere can always be described around any triangular or regular pyramid.
A sphere can be inscribed in a pyramid if the bisector planes of the internal dihedral angles of the pyramid intersect at one point (a necessary and sufficient condition). This point will be the center of the sphere.
The connection of the pyramid with the cone
A cone is called inscribed in a pyramid if their vertices coincide and the base of the cone is inscribed in the base of the pyramid.
A cone can be inscribed in a pyramid if the apothems of the pyramid are equal.
A cone is said to be circumscribed around a pyramid if their vertices coincide and the base of the cone is circumscribed around the base of the pyramid.
A cone can be described around a pyramid if all side edges of the pyramid are equal to each other.
Connection of a pyramid with a cylinder
A pyramid is said to be inscribed in a cylinder if the top of the pyramid lies on one base of the cylinder, and the base of the pyramid is inscribed in another base of the cylinder.
A cylinder can be circumscribed around a pyramid if a circle can be circumscribed around the base of the pyramid.
A tetrahedron has four faces and four vertices and six edges, where any two edges have no common vertices but do not touch.
Each vertex consists of three faces and edges that form trihedral angle.
The segment connecting the vertex of the tetrahedron with the center of the opposite face is called median of the tetrahedron(GM).
Bimedian is called a segment connecting the midpoints of opposite edges that do not touch (KL).
All bimedians and medians of a tetrahedron intersect at one point (S). In this case, the bimedians are divided in half, and the medians in a ratio of 3: 1 starting from the top.
Definition. Acute Angled Pyramid is a pyramid in which the apothem is more than half the length of the side of the base.
Definition. obtuse pyramid is a pyramid in which the apothem is less than half the length of the side of the base.
Definition. regular tetrahedron- a tetrahedron with all four faces - equilateral triangles. It is one of five regular polygons. In a regular tetrahedron, all dihedral angles (between faces) and trihedral angles (at a vertex) are equal.
Definition. Rectangular tetrahedron a tetrahedron is called which has a right angle between three edges at the vertex (the edges are perpendicular). Three faces form rectangular trihedral angle and the faces are right triangles, and the base is an arbitrary triangle. The apothem of any face is equal to half the side of the base on which the apothem falls.
Definition. Isohedral tetrahedron A tetrahedron is called in which the side faces are equal to each other, and the base is a regular triangle. The faces of such a tetrahedron are isosceles triangles.
Definition. Orthocentric tetrahedron a tetrahedron is called in which all the heights (perpendiculars) that are lowered from the top to the opposite face intersect at one point.
Definition. star pyramid A polyhedron whose base is a star is called.
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 area of the lateral surface of a regular quadrangular pyramid.
What is a pyramid?
We give a strict geometric definition pyramids. Suppose there is some polygon with n sides and n corners. Let's choose arbitrary point 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. The perpendicular dropped from the top to the base is called the height of the figure. 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 ground is regular polygon, then the whole pyramid is called correct. The figure below shows what regular pyramids look like with triangular, quadrangular, pentagonal, and hexagonal bases.
The surface of the pyramid
Before turning to the question of the area of the lateral surface of a regular quadrangular pyramid, one 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 whole figure is the sum of the areas of each of its sides.
It is convenient to study the surface using the example of a figure unfolding. Sweep for correct quadrangular pyramid 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 lateral surface. Next, we show how to calculate it for a regular quadrangular pyramid.
Lateral surface area of a rectangular regular pyramid
To calculate the lateral surface area of the specified figure, we again turn to the above sweep. 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. From the course of geometry it is known that the area of \u200b\u200bthe triangle S t is equal to the product of the base and the height, which should be divided in half. That is:
Where h b - height 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 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 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 \u003d √ (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 apotema 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.
We will 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 a quadrangular is 108 cm 2 . It is necessary to calculate the value of the length of its apothem h bif the height of the pyramid is 7 cm.
We write the formula for the area S b of the lateral surface through the height. We have:
S b = 2*√(h 2 + a 2 /4) *a.
Here we have simply substituted the corresponding apotema formula into the expression for S b . Let's square both sides of the equation:
S b 2 \u003d 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.
Plug in the known values and solve quadratic equation:
t 2 + 196*t - 11664 = 0.
We have written 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 apotema, just use the formula:
h b \u003d √ (h 2 + a 2 / 4) \u003d √ (7 2 + 6.916 2 / 4) ≈ 7.808 cm.
Lateral surface of the pyramid of Cheops
Let's 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 originally 146.5 meters. Substitute these numbers into the corresponding formula for S b , we get:
S b \u003d 2 * √ (h 2 + a 2 / 4) * a \u003d 2 * √ (146.5 2 + 230.363 2 / 4) * 230.363 ≈ 85860 m 2.
The found value is slightly larger than the area of 17 football fields.
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 a regular pyramid be given, at the base of which lies a regular n-gon with a side equal to a. Let h be the height of the side face, also called apothema pyramids. The area of one side face is 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 as follows:
Lateral surface area of a regular pyramid is equal to the product of its apothem by half the perimeter of the base.
Concerning total surface area, then simply add the area of \u200b\u200bthe base to the side.
