How to find the volume of a pyramid. The volume of the triangular pyramid. Formulas and an example of solving the problem. triangular base
Theorem.
The volume of the pyramid is equal to one third of the product of the base area by the height.
Proof:
First, we prove the theorem for triangular pyramid, then for arbitrary.1. Consider a triangular pyramidOABSwith volume V, base areaS and height h... Let's draw an axis oh (OM2- height), consider the sectionA1 B1 C1pyramid plane perpendicular to the axisOhand, therefore, parallel to the plane of the base. Let us denote byNS point abscissa M1 the intersection of this plane with the oh axis, and throughS (x)- cross-sectional area. Let us express S (x) across S, h and NS... Note that triangles A1 IN1 WITH1 and ABC are similar. Indeed, A1 IN1 II AB, therefore the triangle OA 1 IN 1 is similar to the triangle OAB. WITH therefore BUT1 IN1 : BUTB = OA 1: OA .
Rectangular triangles OA 1 IN 1 and OAV are also similar (they have a common acute angle with the vertex O). Therefore, OA 1: OA = O 1 M1 : ОМ = x: h. Thus BUT 1 IN 1 : A B = x: h.It can be proved similarly thatB1 C1:Sun = NS: h and A1 C1:AC = NS: h.So the triangleA1 B1 C1 and ABCare similar with the coefficient of similarity NS: h.Therefore, S (x): S = (x: h)², or S (x) = S x ² / h².
Let us now apply the basic formula for calculating the volumes of bodies ata= 0, b =h we get
Thus, the volume of the original pyramid is 1 / 3Sh... The theorem is proved.
Corollary:
The volume V of the truncated pyramid, the height of which is h, and the base areas are equal to S and S1 , are calculated by the formula
h - height of the pyramid
S top. - area of the upper base
S bottom - area of the lower base
A pyramid is a polyhedron with a polygon at its base. All faces, in turn, form triangles that converge at one vertex. Pyramids are triangular, quadrangular, and so on. In order to determine which pyramid is in front of you, it is enough to count the number of corners at its base. The definition of "pyramid height" is very common in geometry problems in school curriculum... In the article we will try to consider different ways of finding it.
Parts of the pyramid
Each pyramid consists of the following elements:
- side faces, which have three corners and converge at the top;
- apothem is the height that descends from its top;
- the top of the pyramid is a point that connects the side edges, but does not lie in the plane of the base;
- base is a polygon that does not have a vertex;
- the height of the pyramid is a segment that crosses the top of the pyramid and forms a right angle with its base.
How to find the height of a pyramid if its volume is known
Through the formula V = (S * h) / 3 (in the formula V is the volume, S is the area of the base, h is the height of the pyramid), we find that h = (3 * V) / S. To consolidate the material, let's solve the problem right away. The triangular base is 50 cm 2, while its volume is 125 cm 3. The height of the triangular pyramid is unknown, which we need to find. Everything is simple here: we insert data into our formula. We get h = (3 * 125) / 50 = 7.5 cm.
How to find the height of a pyramid if you know the length of the diagonal and its edges
As we remember, the height of the pyramid forms a right angle with its base. And this means that the height, edge and half of the diagonal together form. Many, of course, remember the Pythagorean theorem. Knowing two measurements, it will not be difficult to find the third quantity. Recall the well-known theorem a² = b² + c², where a is the hypotenuse, and in our case the edge of the pyramid; b - the first leg or half of the diagonal and c - respectively, the second leg, or the height of the pyramid. From this formula, c² = a² - b².
Now the problem: in a regular pyramid, the diagonal is 20 cm, while the length of the rib is 30 cm. It is necessary to find the height. We solve: c² = 30² - 20² = 900-400 = 500. Hence c = √ 500 = about 22.4.
How to find the height of a truncated pyramid
It is a polygon that has a section parallel to its base. The height of a truncated pyramid is a line segment that connects its two bases. The height can be found for a regular pyramid if the lengths of the diagonals of both bases are known, as well as the edge of the pyramid. Let the diagonal of the larger base be d1, while the diagonal of the smaller base is d2, and the edge has length l. To find the height, you can lower the heights from the two upper opposite points of the diagram to its base. We see that we have got two right-angled triangles, it remains to find the lengths of their legs. To do this, subtract the smaller one from the larger diagonal and divide by 2. So we find one leg: a = (d1-d2) / 2. After that, according to the Pythagorean theorem, we only have to find the second leg, which is the height of the pyramid.
Now let's look at the whole thing in practice. We have a task before us. The truncated pyramid has a square at the base, the length of the diagonal of the larger base is 10 cm, while the smaller one is 6 cm, and the edge is 4 cm. It is required to find the height. To begin with, we find one leg: a = (10-6) / 2 = 2 cm. One leg is 2 cm, and the hypotenuse is 4 cm.It turns out that the second leg or height will be 16-4 = 12, that is, h = √12 = about 3.5 cm.
