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 "height of the pyramid" is very often found in geometry problems in the school curriculum. In the article we will try to consider different ways to find it.
Parts of the pyramid
Each pyramid consists of the following elements:
- side faces that have three corners and converge at the top;
- apothem represents 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;
- a base is a polygon that does not contain a vertex;
- the height of the pyramid is a segment that intersects 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 \u003d (S * h) / 3 (in the formula V is the volume, S is the base area, h is the height of the pyramid), we find that h \u003d (3 * V) / S. To consolidate the material, let's immediately solve the problem. 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 the data into our formula. We get h \u003d (3 * 125) / 50 \u003d 7.5 cm.
How to find the height of a pyramid if the length of the diagonal and its edge are known
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 dimensions, it will not be difficult to find the third value. 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 edge is 30 cm. You need to find the height. We solve: c² \u003d 30² - 20² \u003d 900-400 \u003d 500. Hence c \u003d √ 500 \u003d 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 the segment that connects its two bases. The height can be found at a regular pyramid if the lengths of the diagonals of both bases, as well as the edge of the pyramid, are known. 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 diagonal from the larger diagonal and divide by 2. So we will find one leg: a \u003d (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 this whole thing in practice. We have a task ahead of us. The truncated pyramid has a square at the base, the diagonal length 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 \u003d (10-6) / 2 \u003d 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 \u003d 12, that is, h \u003d √12 = about 3.5 cm.
Theorem. The volume of a pyramid is equal to the product of the area of its base and a third of its height.
First, we prove this theorem for a triangular pyramid, and then for a polygonal one.
1) Based on the triangular pyramid SABC (Fig. 102), we construct a prism SABCDE whose 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. Separate this pyramid from the prism. This leaves the quadrangular pyramid SADEC (which is shown separately for clarity). Let's 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 lemma proved above, these pyramids are equal. 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 prism ABCDES is divided by us into three equal-sized pyramids: SABC, SDEC and SDAC. (Obviously, any triangular prism can be subjected to such a partition. This is one of the important properties of a triangular prism.) Thus, the sum of the volumes of three pyramids that are equal in size to a given one is the volume of the prism; hence,
$$ V_(SABC) = \frac(1)(3) V_(SDEABC) = \frac(S_(ABC)\cdot H)(3) = S_(ABC)\frac(H)(3) $$
where H is the height of the pyramid.
2) Through some vertex E (Fig. 103) of the base of the polygonal pyramid SABCDE we 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 be divided into several triangular ones having a height common with the given pyramid. Denoting the areas of the bases of triangular pyramids through b 1 ,b 2 ,b 3 and height through H, we will have:
volume SABCDE = 1 / 3 b 1H+1/3 b 2H+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 in appropriate units the volume, base area and height of any pyramid, then
Theorem. The volume of a truncated pyramid is equal to the sum of the volumes of three pyramids that have the same height as the height of the truncated pyramid, and 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 geometric mean of the areas of the 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= 1/3H(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 that complements this truncated pyramid to a complete one. Then we can consider the volume of the truncated pyramid V as the difference of two volumes - the full pyramid and the upper additional one.
Denoting the height of the additional pyramid with the letter X, we will find that
V = 1/3 B (H + X) - 1 / 3 bx= 1 / 3 (BH + B x - bx) = 1 / 3 [ВH + (В - b)X].
To find the height X 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 its arithmetic square root from both sides:
$$ \frac(\sqrt(B))(\sqrt(b)) = \frac(H + x)(x) $$
From this equation (which can be thought of as a proportion) we get:
$$ x\sqrt(B) = H\sqrt(b) + x\sqrt(b) $$
$$ (\sqrt(B) - \sqrt(b))x = H\sqrt(b) $$
and hence
$$ x = \frac(H\sqrt(b))(\sqrt(B) - \sqrt(b)) $$
Substituting this expression into the formula we derived for the volume V, we find:
$$ V = \frac(1)(3)\left $$
Since V- b= (√B + √ b) (√B - √ b), then by reducing the fraction by the difference √B - √ b we get:
$$ V = \frac(1)(3) BH +(\sqrt(B) + \sqrt(b))H\sqrt(b) =\\= \frac(1)(3)(BH+H\ sqrt(Bb)+Hb) =\\= \frac(1)(3)H(B+b+\sqrt(Bb)) $$
i.e., we obtain the formula that was required to be proved.
