Definition of a tetrahedron

Tetrahedron- the simplest polyhedral body, the faces and base of which are triangles.

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A tetrahedron has four faces, each of which is formed by three sides. The tetrahedron has four vertices, each with three edges.

This body is divided into several types. Below is their classification.

1. Isohedral tetrahedron- all its faces are the same triangles;
2. Orthocentric tetrahedron- all heights drawn from each vertex to the opposite face are the same in length;
3. Rectangular tetrahedron- edges emanating from one vertex form an angle of 90 degrees with each other;
4. frame;
5. Proportionate;
6. incentric.

Tetrahedron volume formulas

The volume of a given body can be found in several ways. Let's analyze them in more detail.

Through the mixed product of vectors

If the tetrahedron is built on three vectors with coordinates:

A ⃗ = (a x , a y , a z) \vec(a)=(a_x, a_y, a_z)a= (a x​ , a y​ , a z​ )
b ⃗ = (b x , b y , b z) \vec(b)=(b_x, b_y, b_z)b= (b x​ , b y​ , b z​ )
c ⃗ = (c x , c y , c z) \vec(c)=(c_x, c_y, c_z)c= (c x​ , c y​ , c z​ ) ,

then the volume of this tetrahedron is the mixed product of these vectors, that is, such a determinant:

The volume of a tetrahedron through the determinant

V = 1 6 ⋅ ∣ axayazbxbybzcxcycz ∣ V=\frac(1)(6)\cdot\begin(vmatrix) a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \\ \end(vmatrix )V =6 1 ​ ⋅ ∣ ∣ ∣ ∣ ∣ ∣ ​ a x​ b x​ c x​ ​ a y​ b y​ c y​ ​ a z​ b z​ c z​ ​ ∣ ∣ ∣ ∣ ∣ ∣ ​

Task 1

The coordinates of the four vertices of the octahedron are known. A (1 , 4 , 9) A(1,4,9) A (1 , 4 , 9 ), B(8 , 7 , 3) ​​B(8,7,3) B(8, 7, 3), C (1 , 2 , 3) ​​C(1,2,3) C (1 , 2 , 3 ), D(7, 12, 1) D(7,12,1) D (7 , 1 2 , 1 ). Find its volume.

Solution

A (1 , 4 , 9) A(1,4,9) A (1 , 4 , 9 )
B(8 , 7 , 3) ​​B(8,7,3) B(8, 7, 3)
C (1 , 2 , 3) ​​C(1,2,3) C (1 , 2 , 3 )
D(7, 12, 1) D(7,12,1) D (7 , 1 2 , 1 )

The first step is to determine the coordinates of the vectors on which the given body is built.
To do this, you need to find each coordinate of the vector by subtracting the corresponding coordinates of two points. For example, vector coordinates A B → \overrightarrow(AB) A B, that is, a vector directed from a point A A A to the point B B B, these are the differences of the corresponding coordinates of the points B B B and A A A:

AB → = (8 − 1 , 7 − 4 , 3 − 9) = (7 , 3 , − 6) \overrightarrow(AB)=(8-1, 7-4, 3-9)=(7, 3, -6)A B= (8 − 1 , 7 − 4 , 3 − 9 ) = (7 , 3 , − 6 )

AC → = (1 − 1 , 2 − 4 , 3 − 9) = (0 , − 2 , − 6) \overrightarrow(AC)=(1-1, 2-4, 3-9)=(0, - 2, -6)A C= (1 − 1 , 2 − 4 , 3 − 9 ) = (0 , − 2 , − 6 )
AD → = (7 − 1 , 12 − 4 , 1 − 9) = (6 , 8 , − 8) \overrightarrow(AD)=(7-1, 12-4, 1-9)=(6, 8, -eight)A D= (7 − 1 , 1 2 − 4 , 1 − 9 ) = (6 , 8 , − 8 )

Now let's find the mixed product of these vectors, for this we compose a third-order determinant, while assuming that A B → = a ⃗ \overrightarrow(AB)=\vec(a)A B= a, A C → = b ⃗ \overrightarrow(AC)=\vec(b)A C= b, A D → = c ⃗ \overrightarrow(AD)=\vec(c)A D= c.

