Determining the distance between: 1 - point and plane; 2 - straight and flat; 3 - planes; 4 - crossing straight lines are considered together, since the solution algorithm for all these problems is essentially the same and consists of geometric constructions, which must be performed to determine the distance between given point A and plane α. If there is any difference, it consists only in the fact that in cases 2 and 3, before starting to solve the problem, you should mark an arbitrary point A on the straight line m (case 2) or plane β (case 3). distances between intersecting straight lines, we first enclose them in parallel planes α and β and then determine the distance between these planes.

Let us consider each of the noted cases of problem solving.

1. Determining the distance between a point and a plane.

The distance from a point to a plane is determined by the length of a perpendicular segment drawn from a point to the plane.

Therefore, the solution to this problem consists of sequentially performing the following graphical operations:

1) from point A we lower the perpendicular to the plane α (Fig. 269);

2) find the point M of intersection of this perpendicular with the plane M = a ∩ α;

3) determine the length of the segment.

If the plane α general position, then in order to lower a perpendicular onto this plane, it is necessary to first determine the direction of the horizontal and frontal projections of this plane. Finding the meeting point of this perpendicular with the plane also requires additional geometric constructions.

The solution to the problem is simplified if the plane α occupies a particular position relative to the projection planes. In this case, both the projection of the perpendicular and the finding of the point of its meeting with the plane are carried out without any additional auxiliary constructions.

EXAMPLE 1. Determine the distance from point A to the frontally projecting plane α (Fig. 270).

SOLUTION. Through A" we draw the horizontal projection of the perpendicular l" ⊥ h 0α, and through A" - its frontal projection l" ⊥ f 0α. We mark the point M" = l" ∩ f 0α . Since AM || π 2, then [A" M"] == |AM| = d.

From the example considered, it is clear how simply the problem is solved when the plane occupies a projecting position. Therefore, if a general position plane is specified in the source data, then before proceeding with the solution, the plane should be moved to a position perpendicular to any projection plane.

EXAMPLE 2. Determine the distance from point K to the plane specified by ΔАВС (Fig. 271).

1. We transfer the plane ΔАВС to the projecting position *. To do this, we move from the system xπ 2 /π 1 to x 1 π 3 /π 1: the direction of the new x 1 axis is chosen perpendicular to the horizontal projection of the horizontal plane of the triangle.

2. Project ΔABC onto a new plane π 3 (the ΔABC plane is projected onto π 3, in [ C " 1 B " 1 ]).

3. Project point K onto the same plane (K" → K" 1).

4. Through the point K" 1 we draw (K" 1 M" 1)⊥ the segment [C" 1 B" 1 ]. The required distance d = |K" 1 M" 1 |

The solution to the problem is simplified if the plane is defined by traces, since there is no need to draw projections of level lines.

EXAMPLE 3. Determine the distance from point K to the plane α, specified by the tracks (Fig. 272).

* Most rational way transferring the triangle plane to the projecting position is a way to replace projection planes, since in this case it is enough to construct only one auxiliary projection.

SOLUTION. We replace the plane π 1 with the plane π 3, for this we draw a new axis x 1 ⊥ f 0α. On h 0α we mark an arbitrary point 1" and determine its new horizontal projection on the plane π 3 (1" 1). Through the points X α 1 (X α 1 = h 0α 1 ∩ x 1) and 1" 1 we draw h 0α 1. We determine the new horizontal projection of the point K → K" 1. From point K" 1 we lower the perpendicular to h 0α 1 and mark the point of its intersection with h 0α 1 - M" 1. The length of the segment K" 1 M" 1 will indicate the required distance.

2. Determining the distance between a straight line and a plane.

The distance between a line and a plane is determined by the length of a perpendicular segment dropped from an arbitrary point of the line onto the plane (see Fig. 248).

Therefore, the solution to the problem of determining the distance between straight line m and plane α is no different from the examples discussed in paragraph 1 for determining the distance between a point and a plane (see Fig. 270 ... 272). As a point, you can take any point belonging to line m.

3. Determination of the distance between planes.

The distance between the planes is determined by the size of the perpendicular segment dropped from a point taken on one plane to another plane.

