How to construct a section in a parallelepiped perpendicular to a straight line. Problems on constructing sections in a parallelepiped
The task itself usually sounds like this: "build a natural view of a section figure". Of course, we decided not to leave this issue aside and try, if possible, to explain how the inclined section is constructed.
In order to explain how an inclined section is constructed, I will give several examples. I will, of course, start with the elementary ones, gradually increasing the complexity of the examples. I hope that after analyzing these examples of section drawings, you will understand how it is done and will be able to complete your study assignment yourself.
Let's consider a “brick” with dimensions 40x60x80 mm and an arbitrary inclined plane. The cutting plane cuts it at points 1-2-3-4. I think everything is clear here.
Let's move on to constructing a natural view of the section figure.
1. First of all, let's draw the section axis. The axis should be drawn parallel to the section plane - parallel to the line into which the plane is projected in the main view - usually it is in the main view that the task for construction of an inclined section(Further I will always mention the main view, keeping in mind that this almost always happens in educational drawings).
2. On the axis we plot the length of the section. In my drawing it is designated as L. The size L is determined in the main view and is equal to the distance from the point of entry of the section into the part to the point of exit from it.
3. From the resulting two points on the axis, perpendicular to it, we plot the width of the section at these points. The width of the section at the point of entry into the part and at the point of exit from the part can be determined in the top view. In this case, both segments 1-4 and 2-3 are equal to 60 mm. As you can see from the picture above, the edges of the section are straight, so we simply connect our two resulting segments, obtaining a rectangle 1-2-3-4. This is the natural appearance of the cross section of our brick by an inclined plane.
Now let's complicate our part. Let's place a brick on a base 120x80x20 mm and add stiffening ribs to the figure. Let's draw a cutting plane so that it passes through all four elements of the figure (through the base, brick and two stiffeners). In the picture below you can see three views and a realistic image of this part.
Let's try to build a natural view of this inclined section. Let's start again with the section axis: draw it parallel to the section plane indicated in the main view. Let us plot the length of the section on it equal to A-E. Point A is the entry point of the section into the part, and in a particular case, the entry point of the section into the base. The exit point from the base is point B. Mark point B on the section axis. In a similar way, we mark the entry and exit points to the edge, to the “brick” and to the second edge. From points A and B, perpendicular to the axis, we will lay out segments equal to the width of the base (40 in each direction from the axis, 80 mm in total). Let's connect extreme points- we get a rectangle, which is a natural cross-section of the base of the part.
Now it’s time to construct a piece of the section, which is a section of the edge of the part. From points B and C we will put perpendiculars of 5 mm in each direction - we will get segments of 10 mm. Let's connect the extreme points and get a section of the rib.
From points C and D we lay out perpendicular segments equal to the width of the “brick” - completely similar to the first example of this lesson.
By setting aside perpendiculars from points D and E equal to the width of the second edge and connecting the extreme points, we obtain a natural view of its section.
All that remains is to erase the jumpers between the individual elements of the resulting section and apply shading. It should look something like this:
If we divide the figure along a given section, we will see next view:
I hope that you are not intimidated by the tedious paragraphs describing the algorithm. If you have read all of the above and still do not fully understand, how to draw an inclined section, I strongly advise you to pick up a piece of paper and a pencil and try to repeat all the steps after me - this will almost 100% help you learn the material.
I once promised a continuation of this article. Finally, I am ready to present you with a step-by-step construction of an inclined section of a part, closer to the level of homework. Moreover, the inclined section is defined in the third view (the inclined section is defined in the left view)
or write down our phone number and tell your friends about us - someone is probably looking for a way to complete the drawings
or Create a note about our lessons on your page or blog - and someone else will be able to master drawing.
Yes, everything is fine, but I would like to see how to do the same thing on a more complex part, with chamfers and a cone-shaped hole, for example.
Thank you. Aren't the stiffening ribs hatched on the sections?
Exactly. They are the ones who do not hatch. Because that's how they are general rules making cuts. However, they are usually shaded when making cuts in axonometric projections - isometry, dimetry, etc. When making inclined sections, the area related to the stiffener is also shaded.
Thank you, very accessible. Tell me, can an inclined section be done in the top view, or in the left view? If so, I would like to see a simple example. Please.
It is possible to make such sections. But unfortunately I don’t have an example at hand right now. And there is another interesting point: on the one hand, there is nothing new there, but on the other hand, in practice, such sections are actually more difficult to draw. For some reason, everything starts to get confused in the head and most students have difficulties. But don't give up!
Yes, everything is fine, but I would like to see how the same thing is done, but with holes (through and not through), otherwise they never turn into an ellipse in the head
help me with a complex problem
It's a pity that you wrote here. If you could write to us by email, maybe we could have time to discuss everything.
You explain well. What if one of the sides of the part is semicircular? There are also holes in the part.
Ilya, use the lesson from the section on descriptive geometry “Section of a cylinder by an inclined plane.” With its help you can figure out what to do with the holes (they are essentially cylinders too) and with the semicircular side.
I thank the author for the article! It’s brief and easy to understand. About 20 years ago I was gnawing on the granite of science, now I’m helping my son. I forgot a lot, but your article brought me back fundamental understanding topics. I’ll go figure out the inclined section of the cylinder)
Add your comment.
Let's look at how to construct a section of a pyramid, using specific examples. Since there are no parallel planes in the pyramid, constructing the line of intersection (trace) of the cutting plane with the plane of the face most often involves drawing a straight line through two points lying in the plane of this face.
In the simplest problems, you need to construct a section of a pyramid with a plane passing through given points that already lie on the same face.
Example.
Construct plane section (MNP)
Triangle MNP - pyramid section
Points M and N lie in the same ABS plane, therefore, we can draw a straight line through them. The trace of this line is the segment MN. It is visible, which means we connect M and N with a solid line.
Points M and P lie in the same ACS plane, so we draw a straight line through them. Trace is a segment MP. We don’t see it, so we draw the segment MP with a stroke. We construct the trace PN in the same way.
Triangle MNP is the required section.
If the point through which you want to draw a section lies not on an edge, but on a face, then it will not be the end of the trace-segment.
Example. Construct a section of the pyramid by a plane passing through points B, M and N, where points M and N belong, respectively, to the faces ABS and BCS.
Here points B and M lie on the same face of ABS, so we can draw a straight line through them.
Similarly, we draw a straight line through points B and P. We have obtained traces BK and BL, respectively.
Points K and L lie on the same face of ACS, so we can draw a straight line through them. Its trace is the segment KL.
Triangle BKL is the required section.
However, it is not always possible to draw a straight line through the data in the point condition. In this case, you need to find a point lying on the intersection line of the planes containing the faces.
Example. Construct a section of the pyramid with a plane passing through points M, N, P.
Points M and N lie in the same ABS plane, so a straight line can be drawn through them. We get the trace MN. Likewise - NP. Both marks are visible, so we connect them with a solid line.
Points M and P lie in different planes. Therefore, we cannot connect them with a straight line.
Let's continue the straight line NP.
It lies in the plane of the BCS face. NP intersects only with lines lying in the same plane. We have three such direct lines: BS, CS and BC. The lines BS and CS already have points of intersection - these are just N and P. This means that we are looking for the intersection of NP with the line BC.
The intersection point (let's call it H) is obtained by continuing the lines NP and BC to the intersection.
This point H belongs both to the plane (BCS), since it lies on the line NP, and to the plane (ABC), since it lies on the line BC.
Thus, we received another point of the cutting plane lying in the plane (ABC).
We can draw a straight line through H and a point M lying in the same plane.
We get the MT trace.
T is the point of intersection of lines MH and AC.
Since T belongs to the line AC, we can draw a line through it and point P, since they both lie in the same plane (ACS).
The 4-gon MNPT is the desired section of the pyramid by a plane passing through the given points M,N,P.
We worked with the line NP, extending it to find the point of intersection of the cutting plane with the plane (ABC). If we work with direct MN, we arrive at the same result.
We reason like this: line MN lies in the plane (ABS), therefore it can intersect only with lines lying in the same plane. We have three such lines: AB, BS and AS. But with straight lines AB and BS there are already intersection points: M and N.
