In a regular tetrahedron, all dihedral angles at the edges and all trihedral angles at the 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 - circumradius
a - edge 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 regular. 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 equal to 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 / 3 a
AO = 4√3 / 3
Thus the height of the pyramid OM can be found from 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
We find the volume of the pyramid using the formula V = 1/3 Sh
In this case, we find the area of the base using the formula S = √3/4 a 2
V = 1/3 (√3 / 4 * 16) (4√2 / √3)
V = 16√2/3
Answer: 16√2 / 3 cm
From the basic formula for the volume of a tetrahedron
Where S is the area of any face, and H– the height lowered by it, a whole series of formulas can be derived that express the volume through various elements of the tetrahedron. Let us present these formulas for the tetrahedron ABCD.
(2) 
,
where ∠ ( AD,ABC) – angle between edge AD and the plane of the face ABC;
(3) 
,
where ∠ ( ABC,ABD) – angle between 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 straight lines and planes; formula (4) is especially useful, with which you can find the distance between crossing lines AB And CD.
Formulas (2) and (3) are similar to formula S = (1/2)ab sin C for the area of the 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 connecting the volume of a tetrahedron with the radius R its described sphere ( Crellet formula):
where Δ is the area of a triangle whose sides are numerically equal to the products of opposite edges ( AB× CD, A.C.× BD,AD× B.C.). From formula (2) and the cosine theorem for trihedral angles (see Spherical trigonometry), we can derive a formula similar to Heron’s formula for triangles.
Definition of tetrahedron
Tetrahedron– the simplest polyhedral body, the faces and base of which are triangles.
Online calculator
A tetrahedron has four faces, each of which is formed by three sides. The tetrahedron has four vertices, with three edges coming out of each.
This body is divided into several types. Below is their classification.
- Isohedral tetrahedron- all its faces are identical triangles;
- Orthocentric tetrahedron- all heights drawn from each vertex to the opposite face are equal in length;
- Rectangular tetrahedron- edges emanating from one vertex form an angle of 90 degrees with each other;
- Frame;
- Proportionate;
- Incentric.
Tetrahedron volume formulas
The volume of a given body can be found in several ways. Let's look at them in more detail.
Through the mixed product of vectors
If a 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, the following determinant:
Volume of a tetrahedron through the determinantV = 1 6 ⋅ ∣ a x a y a z b x b y b z c x c y c z ∣ 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 ∣ ∣ ∣ ∣ ∣ ∣
Problem 1The 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 this body is built.
To do this, you need to find each vector coordinate by subtracting the corresponding coordinates of the two points. For example, the vector coordinates A B → \overrightarrow(AB) A B, that is, a vector directed from the point A A A to the point B B B, these are the differences between the corresponding coordinates of the points B B B And A A A:
A B → = (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 )
A C → = (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
)
A D → = (7 − 1 , 12 − 4 , 1 − 9) = (6 , 8 , − 8) \overrightarrow(AD)=(7-1, 12-4, 1-9)=(6, 8, -8)A D=
(7
−
1
,
1
2
−
4
,
1
−
9
)
=
(6
,
8
,
−
8
)
Now let’s find the mixed product of these vectors; to do this, we’ll compose a third-order determinant, while accepting 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.
∣ a x a y a z b x b y b z c x c y c z ∣ = ∣ 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 the tetrahedron is equal to:
V = 1 6 ⋅ ∣ a x a y a z b x b y b z c x c y c z ∣ = 1 6 ⋅ ∣ 7 3 − 6 0 − 2 − 6 6 8 − 8 ∣ = 1 6 ⋅ 268 ≈ 44.8 cm 3 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)=\frac(1)(6)\cdot \begin(vmatrix) 7 & 3 & - 6 \\ 0 & -2 & -6 \\ 6 & 8 & -8 \\ \end(vmatrix)=\frac(1)(6)\cdot268\approx44.8\text( cm)^3
Answer
44.8 cm3. 44.8\text( cm)^3.
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 tetrahedronV = 2 ⋅ a 3 12 V=\frac(\sqrt(2)\cdot a^3)(12)
a a
Problem 2Determine the volume of a tetrahedron given its side equal to 11 cm 11\text( cm)
Solution
a=11 a=11
Let's substitute a a
V = 2 ⋅ a 3 12 = 2 ⋅ 1 1 3 12 ≈ 156.8 cm 3 V=\frac(\sqrt(2)\cdot a^3)(12)=\frac(\sqrt(2)\cdot 11^ 3)(12)\approx156.8\text( cm)^3
Answer
156.8 cm3. 156.8\text( cm)^3.