Pyramid is called a polyhedron, the base of which is an arbitrary polygon, and all faces are triangles with a common vertex, which is the apex of the pyramid.
A pyramid is a three-dimensional figure. That is why quite often it is required to find not only its area, but also its volume. The formula for the volume of a pyramid is very simple:
where S is the area of the base and h is the height of the pyramid.
Height a pyramid is called a straight line, lowered from its top to the base at a right angle. Accordingly, in order to find the volume of a pyramid, it is necessary to determine which polygon lies at the base, calculate its area, find out the height of the pyramid and find its volume. Let's consider an example of calculating the volume of a pyramid.
Problem: given a regular quadrangular pyramid. 
The sides of the base are a = 3 cm, all side edges are b = 4 cm. Find the volume of the pyramid.
To begin with, remember that to calculate the volume, you need the height of the pyramid. We can find it by the Pythagorean theorem. To do this, we need the length of the diagonal, or rather, half of it. Then knowing two of the sides right triangle, we can find the height. First, find the diagonal: 
Let's substitute the values into the formula: 
We find the height h using d and edge b: 
Now we will find
Theorem. The volume of the pyramid is equal to the product of the area of its base by a third of its height.
First, we prove this theorem for a triangular pyramid, and then a polygonal one.
1) On the basis of the triangular pyramid SABC (Fig. 102), we construct such a prism SABCDE, in which the height is equal to the height of the pyramid, and one side edge coincides with the edge SB. Let us prove that the volume of the pyramid is one third of the volume of this prism. Let's separate this pyramid from the prism. Then there will be quadrangular pyramid SADEC (shown separately for clarity). Let us draw a cutting plane in it through the vertex S and the diagonal of the base DC. The resulting two triangular pyramids have a common vertex S and equal bases DEC and DAC, lying in the same plane; hence, according to the pyramid lemma proved above, these are equal sizes. Let's compare one of them, namely SDEC, with this pyramid. For the base of the SDEC pyramid, you can take \ (\ Delta \) SDE; then its top will be at point C and the height is equal to the height of this pyramid. Since \ (\ Delta \) SDE = \ (\ Delta \) ABC, then, according to the same lemma, the pyramids SDEC and SABC are equal.
The ABCDES prism is divided by us into three equal pyramids: SABC, SDEC and SDAC. (Obviously, any triangular prism can be subjected to such a division. This is one of the important properties of a triangular prism.) Thus, the sum of the volumes of three pyramids, equal in size to a given one, is the volume of the prism; Consequently,
$$ V_ (SABC) = \ frac (1) (3) V_ (SDEABC) = \ frac (S_ (ABC) \ cdot H) (3) = S_ (ABC) \ frac (H) (3) $$
where H is the height of the pyramid.
2) Through some vertex E (Fig. 103) of the base of the polygonal pyramid SABCDE, draw the diagonals EB and EC.
Then we draw cutting planes through the edge SE and each of these diagonals. Then the polygonal pyramid will split into several triangular ones having a height in common with this pyramid. Denoting the areas of the bases of the triangular pyramids through b 1 , b 2 , b 3 and height through H, we will have:
volume SABCDE = 1/3 b 1 H + 1/3 b 2 H + 1/3 b 3 H = ( b 1 + b 2 + b 3) H / 3 =
= (area ABCDE) H / 3.
Consequence. If V, B and H mean numbers expressing the volume, base area and height of any pyramid in appropriate units, then
Theorem. The volume of the truncated pyramid is equal to the sum of the volumes of three pyramids, which have a height equal to the height of the truncated pyramid, and the bases: one is the lower base of this pyramid, the other is the upper base, and the base area of the third pyramid is equal to the average geometric areas upper and lower bases.
Let the areas of the bases of the truncated pyramid (Fig. 104) be B and b, height H and volume V (a truncated pyramid can be triangular or polygonal - it doesn't matter).
It is required to prove that
V = 1/3 BH + 1/3 b H + 1/3 H √B b= 1/3 H (B + b+ √B b ),
where √B b is the geometric mean between B and b.
To prove on a smaller basis, we place a small pyramid complementing the given truncated pyramid to a complete one. Then the volume of the truncated pyramid V we can consider as the difference of two volumes - the full pyramid and the upper additional one.
Designating the height of the additional pyramid with the letter NS, we will find that
V = 1/3 B (H + NS) - 1 / 3 bx= 1/3 (BH + B x - bx) = 1/3 [BH + (B - b)NS].