∣ axayazbxbybzcxcycz ∣ = ∣ 7 3 − 6 0 − 2 − 6 6 8 − 8 ∣ = 7 ⋅ (− 2) ⋅ (− 8) + 3 ⋅ (− 6) ⋅ 6 + (− 6) ⋅ 0 ⋅ 8 − (− 6) ⋅ (− 2) ⋅ 6 − 7 ⋅ (− 6) ⋅ 8 − 3 ⋅ 0 ⋅ (− 8) = 112 − 108 − 0 − 72 + 336 + 0 = 268 \begin(vmatrix) a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \\ \end(vmatrix)= \begin(vmatrix) 7 & 3 & -6 \\ 0 & -2 & -6 \\ 6 & 8 & -8 \\ \end(vmatrix)=7\cdot(-2)\cdot(-8) + 3\cdot(-6)\cdot6 + (-6)\cdot0\cdot8 - (-6)\cdot (-2)\cdot6 - 7\cdot(-6)\cdot8 - 3\cdot0\cdot(-8) = 112 - 108 - 0 - 72 + 336 + 0 = 268∣ ∣ ∣ ∣ ∣ ∣ ​ a x​ b x​ cx​ ​ ay​ by​ cy​ ​ az​ bz​ cz​ ​ ∣ ∣ ∣ ∣ ∣ ∣ ​ = ∣ ∣ ∣ ∣ ∣ ∣ ​ 7 0 6 ​ 3 − 2 8 ​ − 6 − 6 − 8 ​ ∣ ∣ ∣ ∣ ∣ ∣ ​ = 7 ⋅ (− 2 ) ⋅ (− 8 ) + 3 ⋅ (− 6 ) ⋅ 6 + (− 6 ) ⋅ 0 ⋅ 8 − (− 6 ) ⋅ (− 2 ) ⋅ 6 − 7 ⋅ (− 6 ) ⋅ 8 − 3 ⋅ 0 ⋅ (− 8 ) = 1 1 2 − 1 0 8 − 0 − 7 2 + 3 3 6 + 0 = 2 6 8

That is, the volume of a tetrahedron is:

$V=61\cdot |$

Answer

$44.8\text{cm3 .}$

The formula for the volume of an isohedral tetrahedron along its side

This formula is valid only for calculating the volume of an isohedral tetrahedron, that is, a tetrahedron in which all faces are identical regular triangles.

Volume of an isohedral tetrahedron

$V=12\sqrt{2\cdot a^3}$

a aa is the length of the edge of the tetrahedron.

Task 2

Find the volume of a tetrahedron if its side is given equal to $11\text{cm.}$

Solution

$a=11$

Substitute a aa into the formula for the volume of a tetrahedron:

$V=12\sqrt{2\cdot a^{3=12\sqrt{2\cdot 11^{3\approx 156.8\text{cm3}}}}}$

Answer

$156.8\text{cm3 .}$

Note. This is part of the lesson with problems in geometry (section solid geometry, problems about the pyramid). If you need to solve a problem in geometry, which is not here - write about it in the forum. In tasks, 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 brackets.For simple radical expressions, the sign "√" can be used. regular tetrahedron is a regular triangular pyramid in which all faces are equilateral triangles.

For a regular tetrahedron, all dihedral angles at edges and all trihedral angles at vertices are equal

A tetrahedron has 4 faces, 4 vertices and 6 edges.

The basic formulas for a regular tetrahedron are given in the table.

Where:
S - Surface area of ​​a regular tetrahedron
V - volume
h - height lowered to the base
r - radius of the circle inscribed in the tetrahedron
R - radius of the circumscribed circle
a - rib length

Practical examples

Task.
Find the surface area of ​​a triangular pyramid with each edge equal to √3

Solution.
Since all the edges of a triangular pyramid are equal, it is correct. The surface area of ​​a regular triangular pyramid is S = a 2 √3.
Then
S = 3√3

Answer: 3√3

Task.
All edges of a regular triangular pyramid are 4 cm. Find the volume of the pyramid

Solution.
Since in a regular triangular pyramid the height of the pyramid is projected to the center of the base, which is also the center of the circumscribed circle, then

AO = R = √3 / 3a
AO = 4√3 / 3

Thus the height of the pyramid OM can be found from the right triangle AOM

AO 2 + OM 2 = AM 2
OM 2 = AM 2 - AO 2
OM 2 = 4 2 - (4√3 / 3) 2
OM 2 = 16 - 16/3
OM = √(32/3)
OM = 4√2 / √3

The volume of the pyramid is found by the formula V = 1/3 Sh
In this case, we find the area of ​​\u200b\u200bthe base by the formula S \u003d √3/4 a 2

V = 1/3 (√3 / 4 * 16) (4√2 / √3)
V=16√2/3

Answer: 16√2/3cm

From the basic formula for the volume of a tetrahedron

where S is the area of ​​any face, and H- the height lowered onto it, you can derive a whole series of formulas expressing the volume in terms of various elements of the tetrahedron. We give these formulas for the tetrahedron ABCD.

