Exercise.
In a regular triangular pyramid SABC with base ABC, all edges are equal to 6.
a) Construct a section of the pyramid with a plane passing through the vertex S and perpendicular to the segment connecting the midpoints of the edges AB and BC.
b) Find the distance from the plane of this section to the center of face SAB.
Solution:
a) Construct a section of the pyramid with a plane passing through the vertexSand perpendicular to the segment connecting the midpoints of the edges AB and BC.
Let point M be the midpoint of edge BC, and point N be the midpoint of edge AB, then MN is midline triangle ∆ABC. This means that MN is parallel to AC. Since the SABC pyramid is correct, then at the base lies regular triangle∆ABC, therefore, BD is the median and height of the triangle ∆ABC, i.e. BD is perpendicular to AC and BD is perpendicular to MN. Let us connect points B, D and S in series. We obtain the required section SBD passing through vertex S and perpendicular to the segment connecting the midpoints of edges AB and BC.
b) Find the distance from the plane of this section to the center of the faceSAB.
The distance from a point to a plane is the perpendicular drawn from a given point to the plane. Let's construct the center of the face SAB; to do this, find the point of intersection of the medians of the triangle ∆SAB. Since the triangle ∆SAB is regular, the point of intersection of the medians F is the center of the face SAB.
Let's draw FE parallel to MN. Since MN is perpendicular to the section plane SBD, FE is perpendicular to the section plane SBD. Therefore, FE is the distance from the section plane SBD to the center of the face SAB.
Since points M and N are the midpoints of the edges AB and BC, then MN is the midline of the triangle ∆ABC.
Since BD is the median and height of the triangle ∆ABC, then BP is the median and height of the triangle ∆BMN. Therefore, NP = MP = 1.5.
In a regular pyramid, the apothems SN and SM are equal, which means that the triangle ∆SMN is isosceles, SP is the height of the triangle ∆SMN.
In a regular triangular pyramid SABC, N is the middle of edge BC, S is the vertex. It is known that SN=6, and the lateral surface area is 72. Find the length of segment AB.
Problem solution
IN this lesson demonstrated geometric problem, the solution of which is based on the definition and properties of the correct triangular pyramid. It is stated that everything side faces regular pyramid are isosceles triangles. This means that the lateral surface area of this pyramid can be defined as the side. pov =. Next, during the solution, we consider a triangle, the area of which is equal to half the product of the length of the side and the length of the height drawn to this side. By property isosceles triangle a segment is both a median and a height, therefore the following equality is true: . Having made the appropriate replacement in the formula for the area of the lateral surface of the pyramid, the values known by condition are substituted. Since, by definition of a regular triangular pyramid, there is a regular triangle at its base, the found value is equal to the required length of the segment.
This task is similar to problems of type B13, so it can be successfully used as preparation for the Unified State Exam in mathematics.
We continue to consider the tasks included in the Unified State Examination in mathematics. We have already studied problems where the condition is given and it is required to find the distance between two given points or an angle.
A pyramid is a polyhedron, the base of which is a polygon, the remaining faces are triangles, and they have a common vertex.
A regular pyramid is a pyramid at the base of which lies regular polygon, and its top is projected into the center of the base.
A regular quadrangular pyramid - the base is a square. The top of the pyramid is projected at the intersection point of the diagonals of the base (square).
ML - apothem
∠MLO - dihedral angle at the base of the pyramid
∠MCO - angle between the lateral edge and the plane of the base of the pyramid
In this article we will look at problems to solve a regular pyramid. You need to find some element, lateral surface area, volume, height. Of course, you need to know the Pythagorean theorem, the formula for the area of the lateral surface of a pyramid, and the formula for finding the volume of a pyramid.
In the article "" presents the formulas that are necessary to solve problems in stereometry. So, the tasks:
SABCD dot O- center of the base,S vertex, SO = 51, A.C.= 136. Find side rib S.C..
IN in this case the base is a square. This means that the diagonals AC and BD are equal, they intersect and are bisected by the intersection point. Note that in a regular pyramid the height dropped from its top passes through the center of the base of the pyramid. So SO is the height and the triangleSOCrectangular. Then according to the Pythagorean theorem:
How to extract the root from large number.
Answer: 85
Decide for yourself:
In a regular quadrangular pyramid SABCD dot O- center of the base, S vertex, SO = 4, A.C.= 6. Find the side edge S.C..
In a regular quadrangular pyramid SABCD dot O- center of the base, S vertex, S.C. = 5, A.C.= 6. Find the length of the segment SO.
In a regular quadrangular pyramid SABCD dot O- center of the base, S vertex, SO = 4, S.C.= 5. Find the length of the segment A.C..
SABC R- middle of the rib B.C., S- top. It is known that AB= 7, a S.R.= 16. Find the lateral surface area.
The area of the lateral surface of a regular triangular pyramid is equal to half the product of the perimeter of the base and the apothem (apothem is the height of the lateral face of a regular pyramid drawn from its vertex):
Or we can say this: the area of the lateral surface of the pyramid is equal to the sum three squares side edges. The lateral faces in a regular triangular pyramid are triangles of equal area. In this case:
Answer: 168
Decide for yourself:
In a regular triangular pyramid SABC R- middle of the rib B.C., S- top. It is known that AB= 1, a S.R.= 2. Find the lateral surface area.
In a regular triangular pyramid SABC R- middle of the rib B.C., S- top. It is known that AB= 1, and the area of the lateral surface is 3. Find the length of the segment S.R..
In a regular triangular pyramid SABC L- middle of the rib B.C., S- top. It is known that SL= 2, and the area of the lateral surface is 3. Find the length of the segment AB.
In a regular triangular pyramid SABC M. Area of a triangle ABC is 25, the volume of the pyramid is 100. Find the length of the segment MS.
The base of the pyramid is an equilateral triangle. That's why Mis the center of the base, andMS- height of a regular pyramidSABC. Volume of the pyramid SABC equals: view solution
In a regular triangular pyramid SABC the medians of the base intersect at the point M. Area of a triangle ABC equals 3, MS= 1. Find the volume of the pyramid.
In a regular triangular pyramid SABC the medians of the base intersect at the point M. The volume of the pyramid is 1, MS= 1. Find the area of the triangle ABC.
Let's finish here. As you can see, problems are solved in one or two steps. In the future, we will consider other problems from this part, where bodies of revolution are given, don’t miss it!
Good luck to you!
Sincerely, Alexander Krutitskikh.
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