This means that, extending MN, we look for the point of intersection of it with the straight line AS. Let's call this point R.
Point R lies on line AS, which means it also lies in the plane (ACS) to which line AS belongs.
Since point P lies in the plane (ACS), we can draw a straight line through R and P. We get a trace of PT.
Point T lies in the plane (ABC), so we can draw a straight line through it and point M.
Thus, we obtained the same MNPT cross section.
Let's look at another example of this kind.
Construct a section of the pyramid with a plane passing through points M, N, P.
Draw a straight line through points M and N lying in the same plane (BCS). We get the trace MN (visible).
Draw a straight line through points N and P lying in the same plane (ACS). We get a PN (invisible) trace.
We cannot draw a straight line through points M and P.
1) Line MN lies in the plane (BCS), where there are three more lines: BC, SC and SB. The lines SB and SC already have intersection points: M and N. Therefore, we are looking for the intersection point MN with BC. Continuing these lines, we get point L.
Point L belongs to line BC, which means it lies in the plane (ABC). Therefore, we can draw a straight line through L and P, which also lies in the plane (ABC). Her trail is PF.
F lies on the line AB, and therefore in the plane (ABS). Therefore, through F and point M, which also lies in the plane (ABS), we draw a straight line. Her trail is FM. The quadrilateral MNPF is the required section.
2) Another way is to continue straight PN. It lies in the plane (ACS) and intersects the lines AC and CS lying in this plane at points P and N.
This means that we are looking for the point of intersection of PN with the third straight line of this plane - with AS. We continue AS and PN, at the intersection we get point E. Since point E lies on the line AS, belonging to the plane (ABS), we can draw a straight line through E and point M, which also lies in (ABS). Her trail is FM. Points P and F lie on the water plane (ABC), draw a straight line through them and get a trace PF (invisible).
There are 2 main methods for constructing sections of polyhedra:
Axiomatic method for constructing sections
1. Trace method
Example 1.
On the edges AA" and B"C" of the prism ABCA"B"C" we define points P and Q, respectively. We construct a section of the prism by a plane (PQR), the point R of which we define in one of the following faces:
a) VSSV"S";
b) A"B"C";
c) ABC
Solution.
A) 1) Since the points Q and R lie in the plane (ВСС"), then the straight line QR lies in this plane. Let's draw it. This is the trace of the plane (PQR) onto the plane (ВСС"). (Fig.1)
2) Find points B" and C" at which straight line QR intersects straight lines BB" and SS, respectively. Points B" and C" are traces of the plane (PQR) on straight lines BB" and SS, respectively.
3) Since points B"" and P lie in the plane (ABV"), then the straight line B""P lies in this plane. Let's draw it. The segment B**P is the trace of the plane (PQR) on the face ABC"A".
4) Since points P and C lie in the plane (ACC"), then the straight line PC"" lies in this plane. Let's draw it. This is the trace of the plane (PQR) on the plane (ACC").
5) Find the point V at which the straight line PC"" intersects the edge A"C". This is the plane trace (PQR) on edge A"C".
6) Wheelbarrow since the points Q and V lie in the plane (A"B"C"), then the straight line QV lies in this plane. Let's draw the straight line QV. The segment QV is the trace of the plane (PQR) on the face ABC. So, we get the polygon QB ""PV - the required section.
b) 1) Since the points Q and R lie in the plane (A"B"C"), then the line QR lies in this plane. Let's draw it. This is the trace of the plane (PQR) on the plane (A"B"C"). (Fig. .2)
2) Find points D" and E" at which straight line QR intersects straight lines A"B" and B"C", respectively. Since point D" lies on the edge A"B", the segment QD" is a trace of the plane (PQR) on the edge A"B"C".
3) Since points D" and P lie in the plane (ABB"), then the straight line D"P lies in this plane. Let's draw it. This is the trace of the plane (PQR) on the plane (ABB"), and the segment D"P is the trace plane (PQR) on the face АВВ "А".
4) Since the points P and E "lie in the plane (ACC"), then the straight line PE lies in this plane. Let's draw it. This is the trace of the plane (PQR) on the plane (ACC").
5) Find point C""=PE""CC". Since point C"" lies on the edge CC", then the segment PC"" is the trace of the plane (PQR) on the face ACC"A".
6) Since points Q and C"" lie in the plane (ВСС"), then the straight line QC"" lies in this plane. Let's draw it. This is the trace of the plane (PQR) on the plane (ВСС"), and the segment QC"" is plane trace (PQR) on the face ВСС "В". So, we have obtained the polygon QD "PC" - this is the desired section.
V) 1) Of the three given points P, Q and R, no two lie in any one of the planes of the prism faces, so we find the main trace of the plane (PQR) (i.e., the line of intersection of the plane (PQR) with the plane (ABC), selected as the main one). To do this, we first find the projections of points P, Q and R onto the plane (ABC) in the direction parallel to the side edge of the prism. Since point P lies on edge AA, then point P coincides with point A. Since point Q lies in the plane (BCC), then in this plane through point Q we draw a line parallel to straight line BB" and find point Q ", in which the drawn straight line intersects the straight line BC. Since point R, by condition, lies in the plane chosen as the main one, then point R" coincides with point R. (Fig. 3)
2) The plane is defined by parallel straight lines PP" and QQ". Let's draw straight lines PQ and P"Q" in this plane and find the point S=PQ intersects P"Q". Since the point S" lies on the straight line PQ, then it lies in the plane (PQR), and since the point S" lies on the straight line P"Q, then it lies in the plane (ABC). Thus, the point S" is a common point of the planes (PQR) and (ABC). This means that the planes (PQR) and (ABC) intersect along a straight line passing through the point S".
3) Since point R coincides with point R", then point R is another common point of the planes (PQR) and (ABC). Thus, straight line S"R is the main trace of the plane (PQR). Let's draw this line. As we see from the figure, the straight line S"R intersects the edges AB and BC of the base of the prism, respectively, at points S" "and S""".
4) Since the points S""" and Q lie in the plane (ВСС"), then the straight line S""" Q lies in this plane. Let's draw it. This is the trace of the plane (PQR) on the plane (ВСС"). And the segment S""" Q, is the trace of the plane (PQR) on the face ВСС"В".
5) Similarly, we find the segment S"" P - the trace of the plane (PQR) on the face ABC"A".
7) Find the point F=PC"" intersects A"C" and then obtain the segment PF - the trace of the plane (PQR) on the face ACC"A".
8) Points Q and F lie in the plane A"B"C", therefore straight line QF lies in the plane (A"B"C"). Let's draw the straight line QF and get the segment QF - the trace of the plane (PQR) on the face A "B" C. So, we have obtained the polygon QS"""S""PF - the desired section.
3 note. Let us show another way to find the point C"", in which we do not find the point of intersection of the line S""" Q with the line C"C". We will reason as follows. If the trace of the plane (PQR) on the straight line CC" is a certain point V, then its projection onto the plane (ABC) coincides with the point C. Then the point S""""= V"P" intersects VP lies on the main trace S"R of the plane (PQR). We construct this point S"""" as the point of intersection of the lines V"P" (this is line SA) and S"R. And then we draw the line S"""P. It intersects the line SS" at point V.
Example 2.
On the edge MB of the pyramid МАВСD we define point P, on its face MCD we define point Q. Let us construct a section of the pyramid by plane (PQR), the point R of which we define:
a) on the edge of the MS;
b) on the verge of MAD;
c) in the plane (MAS), outside the pyramid.
Solution.
a) The trace of the plane (PQR) on the face MBC is the segment PR, and its trace on the face MCD is the segment RD", where point D" is the point of intersection of the straight line RQ with the edge MD. It is clear that the plane (PQR) has traces on the faces MAD and MAB (since the plane (PQR) has common points with these faces). Let's find the trace of the plane (PQR) on the straight line MA. Let's do it like this:
1) Let's construct points P", Q" and R" - projections of points P, Q and R from the center M onto the plane (ABC), thus taken as the main plane. (Fig. 4)
3) If the plane (PQR) intersects the straight line MA at some point V, then the point V" coincides with point A and the point S""" = VQ intersects V"Q" lies on the straight line S" S"". In other words, at the point S""" three lines intersect: VQ, V"Q"" and S" S"". The last two straight lines of these three are already in the drawing. Therefore, we will construct the point S""" as the point of intersection of the lines V"Q" and SS"".