(2) ,

where ∠ ( AD,ABC) is the angle between the edge AD and face plane ABC;

(3) ,

where ∠ ( ABC,ABD) is the angle between the faces ABC and ABD;

where | AB,CD| - distance between opposite ribs AB and CD, ∠ (AB,CD) is the angle between these edges.

Formulas (2)–(4) can be used to find the angles between lines and planes; formula (4) is especially useful, with which you can find the distance between skew lines AB and CD.

Formulas (2) and (3) are similar to the formula S = (1/2)ab sin C for the area of ​​a triangle. Formula S = rp similar formula

where r is the radius of the inscribed sphere of the tetrahedron, Σ is its total surface (the sum of the areas of all faces). There is also a beautiful formula that connects the volume of a tetrahedron with a radius R its described scope ( Crelle formula):

where Δ is the area of ​​a triangle whose sides are numerically equal to the products of opposite edges ( AB× CD, AC× BD,AD× BC). From formula (2) and the cosine theorem for trihedral angles (see Spherical trigonometry), one can derive a formula similar to Heron's formula for triangles.

Consider an arbitrary triangle ABC and a point D that does not lie in the plane of this triangle. Connect this point with segments to the vertices of the triangle ABC. As a result, we get triangles ADC , CDB , ABD . The surface bounded by four triangles ABC , ADC , CDB and ABD is called a tetrahedron and is denoted DABC .
The triangles that make up a tetrahedron are called its faces.
The sides of these triangles are called edges of the tetrahedron. And their vertices are the vertices of a tetrahedron

The tetrahedron has 4 faces, 6 ribs and 4 peaks.
Two edges that do not have a common vertex are called opposite.
Often, for convenience, one of the faces of the tetrahedron is called basis, and the remaining three faces are side faces.

Thus, the tetrahedron is the simplest polyhedron, the faces of which are four triangles.

But it is also true that any arbitrary triangular pyramid is a tetrahedron. Then it is also true that a tetrahedron is called a pyramid with a triangle at its base.

The height of the tetrahedron called a segment that connects a vertex to a point located on the opposite face and perpendicular to it.
Median of a tetrahedron called a segment that connects the vertex with the point of intersection of the medians of the opposite face.
Bimedian tetrahedron is called a segment that connects the midpoints of the crossing edges of the tetrahedron.

Since a tetrahedron is a pyramid with a triangular base, the volume of any tetrahedron can be calculated using the formula

• S is the area of ​​any face,
• H- the height lowered on this face

Regular tetrahedron - a special type of tetrahedron

A tetrahedron in which all faces are equilateral triangles is called correct.
Properties of a regular tetrahedron:

• All edges are equal.
• All plane angles of a regular tetrahedron are 60°
• Since each of its vertices is the vertex of three regular triangles, the sum of the plane angles at each vertex is 180°
• Any vertex of a regular tetrahedron is projected to the orthocenter of the opposite face (to the intersection point of the heights of the triangle).

Let us be given a regular tetrahedron ABCD with edges equal to a . DH is its height.
Let's make additional constructions BM - the height of the triangle ABC and DM - the height of the triangle ACD .
Height BM equals BM and equals
Consider triangle BDM , where DH , which is the height of the tetrahedron, is also the height of this triangle.
The height of a triangle dropped to the side MB can be found using the formula

, where
BM=, DM=, BD=a,
p=1/2 (BM+BD+DM)=
Substitute these values ​​into the height formula. Get


Let's take out 1/2a. Get

Apply the formula difference of squares

After some minor transformations, we get


The volume of any tetrahedron can be calculated using the formula
,
where ,

Substituting these values, we get

Thus the volume formula for a regular tetrahedron is

where a–tetrahedron edge

Calculating the volume of a tetrahedron if the coordinates of its vertices are known

Let us be given the coordinates of the vertices of the tetrahedron

Draw vectors from the vertex , , .
To find the coordinates of each of these vectors, subtract the corresponding start coordinate from the end coordinate. Get