From this definition it follows that the algorithm for solving the problem of finding the distance between planes α and β differs from a similar algorithm for solving the problem of determining the distance between line m and plane α only in that line m must belong to plane α, i.e., in order to determine the distance between planes α and β follows:

1) take a straight line m in the α plane;

2) select an arbitrary point A on line m;

3) from point A, lower the perpendicular l to the plane β;

4) determine point M - the meeting point of the perpendicular l with the plane β;

5) determine the size of the segment.

In practice, it is advisable to use a different solution algorithm, which will differ from the one given only in that, before proceeding with the first step, the planes should be transferred to the projection position.

Including this additional operation in the algorithm simplifies the execution of all other points without exception, which ultimately leads to a simpler solution.

EXAMPLE 1. Determine the distance between planes α and β (Fig. 273).

SOLUTION. We move from the system xπ 2 /π 1 to x 1 π 1 /π 3. In relation to new planeπ 3 planes α and β occupy a projecting position, therefore the distance between the new frontal traces f 0α 1 and f 0β 1 is the desired one.

IN engineering practice often you have to solve the problem of constructing a plane parallel to a given one and distant from it by specified distance. Example 2 below illustrates the solution to such a problem.

EXAMPLE 2. It is required to construct projections of a plane β parallel to a given plane α (m || n), if it is known that the distance between them is d (Fig. 274).

1. In the α plane we draw arbitrary horizontal lines h (1, 3) and front lines f (1,2).

2. From point 1 we restore the perpendicular l to the plane α(l" ⊥ h", l" ⊥ f").

3. On the perpendicular l we mark an arbitrary point A.

4. Determine the length of the segment - (the position indicates on the diagram the metrically undistorted direction of the straight line l).

5. Lay out the segment = d on the straight line (1"A 0) from point 1".

6. Mark on the projections l" and l" points B" and B", corresponding to point B 0.

7. Through point B we draw the plane β (h 1 ∩ f 1). To β || α, it is necessary to comply with the condition h 1 || h and f 1 || f.

4. Determining the distance between intersecting lines.

The distance between intersecting lines is determined by the length of the perpendicular enclosed between the parallel planes to which the intersecting lines belong.

In order to draw mutually parallel planes α and β through intersecting straight lines m and f, it is sufficient to draw through point A (A ∈ m) a straight line p parallel to straight line f, and through point B (B ∈ f) a straight line k parallel to straight m . The intersecting lines m and p, f and k define the mutually parallel planes α and β (see Fig. 248, e). The distance between the planes α and β is equal to the required distance between the crossing lines m and f.

Another way can be proposed for determining the distance between intersecting lines, which consists in the fact that, using some method of transforming orthogonal projections, one of the intersecting lines is transferred to the projecting position. In this case, one projection of the line degenerates into a point. The distance between the new projections of crossing lines (point A" 2 and segment C" 2 D" 2) is the required one.

In Fig. 275 shows a solution to the problem of determining the distance between crossing lines a and b, given segments [AB] and [CD]. The solution is performed in the following sequence:

1. Transfer one of the crossing lines (a) to a position parallel to the plane π 3; to do this, they move from the system of projection planes xπ 2 /π 1 to the new x 1 π 1 /π 3, the x 1 axis is parallel to the horizontal projection of straight line a. Determine a" 1 [A" 1 B" 1 ] and b" 1.

2. By replacing the plane π 1 with the plane π 4, we translate the straight line

and to position a" 2, perpendicular to the plane π 4 (the new x 2 axis is drawn perpendicular to a" 1).

3. Construct a new horizontal projection of straight line b" 2 - [ C" 2 D" 2 ].

4. The distance from point A" 2 to straight line C" 2 D" 2 (segment (A" 2 M" 2 ] (is the required one.

It should be borne in mind that the transfer of one of the crossing lines to the projecting position is nothing more than the transfer of the planes of parallelism, in which the lines a and b can be enclosed, also to the projecting position.

In fact, by moving line a to a position perpendicular to the plane π 4, we ensure that any plane containing line a is perpendicular to the plane π 4, including the plane α defined by lines a and m (a ∩ m, m || b ). If we now draw a line n parallel to a and intersecting line b, then we obtain a plane β, which is the second plane of parallelism, which contains the intersecting lines a and b. Since β || α, then β ⊥ π 4 .