4) Let's draw a line QS""" (it coincides with the line VQ, since the line VQ must pass through the point S""", i.e. points V, Q and S""" lie on the same line).
5) Find the point V at which the straight line QS"" "intersects the straight line MA. Point V is the trace of the plane (PQR) on the edge of MA. Further, it is clear that the segments PV and VD" are the traces of the plane (PQR) respectively on the faces of MAB and MAD. Thus, the polygon PRD"V is the required section.
b) 1) We take the plane (ABC) as the main plane and construct points P", Q" and R" - projections of points P, Q and R, respectively, onto the plane (ABC). The center of this interior design is point M. (Fig. 5.)
2) We build a straight line S"S"" - the main trace of the plane (PQR).
3) If the plane (PQR) intersects the straight line MA at point V, then point V" - the projection of point V onto the plane (ABC) from the center M - coincides with point A, and the straight lines S" S "", V "R" and the straight line VR, the point V of which we have not yet constructed, intersect at the point S""". We find this point S"""=V"R" intersects S"S"" ."", and find the point V=RS""" MA intersects. Further construction is clear. The required section is the polygon PVD"T.
V)
(Fig. 6.) Let point R be located in the plane (MAC) as shown in Fig. 6.
1) We take the plane (ABC) as the main plane and construct points P", Q" and R" - projections of points P, Q and R, respectively, onto the plane (ABC). (The design center is point M.)
2) We build a straight line S"S"", - the main trace of the plane (PQR).
3) Find the point V - plane trace (PQR) on the MA straight line. Point V" - the projection of point V onto the plane (ABC) from the center M - coincides in this case with point A.
4) Find the point S"""= P"V" intersects S"S"", and then the point V =PS""" intersects MA.
5) We obtain a trace of the PV plane (PQR) on the plane (MAB).
6) Find the point T - trace of the plane (PQR) on the straight line MO. It is clear that point T" in this case coincides with point D. To construct point T, we construct point S""""=Q"T" intersected by S"S"", and then point T = QS""""intersected by MT" .
7) The set of traces PV, VT, TC", and C"P, i.e. the polygon PVTC" is the desired section.
Combined method for constructing sections
The essence of the combined method for constructing sections of polyhedra is the application of theorems on the parallelism of lines and planes in space in combination with the axiomatic method.
Example No. 1.
On the edges AB and AD of the MABCD pyramid, we define points P and Q, respectively, the midpoints of these edges, and on the edge MC we define a point R. Let us construct a section of the pyramid with a plane passing through points P, Q and R.
Solution
(Figure 14):
1). It is clear that the main trace of the plane PQR is the straight line PQ.
2). Let us find the point K at which the MAC plane intersects the straight line PQ. Points K and R belong to both the PQR plane and the MAC plane. Therefore, by drawing the straight line KR, we get the line of intersection of these planes.
3). Let's find the point N=AC BD, draw a straight line MN and find the point F=KR MN.
4). Point F is the common point of the planes PQR and MDB, that is, these planes intersect along a straight line passing through point F. At the same time, since PQ is the midline of the triangle ABD, then PQ is parallel to BD, that is, the line PQ is parallel to the plane MDB. Then the plane PQR passing through the straight line PQ intersects the plane MDB along a straight line parallel to the straight line PQ, that is, parallel to and straight BD. Therefore, in the plane MDB through point F we draw a line parallel to line BD.
5). Further constructions are clear from the figure. As a result, we obtain the polygon PQD"RB" - the desired section.
1. Construction of a section passing through a given line parallel to another given line.
Let, for example, you need to construct a section of a polyhedron with a plane @ passing through a given line p parallel to a second given line q. In general, solving this problem requires some preliminary constructions, which can be carried out according to the following plan:
1). Through the second line q and some point W of the first line p we draw a betta plane (Fig.
2). In the betta plane, through the point W we draw a straight line q" parallel to q.
3). The intersecting lines p and q." The @ plane is determined. At this point, the preliminary constructions end and you can proceed to the construction of the direct section of the polyhedron by the @ plane. In some cases, the peculiarities of a particular problem allow you to implement a shorter solution plan. Let's consider examples.
Example No. 2.
On the edges BC and MA of the pyramid MABC, we define points P and Q, respectively. We construct a section of the pyramid with a plane @ passing through the straight line PQ parallel to the straight line AR, a point R, which we define as follows: a). On edge MB; b). It coincides with point B; V). On the verge of MAB.
Solution:
A)
.(figure) The plane passing through the second line, that is, line AR, and the point Q taken on the first line are already in the image. This is the MAB plane.
2). In the MAB plane, through point Q we draw a straight line QF parallel to AR.
3). The intersecting lines PQ and QF determine the plane @ (this plane PQF) - the plane of the desired section. Let's construct this section using the trace method.
4). Point B coincides with point F" - the projection of point F onto the plane ABC (from the center M), and point A coincides with point Q" - the projection of point Q onto this plane. Then point S"=FQ F"Q" lies on the main trace of the cutting plane @. Since point P lies on the main trace of the cutting plane, the straight line S"P is the main trace of the plane @, and the segment S""P is the trace of the plane @ on the verge of ABC. It is further clear that point P should be connected to point F. As a result, we obtain the quadrilateral PFQS"" - the desired section.
b)
(Figure: A plane passing through line AB and point P of line PQ has already been constructed in the image. This is plane ABC. Let’s continue construction according to the above plan.
2). In the ABC plane, through point P we draw a line PD parallel to the line AB.
3). The intersecting lines PQ and PD determine the alpha plane (this is the PQD plane) - the plane of the desired section. Let's construct this section.
4). It is clear that the trace of the alpha plane on the MAC face is the segment DQ.
5). We will carry out further constructions taking into account the following considerations. Since straight line PD is parallel to straight line AB, then straight line PD is parallel to plane MAB. Then the alpha plane passing through the straight line PD intersects the MAB plane along a straight line parallel to the straight line PD, that is, to the straight AB. So, in the plane MAB through point Q we draw a straight line QE parallel to AB. The segment QE is the trace of the alpha plane on the MAB face.
6). Let's connect point P to point E. The segment PE is the trace of the alpha plane on the face of the MBC. Thus, the quadrilateral PEQD is the required section. coincides with point A, and point L" coincides with R"=MR BC. Then the point S"=LQ L"Q" lies on the main trace of the cutting plane alpha. This main trace is the straight line S"P, and the trace of the alpha plane on the face ABC is the segment S""P. Further, straight line PL is the trace of the alpha plane on the plane MVS, and the segment PN is the trace of the alpha plane on the face of MVS. So, the quadrilateral PS""QN is the required section.
Example 3.
On the diagonals AC and C"E" of the bases of the prism ABCDEA"B"C"D"E" we define points P and Q, respectively. Let us construct a section of the prism with the alpha plane passing through the line PQ parallel to one of the following lines: a).AB; b) .AS"; V). BC" Solution:
A)
(Figure Plane passing through line AB - the second given line and point P, taken on the first line, has already been constructed. This is the plane ABC.
2). In the plane ABC, through point P we draw a line parallel to line AB, and find points K and L at which this line intersects lines BC and AE, respectively. B"C" are also parallel to each other. Taking into account that KL is parallel to AB and A"B" is parallel to AB, let us draw a line parallel to line A"B" in the plane A"B"C" through point Q and find points F and T at which this line intersects, respectively straight lines C"D" and A"E". Next we obtain segment TL - trace of the alpha plane on the face AEE"A", point S"=KL CD, straight line S"F - trace of the alpha plane on the plane CDD", segment FC"" - the trace of the alpha plane on the face CDD"C" and, finally, the segment C""K - the trace of the alpha plane on the face BCC"B". As a result, we obtain the polygon KLTFC"" - the desired section. 
b)
(Figure Let's draw a plane through the line AC" - the second given line, and the point P taken on the first line. This is the plane ACC".
