Tetrahedron volume and area. Right tetrahedron (pyramid). Proper Tetrahedron - Private Type of Tetrahedron

From the main formula for the volume of tetrahedron

where S. - Square any face, and H. - The height lowered to it, one can derive a number of formulas expressing through various elements of the tetrahedron. We give these formulas for tetrahedron Abcd..

(2) ,

where ∠ ( AD,ABC) - the angle between the edge AD and the plane of the edge ABC;

(3) ,

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

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

Formulas (2) - (4) can be used to find the values \u200b\u200bof the angles between the straight and planes; Formula (4) is particularly useful, with which you can find the distance between cross-country straight AB and CD.

Formulas (2) and (3) are similar to the formula S. = (1/2)aBsin. C. For a triangle square. Formula S. = rP. Similar formula

where r. - The radius of the inscribed sphere of tetrahedron, σ is its complete surface (the sum of the areas of all the faces). There is also a beautiful formula that binds the volume of the tetrahedron with a radius R. its described sphere ( formula Krellle):

where δ is the area of \u200b\u200bthe triangle, the parties of which are numerically equal to the works of opposite ribs ( AB× CD, AC× BD.,AD× BC.). From formula (2) and cosine theorems for triangular angles (see spherical trigonometry) can be derived a formula similar to the formula of Geron for triangles.

Definition of tetrahedron

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

Online calculator

The tetrahedron has four faces, each of which is formed by three parties. The vertices in Tetrahedron four, from each coming out three ribs.

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

1. A washing tetrahedron - He has all the faces of the same triangles;
2. Orthocentric tetrahedron - all heights carried out from each vertex on the opposite face are the same in length;
3. Rectangular tetrahedron - Ribs emanating from one vertex form an angle of 90 degrees with each other;
4. Frame;
5. Proportionate;
6. Incenter.

Tetrahedron volume formulas

The volume of this body can be found in several ways. We will analyze them in more detail.

Through a mixed product of vectors

If the tetrahedron is built on three vectors with coordinates:

A ⃗ \u003d (a x, a y, a z) \\ Vec (a) \u003d (A_X, A_Y, A_Z)a.= (a. x.​ , a. y.​ , a. z.​ )
b ⃗ \u003d (b x, b y, b z) \\ Vec (b) \u003d (b_x, b_y, b_z)b.= (b. x.​ , b. y.​ , b. z.​ )
C ⃗ \u003d (C x, C y, c z) \\ Vec (C) \u003d (C_X, C_Y, C_Z)c.= (c. x.​ , c. y.​ , c. z.​ ) ,

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

Tetrahedron volume through the determinant

V \u003d 1 6 | | axayazbxbybzcxcycz | v \u003d \\ 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 \u003d.6 1 ​ ⋅ ∣ ∣ ∣ ∣ ∣ ∣ ​ a. x.​ b. x.​ c. x.​ ​ a. y.​ b. y.​ c. y.​ ​ a. z.​ b. z.​ c. z.​ ​ ∣ ∣ ∣ ∣ ∣ ∣ ​

Task 1.

Known coordinates of four octahedra vertices. 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.

Decision

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 this body is built.
To do this, find each vector coordinate by subtracting the corresponding coordinates of two points. For example, the coordinates of the vector A B → \\ Overrightarrow (AB) A B., that is, the vector directed from the point A A. A. To point B B. B.is the difference between the corresponding coordinates of the points B B. B. and A A. A.:

AB → \u003d (8 - 1, 7 - 4, 3 - 9) \u003d (7, 3, - 6) \\ Overrightarrow (AB) \u003d (8-1, 7-4, 3-9) \u003d (7, 3, -6)A B.= (8 − 1 , 7 − 4 , 3 − 9 ) = (7 , 3 , − 6 )

AC → \u003d (1 - 1, 2 - 4, 3 - 9) \u003d (0, - 2, - 6) \\ OVERRIGHTARROW (AC) \u003d (1-1, 2-4, 3-9) \u003d (0, - 2, -6)A C.= (1 − 1 , 2 − 4 , 3 − 9 ) = (0 , − 2 , − 6 )
AD → \u003d (7 - 1, 12 - 4, 1 - 9) \u003d (6, 8, - 8) \\ Overrightarrow (AD) \u003d (7-1, 12-4, 1-9) \u003d (6, 8, -eight)A D.= (7 − 1 , 1 2 − 4 , 1 − 9 ) = (6 , 8 , − 8 )

Now we will find a mixed product of these vectors, for this we will make a third-order determinant, while taking that A B → \u003d A ⃗ \\ Overrightarrow (AB) \u003d \\ VEC (A)A B.= a., A C → \u003d B ⃗ \\ Overrightarrow (AC) \u003d \\ VEC (B)A C.= b., A D → \u003d C ⃗ \\ Overrightarrow (AD) \u003d \\ VEC (C)A D.= c..