Any plane in Cartesian system coordinates can be given by the equation `Ax + By + Cz + D = 0`, where at least one of the numbers `A`, `B`, `C` is non-zero. Let a point `M (x_0;y_0;z_0)` be given, let’s find the distance from it to the plane `Ax + By + Cz + D = 0`.

Let the line passing through the point `M` perpendicular to the plane `alpha`, intersects it at point `K` with coordinates `(x; y; z)`. Vector `vec(MK)` is perpendicular to the `alpha` plane, as is the vector `vecn` `(A;B;C)`, i.e., the vectors `vec(MK)` and `vecn` collinear, `vec(MK)= λvecn`.

Since `(x-x_0;y-y_0;z-z-0)` and `vecn(A,B,C)`, then `x-x_0=lambdaA`, `y-y_0=lambdaB`, `z-z_0=lambdaC`.

Point `K` lies in the `alpha` plane (Fig. 6), its coordinates satisfy the equation of the plane. We substitute `x=x_0+lambdaA`, `y=y_0+lambdaB`, `z=z_0+lambdaC` into the equation `Ax+By+Cz+D=0`, we get

`A(x_0+lambdaA)+(B(y_0+lambdaB)+C(z_0+lambdaC)+D=0`,

whence `lambda=-(Ax_0+By_0+Cz_0+D)/(A^2+B^2+C^2)`.

Find the length of the vector `vec(MK)`, which is equal to the distance from the point `M(x_0;y_0;z_0)` to the plane `Ax + By + Cz + D` `|vec(MK)|=|lambdavecn|=|lambda|*sqrt(A^2+B^2+C^2)`.

So, the distance `h` from the point `M(x_0;y_0;z_0)` to the plane `Ax + By + Cz + D = 0` is as follows

`h=(|Ax_0+By_0+Cz_0+D|)/(sqrt(A^2+B^2+C^2))`.

At geometric way finding the distance from point `A` to the plane `alpha` find the base of the perpendicular `A A^"` dropped from point `A` to the plane `alpha`. If point `A^"` is located outside the section of the plane `alpha` specified in problem, then through point `A` draw a straight line `c`, parallel to the plane`alpha`, and choose a more convenient point `C` on it, orthographic projection which `C^"` belongs to this section of the `alpha` plane. Length of segment `C C^"`will be equal to the required distance from point `A`to the `alpha` plane.

In the right hexagonal prism`A...F_1`, all edges of which are equal to `1`, find the distance from point `B` to the plane `AF F_1`.

Let `O` be the center of the lower base of the prism (Fig. 7). The straight line `BO` is parallel to the straight line `AF` and, therefore, the distance from the point `B` to the plane `AF F_1` is equal to the distance `OH` from the point `O` to the plane `AF F_1`. In the triangle `AOF` we have `AO=OF=AF=1`. The height `OH` of this triangle is `(sqrt3)/2`. Therefore, the required distance is `(sqrt3)/2`.

Let's show another way (auxiliary volume method) finding the distance from a point to a plane. It is known that the volume of the pyramid `V` , the area of ​​its base `S`and height length `h`are related by the formula `h=(3V)/S`. But the length of the height of a pyramid is nothing more than the distance from its top to the plane of the base. Therefore, to calculate the distance from a point to a plane, it is enough to find the volume and area of ​​the base of some pyramid with the apex at this point and with the base lying in this plane.

Dana correct prism`A...D_1`, in which `AB=a`, `A A_1=2a`. Find the distance from the intersection point of the diagonals of the base `A_1B_1C_1D_1` to the plane `BDC_1`.

Consider the tetrahedron `O_1DBC_1` (Fig. 8). The required distance `h` is the length of the height of this tetrahedron, lowered from the point `O_1` to the plane of the face `BDC_1` . To find it, it is enough to know the volume `V`tetrahedron `O_1DBC_1` and area triangle `DBC_1`. Let's calculate them. Note that straight line `O_1C_1` perpendicular to the plane `O_1DB`, because it is perpendicular to `BD` and `B B_1` . This means that the volume of the tetrahedron is `O_1DBC_1` equals

Maintaining your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please review our privacy practices and let us know if you have any questions.

Collection and use of personal information

Personal information refers to data that can be used to identify or contact a specific person.

You may be asked to provide your personal information any time you contact us.