2). In the ACC plane, through point P we draw a line parallel to line AC" and find point C" at which this line intersects line CC".
3). The intersecting lines PQ and PC"" determine the alpha plane (plane C""PQ) - the plane of the desired section. Let's construct this section, for example, using the trace method. One point belonging to the trace of the alpha plane onto the ABC plane, which we take as the main one, is already in the drawing. This is point P. Let's find another point on this trace.
4). The projection of point C"" onto the plane ABC is point C, and the projection of point Q is point Q" - the point of intersection of straight line CE with a straight line passing in plane CEE" through point Q parallel to straight line EE." Point S"=C""Q CQ " is the second point of the main trace of the alpha plane. So, the main trace of the alpha plane is the straight line S"P. It intersects the sides BC and AE of the base of the prism, respectively, at points S"" and S""". Then the segment S""S""" is the trace of the cutting plane alpha on the face ABCDE. And the segment S""C"" is the trace of the alpha plane on the face BCC"B". It is easy to see that the lines C"" Q and EE" lie in the same plane. Let us find the point E"" = C""Q EE". Then it is clear to obtain further traces of the alpha plane: S"""S"", S"""T, TF and FC"". As a result, we obtain the polygon S""S"""TFC"" - the desired section.
V)
(Figure: Through the second given straight line - straight line BC" - and, for example, through point P lying on the first given straight line, we draw a plane. We will do this using the trace method. It is easy to establish that the main trace of this plane BC "P is straight line BP. Then we find the point S"=BP CD and trace S"C" of plane BC"P and plane CDD".
2).In the plane BC"P, through point P we draw a line parallel to straight line BC". Let us denote the point of intersection of the drawn line with the line S "C" as V.
3). The intersecting lines PQ and PV determine the alpha plane (PQV plane) - the plane of the desired section. Let's construct this section.
4). We find points Q" and V" - projections of points Q and V, respectively, onto the ABC plane, which we take as the main plane. Then we find the point S""=QV Q"V". This is one of the main trace points of the alpha plane. And there is already one more point in this trace. This is a given point P. So, straight line S""P is the main trace of the alpha plane, and the resulting segment S"""S"""" is the trace of the alpha plane on the face ABCDE. The further course of construction is clear: S"""" "=S""P CD, S"""""V, points C""=S"""""V CC" and F=S"""""V C"D", then FQ and point T= FQ A"E" and finally TS"""". As a result, we obtain the polygon S"""C""FTS"""" - the desired section.
Note: Let us briefly outline the process of solving example 3,c, in which point Q was taken on the first given line, and not point P (Figure 22).
1). We construct the plane BC"Q (this is the plane BC"E").
2). The plane BC"Q intersects the plane ABC along a straight line BN parallel to C"E" (for construction, you can use the fact that BN is parallel to CE).
3). In the plane BC"Q, through point Q we draw a straight line QM parallel to BC" (M=QM BN).
4). We construct a section of the prism by the plane defined by the intersecting lines PQ and QM. This can be done in the following order: MP, S"=MP AE and S""=MP BC, S"""=MP CE, C""=S""""Q CC", S"""C" ", F=S"""C"" C"D", FQ, T=FQ A"E", TS. Polygon S""C""FTS" is the required section.
2. Construction of a section passing through a given point parallel to two given intersecting lines.
Let it be necessary to construct a section of a polyhedron with a plane passing through given point To two parallel given crossing lines l and m. At background:#FFCCCC; border:outset #CC33FF 1.5pt">
1. Select some point W. (This point may lie on one of the given intersecting lines, or may coincide with point K.)
2.Draw straight lines l" and m" through point W. (Naturally, if point W lies on one of the lines, for example on line l, then line l" coincides with line l.)
3. The intersecting lines l" and m" determine the betta plane - the plane of the auxiliary section of the polyhedron. We construct a section of the polyhedron by the betta plane.
4. Construct sections of the polyhedron with the alpha plane passing through point K, parallel to the beta plane.
Let's look at examples of application of the outlined plan.
Example 4.
On the edges AD and C"D" of the prism ABCDA"B"C"D", we define points P and Q, respectively, and on the edge DD" we define point K. Let us construct a section of the prism with the alpha plane passing through point K parallel to the straight line PQ and one of the following straight lines: a) AB; b) A"B; c) BR, the point R of which is set on the edge A"D".
Solution. a)
(Fig. 2 Let point W coincide with point P.
2) In the plane ABC through point P we draw a line parallel to line AB. Let us find the point E at which the drawn straight line intersects the straight line BC.
3) The intersecting straight lines PQ and PE determine the betta plane - the plane of the auxiliary section. Let's construct a section of the prism using the beta plane. Direct PE and points C"" and D"" are traces of the betta plane, respectively, on straight lines CC" and DD". Then we build a straight line D""P and get point F on edge A"D". Thus, the cross section of the prism by the beta plane is the polygon PEC""QF.
4) Now we construct a section of the prism with the alpha plane passing through point K parallel to the beta plane. As a result, we obtain triangle KLN - the desired section.
b)
(Fig. Let point W coincide with point Q. In order to draw a line parallel to straight line A"B through point Q, first draw a gamma plane through straight line A"B and point Q. Let's do it like this. Find point Q" - the projection of point Q onto plane ABC and draw line AQ". It is clear that AQ" is parallel to A"Q. Now through point B in the plane ABC we draw line l" parallel to AQ". The intersecting lines A"B and l" determine the gamma plane. In the gamma plane through the point Q draw a straight line l"" parallel to A"B.
3) The intersecting lines PQ and l"" determine the betta plane - the plane of the auxiliary section of the prism. Let's construct this section. For this, we find the point S"=l" intersects l"", and then the straight line PS" - the main trace of the betta plane. Next we find the point s""=PS" intersects CD and draw the straight line S""Q - the trace of the betta plane on the CDD plane ". We get point D" - the trace of the betta plane on the straight line DD". Point D"" and point P lie in the ADD plane". Therefore, straight line PD"" is the trace of the betta plane on the ADD plane", and the segment PF is the trace of the betta plane on the face ADD"A". Thus, the section of the prism by the betta plane is the quadrilateral PS""QF. (Please note: QF is parallel to PS." And this, naturally, is so. After all, the bases of the prism lie in parallel planes. This circumstance could be used when constructing a section of the prism with the betta plane.)
4) Now we construct a section of the prism with the alpha plane passing through point K parallel to the beta plane. This construction is not difficult to complete. As a result, we obtain triangle KLN - the desired section.
V)
(Fig. We choose point Q as point W.
2) Through straight line BR and point Q we draw the gamma plane. The gamma plane intersects the ABC plane along a straight line l" parallel to QR. To construct a straight line l" we construct points R" and Q" - projections of points R and Q onto the ABC plane, respectively - and draw a straight line Q "R", and then in the ABC plane through the point B, draw a straight line l" parallel to Q"R. In the gamma plane, through the point Q, draw a straight line l"" parallel to BR. We obtain the point S"=l" intersects l"".
3) The intersecting straight lines PQ and l"" determine the betta plane - the plane of the auxiliary section of the prism. Let's construct this section. It is clear that the straight line PS" is the main trace of the betta plane. We then find the points S""= PS" intersects CD, S"""= RS" intersects BC and C"" = QS"" intersects CC". We obtain the segments RS"" ", S"""C"" and C""Q are traces of the betta plane, respectively, on the faces ABCD, ВСС"В and CDD"С". Next, we either draw a straight line in the plane A"B"C" parallel to the trace PS" and get point F, or find point D""=S""Q intersects DD" and draw a straight line D""P. This straight line will intersect line A "D" at point F. Thus, we obtain two more traces of the betta plane: QF n FP. So, the polygon PS"""C""QF is a section of the prism by the betta plane.
4) Now let’s construct a section of the prism with the alpha plane passing through point K parallel to the beta plane. As a result, we obtain triangle KLN - the desired section.
Example 5.
