Find the area of a tetrahedron. The volume of the tetrahedron. Regular tetrahedron is a particular type of tetrahedron
Note... This is part of the lesson with geometry problems (stereometry section, pyramid problems). If you need to solve a geometry problem that 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 symbol square root, and the radical expression is indicated in parentheses.For simple radical expressions, the "√" sign can be used. Regular tetrahedron is correct triangular pyramid in which all faces are equilateral triangles.For a regular tetrahedron, all dihedral angles at the edges and all trihedral angles at the vertices are equal
The 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 a circle inscribed in a tetrahedron
R - radius of the circumscribed circle
a - rib length
Practical examples
A 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
A 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 into 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
The volume of the pyramid is found by the formula V = 1/3 Sh
In this case, the area of the base is found by 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
Answer: 6.
Answer: 000
The surface area of a tetrahedron is 1. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
prototype.
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
Answer:
The surface area of a tetrahedron is Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
Answer: 0.8
The surface area of the tetrahedron is 4.6. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
Answer: 2.3
The surface area of a tetrahedron is 6. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
Answer: 3
The surface area of the tetrahedron is 2.8. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
Answer: 000
The surface area of the tetrahedron is 8.8. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of a tetrahedron is 7. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
Answer: 3.5
The surface area of the tetrahedron is 4.8. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of the tetrahedron is 9.6. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of the tetrahedron is 7.8. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of the tetrahedron is 5.6. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of the tetrahedron is 3.2. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of the tetrahedron is 8.6. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of the tetrahedron is 2.2. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of the tetrahedron is 6.8. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
Answer: 3.4
The surface area of the tetrahedron is 10.2. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of the tetrahedron is 3.8. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of a tetrahedron is 4. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of a tetrahedron is 8. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of a tetrahedron is 9. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Answer: 6.
The surface area of the tetrahedron is 2.4. Find the surface area of a polyhedron whose vertices are the midpoints of the sides of this tetrahedron.
Solution.
This task has not been solved yet, we present the prototype solution
The surface area of a tetrahedron is 12. Find the surface area of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.
The sought surface consists of four pairs of equal triangles, each of which has an area equal to a quarter of the area of the face of the original tetrahedron. Therefore, the required area is equal to half the surface area of the tetrahedron and is equal to 6.
Consider an arbitrary triangle ABC and a point D not lying in the plane of this triangle. Let's connect this point with the vertices of the triangle ABC by segments. 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 the edges of the tetrahedron. And their peaks are the peaks of a tetrahedron
The tetrahedron has 4 faces, 6 ribs and 4 vertices.
Two edges that do not have a common vertex are called opposite edges.
Often for convenience, one of the faces of the tetrahedron is called basis, and the remaining three faces are side faces.
Thus, a tetrahedron is the simplest polyhedron with four triangles as its faces.
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.
Tetrahedron height is called a segment that connects a vertex with a point located on the opposite face and perpendicular to it.
Median tetrahedron is called the segment that connects the vertex with the point of intersection of the medians of the opposite face.
Bimedian tetrahedron is called the segment that connects the midpoints of the crossing edges of the tetrahedron.
Since a tetrahedron is a pyramid with triangular base, then the volume of any tetrahedron can be calculated by the formula
- S- the area of any face,
- H- the height lowered to this face
Regular tetrahedron is a particular type of tetrahedron
A tetrahedron with all faces of an equilateral triangle is called correct.
Properties of a regular tetrahedron:
- All faces are equal.
- All planar 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 triangle heights).
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.
The height BM is equal to BM and is equal to
Consider a triangle BDM, where DH, which is the height of the tetrahedron, is also the height of this triangle.
The height of the triangle lowered 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. We get 
Take out 1 / 2a. We get
We apply the formula difference of squares 
After small transformations, we get 
The volume of any tetrahedron can be calculated by the formula
,
where 
,
Substituting these values, we get 
Thus, the volume formula for a regular tetrahedron is
where a- the edge of the tetrahedron
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. We get 