Below are some examples of the types of personal information we may collect and how we may use such information.

What personal information do we collect:

• When you submit a request on the site, we may collect various information, including your name, phone number, address email etc.

How we use your personal information:

• The personal information we collect allows us to contact you with unique offers, promotions and other events and upcoming events.
• From time to time, we may use your personal information to send important notices and communications.
• We may also use personal information for internal purposes such as auditing, data analysis and various studies in order to improve the services we provide and provide you with recommendations regarding our services.
• If you participate in a prize draw, contest or similar promotion, we may use the information you provide to administer such programs.

Disclosure of information to third parties

We do not disclose the information received from you to third parties.

Exceptions:

• If necessary - in accordance with the law, judicial procedure, legal proceedings, and/or based on public requests or requests from government agencies on the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public importance purposes.
• In the event of a reorganization, merger, or sale, we may transfer the personal information we collect to the applicable successor third party.

Protection of personal information

We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as unauthorized access, disclosure, alteration and destruction.

Respecting your privacy at the company level

To ensure that your personal information is secure, we communicate privacy and security standards to our employees and strictly enforce privacy practices.

Construct traces of the plane given by ∆BCD and determine the distance from point A to given plane method right triangle (for coordinates of points A, B, C and D, see Table 1 of the Tasks section);

1.2. Example of task No. 1

The first task presents a set of tasks on the following topics:

1. Orthogonal projection, Monge diagram, point, straight line, plane: by known coordinates of three points B, C, D construct horizontal and frontal projections of the plane given by ∆ BCD;

2. Traces of a straight line, traces of a plane, properties of belonging to a straight plane: construct traces of the plane given by ∆ BCD;

3. General and particular planes, intersection of a line and a plane, perpendicularity of a line and a plane, intersection of planes, right triangle method: determine the distance from a point A to plane ∆ BCD.

1.2.1. Based on the known coordinates of three points B, C, D let's construct horizontal and frontal projections of the plane given by ∆ BCD(Figure 1.1), for which it is necessary to construct horizontal and frontal projections of the vertices ∆ BCD, and then connect the projections of the vertices of the same name.

It is known that following the plane is a straight line obtained as a result of the intersection of a given plane with the projection plane .

A general plane has 3 traces: horizontal, frontal and profile.

In order to construct traces of a plane, it is enough to construct traces (horizontal and frontal) of any two straight lines lying in this plane and connect them to each other. Thus, the trace of the plane (horizontal or frontal) will be uniquely determined, since through two points on the plane (in in this case these points will be traces of straight lines) you can draw a straight line, and only one.

The basis for this construction is property of belonging to a straight plane: if a straight line belongs to a given plane, then its traces lie on similar traces of this plane .

The trace of a line is the point of intersection of this line with the projection plane. .

The horizontal trace of a straight line lies in horizontal plane projections, frontal – in frontal plane projections.

Let's consider the construction horizontal trace direct D.B., for which you need:

1. Continue the frontal projection straight D.B. until it intersects with the axis X, point of intersection M 2 is frontal projection horizontal trace;

2. From a point M 2 restore the perpendicular (projection connection line) until it intersects with the horizontal projection of the straight line D.B. M 1 and will be a horizontal projection of the horizontal trace (Figure 1.1), which coincides with the trace itself M.

The horizontal trace of the segment is constructed in a similar way NE straight: point M'.

To build frontal trace segment C.B. direct, you need:

1. Continue the horizontal projection of the straight line C.B. until it intersects with the axis X, point of intersection N 1 is a horizontal projection of the frontal trace;

2. From a point N 1 restore the perpendicular (projection connection line) until it intersects with the frontal projection of the straight line C.B. or its continuation. Intersection point N 2 and will be a frontal projection of the frontal trace, which coincides with the trace itself N.

Connecting the dots M′ 1 And M 1 line segment, we get horizontal trace plane απ 1. Point α x of intersection of απ 1 with the axis X called vanishing point . To construct the frontal trace of the απ 2 plane, it is necessary to connect the frontal trace N 2 with the vanishing point of traces α x

Figure 1.1 — Construction of plane traces

The algorithm for solving this problem can be presented as follows:

1. (D 2 B 2 ∩ OX) = M 2 ;
2. (MM 1 ∩ D 1 B 1) = M 1 = M;
3. (C 2 B 2 ∩ OX) = M′ 2 ;
4. (M′ 2 M′ 1 ∩ C 1 B 1) = M′ 1 = M′;
5. (CB∩ π 2) = N 2 = N;
6. (MM′) ≡ απ 1 ;
7. (α x N) ≡ απ 2 .