On the edges MB and MA of the pyramid МАВСD we will define points P and K, respectively, and on the segment AC we will define point Q. We will construct a section of the pyramid with the alpha plane passing through point K parallel to the straight line PQ and one of the following straight lines: a) CD; b) MS; c) RV, the points R and V of which will be set respectively on the edges AB and MC of the pyramid.
Solution.
a)
(Fig. 2 In the ABC plane, through point Q, draw a line parallel to straight line CD, and find points S." S"" and S""", at which this straight line intersects lines BC, AD and AB, respectively.
2) The intersecting lines PQ and S"S"" determine the betta plane - the plane of the auxiliary section of the pyramid. Let's construct this section. The main trace of the betta plane is the straight line S"S"". The segment PS" is the trace of the betta plane on the face MBC, the straight line PS""" is its trace on the plane MAB, the segment PA" is on the face MAB, the segment A"S"" is on the face MAD.
b)
(Fig. 27.) Let us construct the given section in the following order:
1) In the MAC plane through
point Q draw a straight line QA parallel to MC
2) Let’s construct an auxiliary section of the pyramid with a plane determined by. To this end, we will find the point S"=PA" intersects AB, draw a straight line S"Q, which is the main trace of the plane PQA", obtain the points S""=S"Q intersects AD and S""=S"Q intersects BC and connect point A" with point S"", and point P with point S""". The quadrilateral PA"S""S""" is an auxiliary section of the pyramid. The plane of this section is parallel to straight lines PQ and MC, but does not pass through point K .
3) Now let's construct a section of the pyramid with a plane passing through point K parallel to the plane PQA." As a result, we obtain the quadrilateral B"KFE - the desired section.
a)
(Fig. 28.) Let us construct a given section of the pyramid by first constructing an auxiliary section of it with a plane passing through the straight line PQ parallel to the straight line RV. Let's do this in the following order:
1) Construct a point S"=PV intersects BC and draw a straight line S"R.
2) The intersecting lines S"V and S"R define the plane. In this plane, through point P we draw a straight line PS"" parallel to RV.
3) The intersecting lines PQ and PS"" determine the plane of the auxiliary section of the pyramid. Let's construct this section. We find sequentially the straight line S""Q - the main trace of the plane of the auxiliary section, then the points T"=S""Q intersects BC, T""=S""Q intersects AB and T""=S""Q intersects CD, Let's carry out then the straight line T"P and find the point E = T"P intersects "MC. Let's connect point P to point T"", and point E to T""". The quadrilateral PT""T""E is an auxiliary section of the pyramid. The plane of this section is parallel to the lines PQ and RV, but does not pass through the point K. Now we construct a section of the pyramid with a plane passing through the point K parallel to the plane of the auxiliary section. As a result, we obtain the quadrilateral KV "C"D" is the required section.
Finding the cross-sectional area in polyhedra.
Task No. 1.
Task No. 2
Task No. 3.
Task No. 4.
Task No. 5.
Task No. 6.
Problem No. 7
Task No. 8.
Using the properties of similar triangles.
Therefore, below are presented several simple problems in which such triangles play the main role - especially since they also need to be constructed (and seen!!!) using a standard stereometric technique: intersect one plane with another plane and construct their intersection line along the two points common to planes.
Task No. 1.
Task No. 2
Task No. 3
Task No. 4
Problem #5
To find the distance between intersecting lines, you can use four main methods:
1) Finding the length of the common perpendicular of two intersecting lines, that is, a segment with ends on these lines and perpendicular to both.
2) Finding the distance from one of the intersecting lines to a plane parallel to it, passing through the other line.
3) Finding the distance between two parallel planes passing through given intersecting lines.
4) Finding the distance from a point - which is a projection of one of the intersecting lines onto a plane perpendicular to it - to the projection of another line onto the same plane.
Problem No. 18
Problem No. 19
Present 4 options for solving this problem and choose the most rational one. Justify your choice.
Problem No. 20
Problem No. 21
Problem No. 22
Finding the distance and angle between intersecting lines in a polyhedron.
Task No. 1.
Task No. 2.
Task No. 3.
passing through side rib and the median of the base intersecting with it, and a plane passing through the same median and the middle of any other lateral edge.
Sections.
Task No. 1.
Task No. 2.
Task No. 3.
Two opposite edges of a tetrahedron are perpendicular, and their lengths are equal to a and b, the distance between them is c. A cube is inscribed in a tetrahedron, the four edges of which are perpendicular to these two edges of the tetrahedron, and on each face of the tetrahedron there are exactly two vertices of the cube. Find the edge of the cube.
Task No. 4.
Task No. 5.
Task No. 6.
Task No. 7.
Task No. 8.
Task No. 9.
Ratio of volumes of parts of a polyhedron.
Task No. 1.
Task No. 2.
Task No. 3.
Task No. 4.
Projections and sections of regular polyhedra.
Task No. 1.
Show that the projections of the dodecahedron and icosahedron onto planes parallel to their faces are regular polygons.
Task No. 2.
Show that the projection of a dodecahedron onto a plane perpendicular to the line passing through its center and the middle of the edge is a hexagon (and not a decagon).
Task No. 3.
a) show that the projection of the icosahedron onto the plane. perpendicular to the line passing through its center and vertex is a regular decagon. b). Prove that the projection of a dodecahedron onto a plane perpendicular to the line passing through its center and vertex is an irregular 12-gon.
Task No. 4.
Is there a section of a cube that is a regular hetagon?
Task No. 5.
Is there a section of the octahedron that is a regular hexagon?
Task No. 6.
Is there a section of the dodecahedron that is a regular hexagon?
Task No. 7.
All faces ABC and ABD of the icosahedron have a common edge AB. A plane parallel to plane ABC is drawn through vertex D. Is it true that the section of the icosahedron by this plane is a regular hexagon?
Answers to problems by topic:
4. Angle between planes.
5. Sections
6. Ratio of volumes of parts of a polyhedron.
7. Projections and sections of regular polyhedra.
1. Finding the cross-sectional area in polyhedra.
The solution of the problem
№1 №2 №3 №4 №5 №6 №7 №8
Task No. 1.
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Task No. 2.
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Task No. 3.
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Task No. 4.
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Task No. 5.
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Task No. 6.
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Task No. 7.
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2. Using the properties of similar triangles.
The solution of the problem
№1 №2 №3 №4 №5
Task No. 1.
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2nd case 
Task No. 2.
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Task No. 3.
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Point C belongs to the plane CB"A"D (since CD" is perpendicular to C"D as the diagonal of a square and since B"C" is perpendicular to the plane CC"D"D, from which it follows that B"C" is perpendicular to CE, we obtain CE is perpendicular to B"C" and CE is perpendicular to C"D). Then we draw EF perpendicular to B"D and then we obtain B"D is perpendicular to CF (by the theorem of three perpendiculars: CF with respect to the plane AB"C"D is inclined, CE - perpendicular and EF - projection of the inclined CF; then it is perpendicular to the inclined CF itself.) Since EF and CF belong respectively to both planes, the angle phi (angle CFE) is the desired one.
This justification is followed by a simple computational part.
"B"EF and D"C"EF), as a result of which the perpendiculars A""M and D""M, drawn in both figures to their line of intersection, will end up at one point M, moreover, inside and not outside the prism , since angles B"A""D and C"D""A are obtuse (B"D and greater than BD=AC=A""C"" and C"A greater than AC=BD=B""D"" Next, having found the diagonals and sides of rhombuses, you can find the segments A""M and D""M using, for example, two formulas for the area of a rhombus 
Note: Of course, in this and similar problems, no dimensions of the polyhedron (for example, “a”) are needed, therefore, when selecting the numerical values of the parameter “k” for various options problem, the content of its condition in the appropriate place should be formulated, for example, like this: “... in a prism, the height of which is so many times greater than the side of the base...”, etc.
3. Finding the distance and angle between intersecting lines in a polyhedron.
The solution of the problem
№1 №2 №3 №4 №5
Task No. 1.