Inscribed and circumscribed sphere and ball. 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 the 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 pyramid to be discarded is homothetic to the large pyramid with the center of the homothety at the apex. The similarity coefficient is equal to the ratio of heights: k=h 2 /h 1, or side edges, or other appropriate linear dimensions both pyramids. We know that the areas of similar figures are related as squares of linear dimensions; so the areas of the bases of both pyramids (i.e. spare 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. In the same relation are side surfaces pyramids. There is a similar rule for volumes.
Volumes of similar bodies are related as cubes of their linear dimensions; for example, the volumes of the pyramids are related as the products of their heights by the area of the bases, from which our rule immediately follows. It has a completely general character and directly follows from the fact that the volume always has the dimension of the third power of length. Using this rule, we derive a formula expressing the volume of a truncated pyramid in terms of the height and areas 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 continued until complete pyramid, then the similarity coefficient of the full pyramid and the small pyramid is easy to find as the root of the ratio S 2 /S 1. The height of the truncated pyramid is expressed as h = h 1 - h 2 = h 1 (1 - k). Now we have for the volume of the truncated pyramid (V 1 and V 2 denote the volumes of the full and small pyramids)
truncated pyramid volume formula
We 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 argue in exactly the same way as when deriving the formula for volume. We supplement the pyramid with the upper part, we have P 2 \u003d kP 1, S 2 \u003d 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 horses of the side surfaces of the entire resulting pyramid and its top, respectively. For the lateral surface we find (a 1 and a 2 - apothems of the pyramids, a \u003d a 1 - a 2 \u003d a 1 (1-k))
formula for the lateral surface area of a regular truncated pyramid
When preparing for the exam in mathematics, students have to systematize their knowledge of algebra and geometry. I would like to combine all known information, for example, how to calculate the area of a pyramid. Moreover, starting from the base and side faces to the entire surface area. If the situation is clear with the side faces, since they are triangles, then the base is always different.
What to do when finding 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 incorrect one. In the USE tasks of interest to schoolchildren, there are only tasks with the correct figures at the base. Therefore, we will only talk about them.
right triangle
That is equilateral. One in which all sides are equal and denoted by the letter "a". In this case, the area of \u200b\u200bthe 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 the side again:
Arbitrary regular n-gon
The side of a polygon has the same designation. For the number of corners, the Latin letter n is used.
S = (n * a 2) / (4 * tg (180º/n)).
How to proceed when calculating the lateral and total surface area?
Since the base is a regular figure, all the 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 \u200b\u200bthe pyramid, you 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". General formula for the lateral surface area looks like this:
S \u003d 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 vertex (α) are given. Then it is supposed to use such a formula to calculate the lateral area of \u200b\u200bthe pyramid:
S = n/2 * in 2 sin α .
Task #1
Condition. Find the total area of the pyramid if its base lies with 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 \u003d 3 * 4 \u003d 12 cm. Since the apothem is known, you can immediately calculate the area of \u200b\u200bthe entire lateral surface: ½ * 12 * √3 = 6√3 cm 2.
For a triangle at the base, the following area value will be obtained: (4 2 * √3) / 4 \u003d 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 cm2.
Task #2
Condition. There is a regular quadrangular pyramid. The length of the side of the base is 7 mm, the side edge is 16 mm. You need to know its surface area.
Solution. Since the polyhedron is quadrangular and regular, then its base is a square. Having learned the areas of the base and side faces, it will be possible to calculate the area of \u200b\u200bthe pyramid. The formula for the square is given above. And at 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 this 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 \u003d 267.576 mm 2.
Answer. The desired value is 267.576 mm 2.
Task #3
Condition. For a regular quadrangular pyramid, you need to calculate the area. In it, the side of the square is 6 cm and the height is 4 cm.
Solution. The easiest way is to use the formula with the product of the perimeter and the apothem. The first value is easy to find. The second is a little more difficult.
We'll 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 into its middle.
The desired apothem (hypotenuse right triangle) is equal to √(3 2 + 4 2) = 5 (cm).
Now you can calculate the desired value: ½ * (4 * 6) * 5 + 6 2 \u003d 96 (cm 2).
Answer. 96 cm2.
Task #4
Condition. The correct side of its base is 22 mm, the side ribs are 61 mm. What is the area of the lateral surface of this polyhedron?
Solution. The reasoning in it is the same as described in problem No. 2. Only there was given a pyramid with a square at the base, and now it is a hexagon.
First of all, the area of \u200b\u200bthe base is calculated using the above formula: (6 * 22 2) / (4 * tg (180º / 6)) \u003d 726 / (tg30º) \u003d 726√3 cm 2.
Now you need to find out the semi-perimeter of an isosceles triangle, which is a lateral face. (22 + 61 * 2): 2 = 72 cm. It remains to calculate the area of \u200b\u200beach such triangle using the Heron formula, and then multiply it by six and add it to the one that turned out for the base.
Calculations using the Heron formula: √ (72 * (72-22) * (72-61) 2) \u003d √ 435600 \u003d 660 cm 2. Calculations that will give the lateral surface area: 660 * 6 \u003d 3960 cm 2. It remains to add them up to find out the entire surface: 5217.47≈5217 cm 2.
Answer. Base - 726√3 cm 2, side surface - 3960 cm 2, entire area - 5217 cm 2.