| Axayazbxbybzcxcycz | \u003d | 7 3 - 6 0 - 2 - 6 6 8 - 8 | \u003d 7 ⋅ (- 2) ⋅ (- 8) + 3 ⋅ (- 6) ⋅ 6 + (- 6) ⋅ 0 ⋅ 8 - (- 6) ⋅ (- 2) ⋅ 6 - 7 ⋅ (- 6) ⋅ 8 - 3 ⋅ 0 ⋅ (- 8) \u003d 112 - 108 - 0 - 72 + 336 + 0 \u003d 268 \\ Begin (VMATRIX) A_X & A_Y & A_Z \\\\ B_X & B_Y & B_Z \\\\ C_X & C_Y & C_Z \\\\ \\ END (VMATRIX) \u003d \\ Begin (VMatrix) 7 & 3 & -6 \\\\ 0 & -2 & -6 \\\\ 6 & 8 & -8 \\\\ \\ END (VMatrix) \u003d 7 \\ Cdot (-2) \\ CDOT (-8) + 3 \\ CDOT (-6) \\ CDOT6 + (-6) \\ CDOT0 \\ CDOT8 - (-6) \\ CDOT (-2) \\ Cdot6 - 7 \\ Cdot (-6) \\ Cdot8 - 3 \\ Cdot0 \\ Cdot (-8) \u003d 112 - 108 - 0 - 72 + 336 + 0 \u003d 268∣ ∣ ∣ ∣ ∣ ∣ ​ a. x.​ b. x.​ c.x.​ ​ a.y.​ b.y.​ c.y.​ ​ a.z.​ b.z.​ c.z.​ ​ ∣ ∣ ∣ ∣ ∣ ∣ ​ = ∣ ∣ ∣ ∣ ∣ ∣ ​ 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 tetrahedron is:

$V.=61\cdot |$

Answer

$44.8\text{cm3 .}$

The formula of the volume of the washed tetrahedron on its side

This formula is valid only for calculating the volume of an anxulated tetrahedron, that is, such a tetrahedron, in which all the faces are the same correct triangles.

Volume of an anxulated tetrahedron

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

A A.a. - The length of the edge of the tetrahedron.

Task 2.

Determine the volume of the tetrahedron if his side is given equal $11\text{cm.}$

Decision

$a.=11$

Substitute A A.a. In the formula for the volume of 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 the tasks of geometry (section stereometry, tasks about the pyramid). If you need to solve the task of geometry, which is not here - write about it in the forum. In tasks, the SQRT () function is used in the tasks instead of the SQRT () function, in which SQRT - symbol square root, and in brackets indicated the guided expression. For simple feeding expressions, the "√" sign can be used.. Right tetrahedron - This is the correct triangular pyramid in which all the faces are equilateral triangles.

At the right tetrahedron, all couphed corners with ribers and all three-headed corners at the tops are equal

Tetrahedron 4 faces, 4 vertices and 6 ribs.

The basic formulas for the correct tetrahedron are shown in the table.

Where:
S - surface area of \u200b\u200bthe correct tetrahedron
V - volume
h - height, lowered to the base
R - the radius inscribed in the tetrahedron of the circle
R - radius of the circumference
A - Rib length

Practical examples

A task.
Find the surface area of \u200b\u200bthe triangular pyramid, in which each edge is equal to √3

Decision.
Since all edges of the triangular pyramid are equal - it is correct. The surface area of \u200b\u200bthe correct triangular pyramid is s \u003d a 2 √3.
Then
S \u003d 3√3

Answer: 3√3

A task.
All edges of the correct triangular pyramid are 4 cm. Find the volume of the pyramid

Decision.
Since in the correct triangular pyramid, the pyramid height is projected into the center of the base, which is simultaneously the center of the described circle,

AO \u003d R \u003d √3 / 3 a
AO \u003d 4√3 / 3

Thus, the height of the OM pyramid can be found from rectangular triangle AOM.

AO 2 + OM 2 \u003d am 2
Om 2 \u003d am 2 - AO 2
Om 2 \u003d 4 2 - (4√3 / 3) 2
Om 2 \u003d 16 - 16/3
Om \u003d √ (32/3)
Om \u003d 4√2 / √3

The volume of the pyramid will find by formula v \u003d 1/3 sh
At the same time, the base area will find according to the formula S \u003d √3 / 4 A 2

V \u003d 1/3 (√3 / 4 * 16) (4√2 / √3)
V \u003d 16√2 / 3

Answer: 16√2 / 3 cm