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№1
Solving the problem in the first way
assumes:
- a difficult justification for the fact that the desired perpendicular (h skr.) with ends on two given intersecting lines is located inside the cube (and not outside it);
- approximate determination of the location of this perpendicular;
- the guess that to find the length of the segment h skr. It is necessary, using the theorem of three perpendiculars, to project it onto the adjacent faces of the cube, to which the crossing straight lines (diagonals) belong, and only then approach a simple solution:
| 2.
Solving the problem in the second way involves
the following actions: |
Construction of another cutting plane passing through the diagonal B"D and intersecting the second of the intersecting straight lines A"C"; it is convenient to select the diagonal section BB"D"D with this plane, this sign of the perpendicularity of two planes, the plane BB"D"D is perpendicular to the plane ACD", since the plane BB"D"D passes through a line (B"D) perpendicular to another plane (ACD"). Next, a line of intersection of both planes is constructed along their 2 common points (D"O) and is fixed by the intersection of this line with the diagonal B"D (point N);
- and finally, according to the theorem that if a plane is perpendicular to one of the parallel lines, then it is also perpendicular to the other, from point O" belongs to A"C" we draw a segment O in the section plane BB"D"D until it intersects with D"O "M is parallel to B"D; in this case O"M will be perpendicular to the plane ACD" and therefore O"M = h skr.;
- then in the computational part of the solution, having considered the section BB"D'D and in it - right triangle OO'D', we find: As we see, both first methods are of little use for tasks that present at least some complexity
| 3.
Solving the problem in the third way involves
: |
And finally, in the computational part of the solution, you can use the technique from the previous solution method or resort to the similarity of triangles:
| 4.
Solving the problem in the fourth way involves: |
Task No. 3.
In this problem, to choose a solution method, the determining factor is the perpendicularity of the straight line AC to the diagonal plane ВB'D'D (since AC is perpendicular to ВD and AC is perpendicular to BB'), to which another straight line B'F belongs, i.e., the cutting plane BB' D'D is convenient for selecting it as the projection plane. And then comes the simple computational part:
1). Based on the similarity between triangle DFT and triangle D'FB', we find DT = kd;
2). From the similarity of triangle NOT and triangle BB'T we find ON: 
Task No. 4.
This problem is presented here to demonstrate the application of the second method (constructing a perpendicular from the first line to a parallel plane containing the second line) to the simplest situations of the location of crossing lines in such a complex polyhedron as a regular hexagonal prism.
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Task No. 5.
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5. Sections.
The solution of the problem
№1 №2 №3 №4 №5 №6
Task No. 1.
In any case, points A, B and C lie in the same plane, and therefore we can consider a section by the plane containing these points. Since the section plane passes through the point of contact of the spheres (spheres of the plane), and the sections are tangent to the circle (circle and straight line). Let O' and 0'' be the centers of the first and second circles. Since O'A || 0''B and points O', C and 0'' lie on the same straight line, angle AO'C = angle BO''C. Therefore, angle ACO' = angle BCO'', i.e. points A, B and C lie on the same straight line.
Task No. 2.
The axial section of this truncated cone is a circumscribed trapezoid ABCD with bases AD = 2R and BC = 2r. Let P be the point of contact of the inscribed circle with side AB, O be the center of the inscribed circle. In triangle ABO, the sum of the angles at vertices A and B is 90°, so it is rectangular. Therefore, AR: RO - RO: BP, i.e. RO'2 = AR*BP. It is also clear that AP = R and BP = r. Therefore, the radius PO of the sphere inscribed in the cone is equal to the square root of the product of R and r, which means S = 4n(R2 + Rr+ r2). Expressing the volume of a given truncated cone using formulas, we find that the area of its total surface is equal to 2n(R2 + Rr+ r2) = S/2 (it must be taken into account that the height of the truncated cone is equal to twice the radius of the sphere around which it is described).
Task No. 3.
The common perpendicular to these edges is divided by the planes of the cube faces parallel to them into segments of length y, x and z (x is the length of the edge of the cube; a segment of length y is adjacent to edge a). The planes of the cube's faces, parallel to these edges, intersect the tetrahedron along two rectangles. The smaller sides of these rectangles are equal to the edge of the cube x. Since the sides of these rectangles are easy to calculate, we get x = bу/с and x = az/с. Consequently, c = x + y + g = x + cx/b + ex/a, i.e. x = abc/(ab + bc + ca).
Task No. 4.
Each side of the resulting polygon belongs to one of the faces of the cube, so the number of its sides does not exceed 6. In addition, the sides belonging to opposite faces of the cube are parallel, since the lines of intersection of a plane with two parallel planes are parallel. Consequently, the cross section of a cube cannot be a regular pentagon, since it does not have parallel sides. It is easy to check that a regular triangle, a square and a regular hexagon can be sections of a cube.
Task No. 5.
Let's consider a circle, which is a section of a given body, and draw a straight line l through its center, perpendicular to its plane. This line intersects the given body along a certain segment AB. All sections passing through line l are circles with diameter AB.
Task No. 6.
Let us consider an arbitrary section passing through vertex A. This section is a triangle ABC, and its sides AB and AC are generators of a cone, i.e. have a constant length. Therefore, the cross-sectional area is proportional to the sine of the angle BAC. The BAC angle changes from 0° to f,
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Task No. 2.
Consider a cube whose vertices are located at the vertices of a dodecahedron. In our problem we are talking about a projection onto a plane parallel to the face of this cube. Now it is easy to verify that the projection of the dodecahedron is indeed a hexagon (Fig. 70).
Task No. 3.
a) The projection of the icosahedron under consideration transforms into itself when rotated by 36° (in this case, the projections of the upper faces transform into projections of the lower faces). Consequently, it is a regular 10-gonal (Fig. 71, a).
b) The projection of the dodecahedron under consideration is a 12-gon, transforming into itself when rotated by 60° (Fig. 71. b). Half of its sides are projections of edges parallel to the projection plane, and the other half of its sides are projections of edges not parallel to the projection plane. Therefore, this 12-gon is irregular.
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Task No. 4.
Exists. The midpoints shown in Fig. The 72 edges of the cube are the vertices of a regular hexagon. This follows from the fact that the sides of this hexagon are parallel to the sides of the regular triangle PQR, and their lengths are half the lengths of the sides of this triangle.
Task No. 6. Exists. Let us take three pentagonal faces with a common vertex A and consider the section by a plane intersecting these faces and parallel to the plane in which the three pairwise common vertices of the faces under consideration lie (Fig. 74). This section is a hexagon with pairs of parallel opposite sides. When rotated by 120° relative to the axis passing through vertex A and perpendicular to the cutting plane, the dodecahedron and cutting plane transform into themselves. Therefore, the section is a convex hexagon with angles of 120°, the lengths of the sides of which, alternating, take on two values. In order for this hexagon to be regular, it is enough that these two values are equal. When the cutting plane moves from one of its extreme positions to the other, moving away from the vertex A, the first of these values increases from 0 to d, and the second decreases from d to a, where a is the length of the edge of the dodecahedron. (d is the length of the diagonal of the face (d is greater than a). Therefore, at some point these values are equal, i.e. the section is a regular hexagon. |
|
Task No. 7.
No, that's not true. Let us consider the projection of the icosahedron onto the ABC plane. It is a regular hexagon (see Fig. 69). Therefore, the section under consideration would be a regular hexagon only if all 6 vertices connected by edges to points A, B and C (and different from A, B and C) lay in the same plane. But, as you can easily see, this is not true (otherwise it would turn out that all the vertices of the icosahedron are located on three parallel planes).
TASKS
|
2. Using the properties of similar triangles.
The solution of the problem
№1 №2 №3 №4 №5
Task No. 1.
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2nd case
Task No. 2.
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Task No. 3.
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Task No. 4.
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Do you know what is called the section of polyhedra by a plane? If you still doubt the correctness of your answer to this question, you can check yourself quite simply. We suggest you take a short test below.
Question. What is the number of the figure that shows the section of a parallelepiped by a plane?
So, the correct answer is in Figure 3.
If you answer correctly, it confirms that you understand what you are dealing with. But, unfortunately, even the correct answer to a test question does not guarantee you the highest grades in lessons on the topic “Sections of polyhedra.” After all, the most difficult thing is not recognizing sections in finished drawings, although this is also very important, but their construction.