Other materialsPyramid called a polyhedron whose base is an arbitrary polygon, and all faces are triangles with a common vertex, which is the top of the pyramid.
The 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 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 the 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. Consider an example of calculating the volume of a pyramid.
Task: given a regular quadrangular pyramid. 
Base sides a = 3 cm, all side edges b = 4 cm. Find the volume of the pyramid.
First, remember that to calculate the volume, you need the height of the pyramid. We can find it using the Pythagorean theorem. To do this, we need the length of the diagonal, or rather, half of it. Then knowing two of the sides of a right triangle, we can find the height. First, find the diagonal: 
Substitute the values in the formula: 
We find the height h using d and edge b : 
Now let's find
The main characteristic of any geometric figure 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 a triangular pyramid?
Everyone has heard of the ancient Egyptian pyramids, yet they are quadrangular regular, not triangular. Let's explain how to get a triangular pyramid.
Let's take an arbitrary triangle and connect all its vertices with some one point located outside the plane of this triangle. The resulting 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 in the general case are different. Each triangle is the sides of the pyramid or its face. This pyramid is often called a tetrahedron, that is, a four-sided three-dimensional figure.
In addition to the sides, the pyramid also has edges (there are 6 of them) and vertices (there are 4 of them).
with triangular base
The figure, which is obtained using an arbitrary triangle and a point in space, will be an irregular inclined pyramid in the general case. Now imagine that the original triangle has the same sides, and a point in space is located exactly above its geometric center at a distance h from the plane of the triangle. The pyramid built using these initial data will be correct.
Obviously, the number of edges, sides and vertices of a regular triangular pyramid will be the same as that of 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 in the geometric center (the point of intersection of the medians);
- the side surface of such a pyramid is formed by three identical triangles that are isosceles or equilateral.
The regular triangular pyramid is not only a purely theoretical geometric object. Some structures in nature have its shape, such as the crystal lattice of diamond, where a carbon atom is connected to four of the same atoms by covalent bonds, or a methane molecule, where the tops of the 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 and the apothem h a lowered 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 a generic 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 \u200b\u200bthe base (an equilateral triangle) for it is equal to:
We substitute 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 identical equilateral triangles. In this case, its volume can be determined only on the basis of knowing the parameter of its edge a. The corresponding expression looks like:
Truncated pyramid
If the upper part containing the vertex is cut off from a regular triangular pyramid, then a truncated figure will be obtained. Unlike the original one, 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 its three 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 in the article clearer for the reader, we will show with a clear example how to use some of the written formulas.
Let the volume of a triangular pyramid be 15 cm 3. It is known that the figure is correct. You should find the apothem a b of the lateral edge 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 \u003d √ (h 2 + a 2 / 12) \u003d √ (16 + 25.98 2 / 12) \u003d 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.
Theorem.
The volume of a pyramid is equal to one third of the product of the area of the base and the height..
Proof:
First we prove the theorem for a triangular pyramid, then for an arbitrary one.1. Consider a triangular pyramidOABCwith volume V, base areaS and height h. Draw an axis oh (OM2- height), consider the sectionA1 B1 C1pyramids with a plane perpendicular to the axisohand, therefore, parallel to the plane of the base. Denote byX abscissa point M1 intersection of this plane with the x-axis, and throughS(x)- cross-sectional area. Express S(x) through S, h and X. Note that triangles A1 AT1 With1 and ABC are similar. Indeed A1 AT1 II AB, so triangle OA 1 AT 1 similar to triangle OAB. With consequently, BUT1 AT1 : BUTB= OA 1: OA .
right triangles OA 1 AT 1 and OAB are also similar (they have a common acute angle with vertex O). Therefore, OA 1: OA = O 1 M1 : OM = x: h. Thus BUT 1 AT 1 : A B = x: h.Similarly, it is proved thatB1 C1:sun = X: h and A1 C1:AC = X: h.So the triangleA1 B1 C1 and ABCsimilar with coefficient of similarity X: h.Therefore, S(x) : S = (x: h)², or S(x) = S x²/ 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 has been proven.
Consequence:
Volume V of a truncated pyramid with height h and base areas S and S1 , are calculated by the formula
h - the height of the pyramid
S top - area of the upper base
S lower - area of the lower base