1.2.2. To solve the second part of the first task you need to know that:

• distance from point A to plane ∆ BCD determined by the length of the perpendicular restored from this point to the plane;
• any line is perpendicular to a plane if it is perpendicular to two intersecting lines lying in this plane;
• on the diagram, the projections of a straight line perpendicular to the plane are perpendicular to the inclined projections of the horizontal and frontal of this plane or the traces of the same name of the plane (Fig. 1.2) (see the Theorem on the perpendicular to the plane in the lectures).

To find the base of a perpendicular, it is necessary to solve the problem of the intersection of a line (in this problem, such a line is a perpendicular to a plane) with a plane:

1. Enclose the perpendicular in an auxiliary plane, which should be a plane of particular position (horizontally projecting or frontally projecting; in the example, horizontally projecting γ is taken as an auxiliary plane, that is, perpendicular to π 1, its horizontal trace γ 1 coincides with a horizontal projection of a perpendicular);

2. Find the line of intersection of the given plane ∆ BCD with auxiliary γ ( MN in Fig. 1.2);

3. Find the point of intersection of the line of intersection of planes MN with a perpendicular (point TO in Fig. 1.2).

4. To determine the true distance from a point A to a given plane ∆ BCD should be used right triangle method: the true size of a segment is the hypotenuse of a right triangle, one leg of which is one of the projections of the segment, and the other is the difference in distances from its ends to the plane of projections in which the construction is being carried out.

5. Determine the visibility of perpendicular sections using the competing point method. For example - points N And 3 to determine visibility on π 1, points 4 , 5 - to determine visibility on π 2.

Figure 1.2 - Construction of a perpendicular to the plane

Figure 1.3 — Design example control task №1

Video example of completing task No. 1

1.3. Task options 1

Instructions

To find the distance from points to plane using descriptive methods: select on plane arbitrary point; draw two straight lines through it (lying in this plane); restore perpendicular to plane passing through this point (construct a line perpendicular to both intersecting lines at the same time); draw a straight line parallel to the constructed perpendicular through a given point; find the distance between the point of intersection of this line with the plane and the given point.

If the position points given by its three-dimensional coordinates, and the position plane – linear equation, then to find the distance from plane to points, use the methods analytical geometry: indicate the coordinates points through x, y, z, respectively (x – abscissa, y – ordinate, z – applicate); denote by A, B, C, D the equations plane(A – parameter at abscissa, B – at , C – at applicate, D – free member); calculate the distance from points to plane according to the formula:s = | (Ax+By+Cz+D)/√(A²+B²+C²) |,where s is the distance between the point and the plane,|| - absolute value(or module).

Example: Find the distance between point A with coordinates (2, 3, -1) and the plane, given by the equation: 7x-6y-6z+20=0. Solution. From the conditions it follows that: x=2,y=3,z=-1,A=7,B=-6,C=-6,D=20. Substitute these values ​​into the above. You get: s = | (7*2+(-6)*3+(-6)*(-1)+20)/√(7²+(-6)²+(-6)²) | = | (14-18+6+20)/11 | = 2.Answer: Distance from points to plane equals 2 (arbitrary units).

Tip 2: How to determine the distance from a point to a plane

Determining the distance from points to plane- one of the common tasks school planimetry. As is known, the smallest distance from points to plane there will be a perpendicular drawn from this points to this plane. Therefore, the length of this perpendicular is taken as the distance from points to plane.

You will need

• plane equation

Instructions

Let the first of the parallel f1 be given by the equation y=kx+b1. Translating the expression into general view, you get kx-y+b1=0, that is, A=k, B=-1. The normal to it will be n=(k, -1).
Now follows an arbitrary abscissa of the point x1 on f1. Then its ordinate is y1=kx1+b1.
Let the equation of the second of the parallel lines f2 be of the form:
y=kx+b2 (1),
where k is the same for both lines, due to their parallelism.