To begin with, let us formulate the definition of a section of a polyhedron. So, a section of a polyhedron is a polygon whose vertices lie on the edges of the polyhedron, and whose sides lie on its faces.
Now let’s practice quickly and accurately constructing intersection points 
a given straight line with a given plane. To do this, let's solve the following problem.
Construct the intersection points of straight line MN with the planes of the lower and upper bases of the triangular prism ABCA 1 B 1 C 1, provided that point M belongs to the side edge CC 1, and point N belongs to edge BB 1.
Let's start by extending straight line MN in both directions in the drawing (Fig. 1). Then, in order to obtain the intersection points required by the problem, we extend the lines lying in the upper and lower bases. And now comes the most difficult moment in solving the problem: which lines in both bases need to be extended, since each of them has three lines.
In order to correctly complete the final step of construction, it is necessary to determine which of the direct bases are in the same plane as the straight line MN of interest to us. In our case, this is straight CB in the lower and C 1 B 1 in the upper bases. And it is precisely these that we extend until they intersect with the straight line NM (Fig. 2).
The resulting points P and P 1 are the points of intersection of the straight line MN with the planes of the upper and lower bases of the triangular prism ABCA 1 B 1 C 1 .
After analyzing the presented problem, you can proceed directly to constructing sections of polyhedra. The key point here will be reasoning that will help you arrive at the desired result. As a result, we will eventually try to create a template that will reflect the sequence of actions when solving problems of this type.
So, let's consider the following problem. Construct a section of a triangular prism ABCA 1 B 1 C 1 by a plane passing through points X, Y, Z belonging to edges AA 1, AC and BB 1, respectively.
Solution: Let's draw a drawing and determine which pairs of points lie in the same plane.
Pairs of points X and Y, X and Z can be connected, because they lie in the same plane.
Let's build additional point, which will lie on the same face as point Z. To do this, extend the lines XY and CC 1, because they lie in the plane of the face AA 1 C 1 C. Let's call the resulting point P.
Points P and Z lie in the same plane - in the plane of the face CC 1 B 1 B. Therefore, we can connect them. The straight line PZ intersects the edge CB at a certain point, let's call it T. Points Y and T lie in the lower plane of the prism, connect them. Thus, the quadrilateral YXZT was formed, and this is the desired section. 
Summarize. To construct a section of a polyhedron with a plane, you must:
1) draw straight lines through pairs of points lying in the same plane.
2) find the lines along which the section planes and faces of the polyhedron intersect. To do this, you need to find the intersection points of a straight line belonging to the section plane with a straight line lying in one of the faces.
The process of constructing sections of polyhedra is complicated because it is different in each specific case. And no theory describes it from beginning to end. There's really only one the right way learning to quickly and accurately construct sections of any polyhedra is a constant practice. The more sections you build, the easier it will be for you to do this in the future.
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Definition
Section is flat figure, which is formed when a spatial figure intersects with a plane and whose boundary lies on the surface of the spatial figure.
Comment
To construct sections of various spatial figures, it is necessary to remember the basic definitions and theorems about the parallelism and perpendicularity of lines and planes, as well as the properties of spatial figures. Let us recall the basic facts.
For a more detailed study, it is recommended to familiarize yourself with the topics “Introduction to Stereometry. Parallelism" and "Perpendicularity. Angles and distances in space”.
Important Definitions
1. Two lines in space are parallel if they lie in the same plane and do not intersect.
2. Two straight lines in space intersect if a plane cannot be drawn through them.
4. Two planes are parallel if they do not have common points.
5. Two lines in space are called perpendicular if the angle between them is equal to \(90^\circ\) .
6. A line is called perpendicular to a plane if it is perpendicular to any line lying in this plane.
7. Two planes are called perpendicular if the angle between them is \(90^\circ\) .
Important axioms
1. Through three points that do not lie on the same line, a plane passes through, and only one.
2. A plane, and only one, passes through a straight line and a point not lying on it.
3. A plane passes through two intersecting lines, and only one.
Important Theorems
1. If a line \(a\) that does not lie in the plane \(\pi\) is parallel to some line \(p\) that lies in the plane \(\pi\) then it is parallel to this plane.
2. Let the straight line \(p\) be parallel to the plane \(\mu\) . If the plane \(\pi\) passes through the line \(p\) and intersects the plane \(\mu\), then the line of intersection of the planes \(\pi\) and \(\mu\) is the line \(m\) - parallel to the straight line \(p\) .
3. If two intersecting lines from one plane are parallel to two intersecting lines from another plane, then such planes will be parallel.
4. If two parallel planes \(\alpha\) and \(\beta\) are intersected by a third plane \(\gamma\), then the lines of intersection of the planes are also parallel:
\[\alpha\parallel \beta, \ \alpha\cap \gamma=a, \ \beta\cap\gamma=b \Longrightarrow a\parallel b\]
5. Let the straight line \(l\) lie in the plane \(\lambda\) . If the line \(s\) intersects the plane \(\lambda\) at a point \(S\) not lying on the line \(l\), then the lines \(l\) and \(s\) intersect.
6. If a line is perpendicular to two intersecting lines lying in a given plane, then it is perpendicular to this plane.
7. Theorem about three perpendiculars.
Let \(AH\) be perpendicular to the plane \(\beta\) . Let \(AB, BH\) be the inclined plane and its projection onto the plane \(\beta\) . Then the line \(x\) in the plane \(\beta\) will be perpendicular to the inclined one if and only if it is perpendicular to the projection.
8. If a plane passes through a line perpendicular to another plane, then it is perpendicular to this plane.
Comment
Another important fact often used to construct sections:
in order to find the point of intersection of a line and a plane, it is enough to find the point of intersection of a given line and its projection onto this plane.
To do this, from two arbitrary points \(A\) and \(B\) of the straight line \(a\) we draw perpendiculars to the plane \(\mu\) – \(AA"\) and \(BB"\) (points \ (A", B"\) are called projections of points \(A,B\) onto the plane). Then the line \(A"B"\) is the projection of the line \(a\) onto the plane \(\mu\) . The point \(M=a\cap A"B"\) is the intersection point of the straight line \(a\) and the plane \(\mu\) .
Moreover, we note that all points \(A, B, A", B", M\) lie in the same plane.
Example 1.
Given a cube \(ABCDA"B"C"D"\) . \(A"P=\dfrac 14AA", \KC=\dfrac15 CC"\). Find the intersection point of the straight line \(PK\) and the plane \(ABC\) .
Solution
1) Because the edges of the cube \(AA", CC"\) are perpendicular to \((ABC)\), then the points \(A\) and \(C\) are projections of the points \(P\) and \(K\). Then the line \(AC\) is the projection of the line \(PK\) onto the plane \(ABC\) . Let us extend the segments \(PK\) and \(AC\) beyond the points \(K\) and \(C\), respectively, and obtain the point of intersection of the lines - the point \(E\) .
2) Find the ratio \(AC:EC\) . \(\triangle PAE\sim \triangle KCE\) at two corners ( \(\angle A=\angle C=90^\circ, \angle E\)- general), means \[\dfrac(PA)(KC)=\dfrac(EA)(EC)\]
If we denote the edge of the cube as \(a\) , then \(PA=\dfrac34a, \KC=\dfrac15a, \AC=a\sqrt2\). Then:
\[\dfrac(\frac34a)(\frac15a)=\dfrac(a\sqrt2+EC)(EC) \Rightarrow EC=\dfrac(4\sqrt2)(11)a \Rightarrow AC:EC=4:11\ ]
Example 2.
Given a regular triangular pyramid \(DABC\) with a base \(ABC\) whose height is equal to the side of the base. Let the point \(M\) divide the side edge of the pyramid in the ratio \(1:4\), counting from the top of the pyramid, and \(N\) - the height of the pyramid in the ratio \(1:2\), counting from the top of the pyramid. Find the point of intersection of the straight line \(MN\) with the plane \(ABC\) .
