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 middle line 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.
Point F is the intersection point of the medians, therefore,
Triangle ∆SEF and ∆SPM – similar triangles, hence,
Answer: 1
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In a regular triangular pyramid SABC, M is the middle of edge AB, S is the vertex. It is known that BC = 3, and the area of the lateral surface of the pyramid is 45. Find the length of the segment SM.
Solution
Since the area of the lateral surface of the pyramid is 45, and it has 3 lateral faces and the areas of these faces are equal, then the area of each lateral face is 45:3 = 15.
That is, S ABS =15.
Since the pyramid is regular, triangle ABS is isosceles. SM is the median, and therefore the height. Then the area of triangle ABS is
S ABS = 1/2 * AB* SM = 1/2 * 3 * SM = 1.5 SM.
Task 12 ( Typical options 2015)
IN rectangular parallelepiped ABCDA1B1C1D1 the lengths of the edges are known: AB = 7, AD = 10, AA1 = 24. Find the cross-sectional area of the parallelepiped by the plane passing through points A, B and C1.
Solution
Rectangle ABC1D1 is a section of a parallelepiped by a plane passing through points A, B and C1.
Let's find the area of this rectangle.
S ABC1D1 = AB*BC1.
From right triangle BCC1 by Pythagorean theorem
BC1 2 = BC 2 +CC1 2,
BC1 2 = 10 2 +24 2,
BC1 2 = 676, BC1=26.
Then S ABC1D1 =7*26=182.
Answer: 182.
Task 12 (Typical options 2015)
The volume of a cube is 12. Find the volume triangular prism, cut off from it by a plane passing through the midpoints of two edges emerging from one vertex, and parallel to the third an edge emerging from the same vertex.
Solution
Let's denote the edge of the cube by a. Then a 3 = 12.
Latest solutions
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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.
The solution of the problem
This lesson demonstrates 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.