To find the height NS we use the theorem from, according to which we can write the equation:
$$ \ frac (B) (b) = \ frac ((H + x) ^ 3) (x ^ 2) $$
To simplify this equation, we extract from both sides its arithmetic Square root:
$$ \ frac (\ sqrt (B)) (\ sqrt (b)) = \ frac (H + x) (x) $$
From this equation (which can be considered as a proportion) we get:
$$ x \ sqrt (B) = H \ sqrt (b) + x \ sqrt (b) $$
$$ (\ sqrt (B) - \ sqrt (b)) x = H \ sqrt (b) $$
and therefore
$$ x = \ frac (H \ sqrt (b)) (\ sqrt (B) - \ sqrt (b)) $$
Substituting this expression into the formula we derived for the volume V, we find:
$$ V = \ frac (1) (3) \ left $$
Since B - b= (√B + √ b) (√B - √ b), then by reducing the fraction by the difference √B - √ b we get:
$$ V = \ frac (1) (3) BH + (\ sqrt (B) + \ sqrt (b)) H \ sqrt (b) = \\ = \ frac (1) (3) (BH + H \ sqrt (Bb) + Hb) = \\ = \ frac (1) (3) H (B + b + \ sqrt (Bb)) $$
that is, we obtain the formula that was required to be proved.
Other materialsThe main characteristic of any geometric shape in space is its volume. In this article, we will consider what a pyramid with a triangle at the base is, and also show how to find the volume of a triangular pyramid - regular full and truncated.
What is this - a triangular pyramid?
Everyone has heard of the ancients Egyptian pyramids, however, they are quadrangular regular, not triangular. Let's explain how to get a triangular pyramid.
Take an arbitrary triangle and connect all its vertices with some one point located outside the plane of this triangle. The formed figure will be called a triangular pyramid. It is shown in the figure below.
As you can see, the figure under consideration is formed by four triangles, which are generally different. Each triangle is a side or face of a pyramid. This pyramid is often called a tetrahedron, that is, a four-sided volumetric figure.
In addition to the sides, the pyramid also has edges (there are 6 of them) and vertices (there are 4 of them).
triangular base
A figure obtained using an arbitrary triangle and a point in space will generally be an irregular inclined pyramid. Now imagine that the original triangle has the same sides, and the point in space is located exactly above its geometric center at a distance h from the plane of the triangle. The pyramid built using this initial data will be correct.
Obviously, the number of edges, sides and vertices for a regular triangular pyramid will be the same as for a pyramid built from an arbitrary triangle.
However, the correct figure has some distinctive features:
- its height, drawn from the top, will exactly intersect the base at the geometric center (the point of intersection of the medians);
- the lateral surface of such a pyramid is formed by three identical triangles, which are isosceles or equilateral.
A regular triangular pyramid is not only a purely theoretical geometric object. Some structures in nature have its form, for example, the crystal lattice of a diamond, where a carbon atom is connected to four of the same atoms by covalent bonds, or a methane molecule, where the tops of a pyramid are formed by hydrogen atoms.
triangular pyramid
You can determine the volume of absolutely any pyramid with an arbitrary n-gon at the base using the following expression:
Here the symbol S o denotes the area of the base, h is the height of the figure drawn to the marked base from the top of the pyramid.
Since the area of an arbitrary triangle is equal to half the product of the length of its side a by the apothem h a dropped to this side, the formula for the volume of a triangular pyramid can be written in the following form:
V = 1/6 × a × h a × h
For general type determining the height is not an easy task. To solve it, the easiest way is to use the formula for the distance between a point (vertex) and a plane (triangular base), represented by a general equation.
For the correct one, it has a specific look. The area of the base (equilateral triangle) for it is equal to:
Substituting it into the general expression for V, we get:
V = √3 / 12 × a 2 × h
A special case is the situation when all sides of a tetrahedron turn out to be the same equilateral triangles. In this case, its volume can be determined only on the basis of knowledge of the parameter of its edge a. The corresponding expression is:
Truncated pyramid
If the upper part containing the vertex is cut off at a regular triangular pyramid, then you get a truncated figure. Unlike the original, it will consist of two equilateral triangular bases and three isosceles trapezoids.
The photo below shows what a regular truncated triangular pyramid made of paper looks like.
To determine the volume of a truncated triangular pyramid, it is necessary to know three of its linear characteristics: each of the sides of the bases and the height of the figure, equal to the distance between the upper and lower bases. The corresponding formula for volume is written as follows:
V = √3 / 12 × h × (A 2 + a 2 + A × a)
Here h is the height of the figure, A and a are the lengths of the sides of the large (lower) and small (upper) equilateral triangles, respectively.
The solution of the problem
To make the information provided in the article clearer for the reader, we will show with an illustrative example how to use some of the written formulas.
Let the volume of the triangular pyramid be 15 cm 3. The figure is known to be correct. The apothem a b of the lateral rib should be found if it is known that the height of the pyramid is 4 cm.
Since the volume and height of the figure are known, you can use the appropriate formula to calculate the length of the side of its base. We have:
V = √3 / 12 × a 2 × h =>
a = 12 × V / (√3 × h) = 12 × 15 / (√3 × 4) = 25.98 cm
a b = √ (h 2 + a 2/12) = √ (16 + 25.98 2/12) = 8.5 cm
The calculated length of the apothem of the figure turned out to be greater than its height, which is true for any type of pyramid.