Next you need to create canonical equation a line perpendicular to both f2 and f1 containing the point M (x1, y1). In this case, it is assumed that x0=x1, y0=y1, S=(k, -1). As a result, you should get the following equality:
(x-x1)/k =(y-kx1-b1)/(-1) (2).

Having solved the system of equations consisting of expressions (1) and (2), you will find the second point that determines the required distance between the parallel ones N(x2, y2). The required distance itself will be equal to d=|MN|=((x2-x1)^2+(y2-y1)^2)^1/2.

Example. Let the equations of given parallel lines on the plane f1 – y=2x +1 (1);
f2 – y=2x+5 (2). Take an arbitrary point x1=1 on f1. Then y1=3. The first point will thus have coordinates M (1,3). General perpendicular equation (3):
(x-1)/2 = -y+3 or y=-(1/2)x+5/2.
Substituting this y value into (1), you get:
-(1/2)x+5/2=2x+5, (5/2)x=-5/2, x2=-1, y2=-(1/2)(-1) +5/2= 3.
The second base of the perpendicular is at the point with coordinates N (-1, 3). The distance between parallel lines will be:
d=|MN|=((3-1)^2+(3+1)^2)^1/2=(4+16)^1/2=4.47.

Sources:

• Lung development athletics in Russia

Top of any flat or volumetric geometric figure uniquely determined by its coordinates in space. In the same way, any arbitrary point in the same coordinate system can be uniquely determined, and this makes it possible to calculate the distance between this arbitrary point and the top of the figure.

You will need

• - paper;
• - pen or pencil;
• - calculator.

Instructions

Reduce the problem to finding the length of a segment between two points, if the coordinates of the point specified in the problem and the vertices of the geometric figure are known. This length can be calculated using the Pythagorean theorem in relation to the projections of a segment on the coordinate axis - it will be equal to square root from the sum of the squares of the lengths of all projections. For example, let in three-dimensional system coordinates are given by point A(X₁;Y₁;Z₁) and vertex C of any geometric figure with coordinates (X₂;Y₂;Z₂). Then the lengths of the projections of the segment between them onto coordinate axes can be as X₁-X₂, Y₁-Y₂ and Z₁-Z₂, and the length of the segment as √((X₁-X₂)²+(Y₁-Y₂)²+(Z₁-Z₂)²). For example, if the coordinates of the point are A(5;9;1), and the vertices are C(7;8;10), then the distance between them will be equal to √((5-7)²+(9-8)²+(1- 10)²) = √(-2²+1²+(-9)²) = √(4+1+81) = √86 ≈ 9.274.

First calculate the coordinates of the vertex if they are not explicitly presented in the problem conditions. The specific method depends on the type of figure and known additional parameters. For example, if known 3D coordinates three vertices A(X₁;Y₁;Z₁), B(X₂;Y₂;Z₂) and C(X₃;Y₃;Z₃), then the coordinates of its fourth vertex ( opposite the vertex B) will be (X₃+X₂-X₁; Y₃+Y₂-Y₁; Z₃+Z₂-Z₁). After determining the coordinates of the missing vertex, calculating the distance between it and an arbitrary point will again be reduced to determining the length of the segment between these two points in given system coordinates - do this in the same way as described in the previous step. For example, for the vertex of the parallelogram described in this step and point E with coordinates (X₄;Y₄;Z₄), the formula for calculating the distance from the previous step can be as follows: √((X₃+X₂-X₁-X₄)²+(Y₃+Y₂-Y₁- Y₄)²+(Z₃+Z₂-Z₁-Z₄)²).

For practical calculations you can use, for example, the built-in search engine Google. So, to calculate the value using the formula obtained in the previous step, for points with coordinates A(7;5;2), B(4;11;3), C(15;2;0), E(7;9; 2), enter the following search query: sqrt((15+4-7-7)^2+(2+11-5-9)^2+(0+3-2-2)^2). The search engine will calculate and display the result of the calculation (5.19615242).

Video on the topic

Recovery perpendicular To plane– one of important tasks in geometry, it underlies many theorems and proofs. To construct a line perpendicular plane, you need to perform several steps sequentially.

You will need

• - given plane;
• - the point from which you want to draw a perpendicular;
• - compass;
• - ruler;
• - pencil.