Solution
1) Let \(DM:MA=1:4, \DN:NO=1:2\) (see figure). Because the pyramid is regular, then the height falls at the point \(O\) of intersection of the medians of the base. Let's find the projection of the straight line \(MN\) onto the plane \(ABC\) . Because \(DO\perp (ABC)\) , then \(NO\perp (ABC)\) . This means that \(O\) is a point belonging to this projection. Let's find the second point. Let us drop the perpendicular \(MQ\) from the point \(M\) to the plane \(ABC\) . The point \(Q\) will lie on the median \(AK\) .
Indeed, because \(MQ\) and \(NO\) are perpendicular to \((ABC)\), then they are parallel (which means they lie in the same plane). Therefore, since points \(M, N, O\) lie in the same plane \(ADK\), then the point \(Q\) will lie in this plane. But also (by construction) the point \(Q\) must lie in the plane \(ABC\), therefore, it lies on the line of intersection of these planes, and this is \(AK\) .
This means that the line \(AK\) is the projection of the line \(MN\) onto the plane \(ABC\) . \(L\) is the point of intersection of these lines.
2) Note that in order to draw the drawing correctly, it is necessary to find the exact position of the point \(L\) (for example, in our drawing the point \(L\) lies outside the segment \(OK\), although it could lie inside it; which is correct?).
Because according to the condition, the side of the base is equal to the height of the pyramid, then we denote \(AB=DO=a\) . Then the median is \(AK=\dfrac(\sqrt3)2a\) . Means, \(OK=\dfrac13AK=\dfrac 1(2\sqrt3)a\). Let's find the length of the segment \(OL\) (then we can understand whether the point \(L\) is inside or outside the segment \(OK\): if \(OL>OK\) then it is outside, otherwise it is inside).
A) \(\triangle AMQ\sim \triangle ADO\) at two corners ( \(\angle Q=\angle O=90^\circ, \\angle A\)- general). Means,
\[\dfrac(MQ)(DO)=\dfrac(AQ)(AO)=\dfrac(MA)(DA)=\dfrac 45 \Rightarrow MQ=\dfrac 45a, \AQ=\dfrac 45\cdot \dfrac 1(\sqrt3)a\]
Means, \(QK=\dfrac(\sqrt3)2a-\dfrac 45\cdot \dfrac 1(\sqrt3)a=\dfrac7(10\sqrt3)a\).
b) Let us denote \(KL=x\) .
\(\triangle LMQ\sim \triangle LNO\) at two corners ( \(\angle Q=\angle O=90^\circ, \\angle L\)- general). Means,
\[\dfrac(MQ)(NO)=\dfrac(QL)(OL) \Rightarrow \dfrac(\frac45 a)(\frac 23a) =\dfrac(\frac(7)(10\sqrt3)a+x )(\frac1(2\sqrt3)a+x) \Rightarrow x=\dfrac a(2\sqrt3) \Rightarrow OL=\dfrac a(\sqrt3)\]
Therefore, \(OL>OK\) means that the point \(L\) really lies outside the segment \(AK\) .
Comment
Do not be alarmed if, when solving a similar problem, you find that the length of the segment is negative. If in the conditions of the previous problem we received that \(x\) is negative, this would mean that we incorrectly chose the position of the point \(L\) (that is, that it is located inside the segment \(AK\)) .
Example 3
Given a regular quadrangular pyramid \(SABCD\) . Find the section of the pyramid by the plane \(\alpha\) passing through the point \(C\) and the middle of the edge \(SA\) and parallel to the line \(BD\) .
Solution
1) Let us denote the middle of the edge \(SA\) by \(M\) . Because the pyramid is regular, then the height \(SH\) of the pyramid falls to the point of intersection of the diagonals of the base. Consider the plane \(SAC\) . The segments \(CM\) and \(SH\) lie in this plane, let them intersect at the point \(O\) .
In order for the plane \(\alpha\) to be parallel to the line \(BD\) , it must contain some line parallel to \(BD\) . The point \(O\) is located together with the line \(BD\) in the same plane - in the plane \(BSD\) . Let us draw in this plane through the point \(O\) the straight line \(KP\parallel BD\) (\(K\in SB, P\in SD\) ). Then, by connecting the points \(C, P, M, K\) , we obtain a section of the pyramid by the plane \(\alpha\) .
2) Let us find the relation in which the points \(K\) and \(P\) are divided by the edges \(SB\) and \(SD\) . This way we will completely define the constructed section.
Note that since \(KP\parallel BD\) , then by Thales’ theorem \(\dfrac(SB)(SK)=\dfrac(SD)(SP)\). But \(SB=SD\) means \(SK=SP\) . Thus, only \(SP:PD\) can be found.
Consider \(\triangle ASC\) . \(CM, SH\) are the medians in this triangle, therefore, the intersection point is divided in the ratio \(2:1\), counting from the vertex, that is, \(SO:OH=2:1\) .
Now according to Thales' theorem from \(\triangle BSD\) : \(\dfrac(SP)(PD)=\dfrac(SO)(OH)=\dfrac21\).
3) Note that according to the theorem of three perpendiculars, \(CO\perp BD\) is like an oblique one (\(OH\) is a perpendicular to the plane \(ABC\), \(CH\perp BD\) is a projection). So, \(CO\perp KP\) . Thus, the section is a quadrilateral \(CPMK\) whose diagonals are mutually perpendicular.
Example 4
Given a rectangular pyramid \(DABC\) with an edge \(DB\) perpendicular to the plane \(ABC\) . The base is a right triangle with \(\angle B=90^\circ\) , and \(AB=DB=CB\) . Draw a plane through the straight line \(AB\) perpendicular to the face \(DAC\) and find the section of the pyramid by this plane.
Solution
1) The plane \(\alpha\) will be perpendicular to the face \(DAC\) if it contains a line perpendicular to \(DAC\) . Let's draw a perpendicular from the point \(B\) to the plane \(DAC\) - \(BH\) , \(H\in DAC\) .
Let us draw auxiliary \(BK\) – median in \(\triangle ABC\) and \(DK\) – median in \(\triangle DAC\) .
Because \(AB=BC\) , then \(\triangle ABC\) is isosceles, which means \(BK\) is the height, that is, \(BK\perp AC\) .
Because \(AB=DB=CB\) and \(\angle ABD=\angle CBD=90^\circ\), That \(\triangle ABD=\triangle CBD\), therefore, \(AD=CD\) , therefore, \(\triangle DAC\) is also isosceles and \(DK\perp AC\) .
Let's apply the theorem about three perpendiculars: \(BH\) – perpendicular to \(DAC\) ; oblique \(BK\perp AC\) , which means projection \(HK\perp AC\) . But we have already determined that \(DK\perp AC\) . Thus, the point \(H\) lies on the segment \(DK\) .
By connecting the points \(A\) and \(H\) we obtain a segment \(AN\) along which the plane \(\alpha\) intersects the face \(DAC\) . Then \(\triangle ABN\) is the desired section of the pyramid by the plane \(\alpha\) .
2) Determine the exact position of the point \(N\) on the edge \(DC\) .
Let's denote \(AB=CB=DB=x\) . Then \(BK\) as the median dropped from the vertex right angle in \(\triangle ABC\) is equal to \(\frac12 AC\) , therefore \(BK=\frac12 \cdot \sqrt2 x\) .
Consider \(\triangle BKD\) . Let's find the ratio \(DH:HK\) .
Note that since \(BH\perp (DAC)\), then \(BH\) is perpendicular to any straight line from this plane, which means \(BH\) is the height in \(\triangle DBK\) . Then \(\triangle DBH\sim \triangle DBK\), hence
\[\dfrac(DH)(DB)=\dfrac(DB)(DK) \Rightarrow DH=\dfrac(\sqrt6)3x \Rightarrow HK=\dfrac(\sqrt6)6x \Rightarrow DH:HK=2:1 \]
Let's now consider \(\triangle ADC\) . The medians of the exact intersection triangle are divided in the ratio \(2:1\), counting from the vertex. This means that \(H\) is the intersection point of the medians in \(\triangle ADC\) (since \(DK\) is the median). That is, \(AN\) is also a median, which means \(DN=NC\) .