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This - oldest geometric problem.
Step-by-step instruction
1st method. - Using the “golden” or “Egyptian” triangle. The sides of this triangle have the aspect ratio 3:4:5, and the angle is strictly 90 degrees. This quality was widely used by the ancient Egyptians and other ancient cultures.
Ill.1. Construction of the Golden, or Egyptian triangle
- We manufacture three measurements (or rope compasses - a rope on two nails or pegs) with lengths 3; 4; 5 meters. The ancients often used the method of tying knots with equal distances between them. Unit of length - " nodule».
- We drive a peg at point O and attach the measure “R3 - 3 knots” to it.
- We stretch the rope along known border– towards the intended point A.
- At the moment of tension on the border line - point A, we drive in a peg.
- Then - again from point O, stretch the measure R4 - along the second border. We don’t drive the peg in yet.
- After this, we stretch the measure R5 - from A to B.
- We drive a peg at the intersection of measurements R2 and R3. - This desired point IN - third vertex of the golden triangle, with sides 3;4;5 and with a right angle at point O.
2nd method. Using a compass.
The compass may be rope or pedometer. Cm:
Our compass pedometer has a step of 1 meter.
Ill.2. Compass pedometer
Construction - also according to Ill. 1.
- From the reference point - point O - the neighbor's corner, draw a segment of arbitrary length - but larger than the radius of the compass = 1m - in each direction from the center (segment AB).
- We place the leg of the compass at point O.
- We draw a circle with radius (compass pitch) = 1 m. It is enough to draw short arcs - 10-20 centimeters each, at the intersection with the marked segment (through points A and B). With this action we found equidistant points from the center- A and B. The distance from the center does not matter here. You can simply mark these points with a tape measure.
- Next you need to draw arcs with centers at points A and B, but several (arbitrarily) larger radius, than R=1m. You can reconfigure our compass to a larger radius if it has an adjustable pitch. But for such a small current task I wouldn’t want to “pull” it. Or when there is no adjustment. Can be done in half a minute rope compass.
- We place the first nail (or the leg of a compass with a radius greater than 1 m) alternately at points A and B. And draw two arcs with the second nail - in a taut state of the rope - so that they intersect with each other. It is possible at two points: C and D, but one is enough - C. And again, short serifs at the intersection at point C will suffice.
- Draw a straight line (segment) through points C and D.
- All! The resulting segment, or straight line, is exact direction on North:). Sorry, - at a right angle.
- The figure shows two cases of boundary discrepancy across a neighbor's property. Ill. 3a shows a case where a neighbor’s fence moves away from the desired direction to its detriment. On 3b - he climbed onto your site. In situation 3a, it is possible to construct two “guide” points: both C and D. In situation 3b, only C.
- Place a peg at corner O, and a temporary peg at point C, and stretch a cord from C to the rear boundary of the site. - So that the cord barely touches peg O. By measuring from point O - in direction D, the length of the side according to the general plan, you will get a reliable rear right corner of the site.
Ill.3. Construction right angle– from the neighbor’s corner, using a pedometer and a rope compass
If you have a compass-pedometer, then you can do without rope altogether. In the previous example, we used the rope one to draw arcs of a larger radius than those of the pedometer. More because these arcs must intersect somewhere. In order for the arcs to be drawn with a pedometer with the same radius - 1m with a guarantee of their intersection, it is necessary that points A and B are inside the circle with R = 1m.
- Then measure these equidistant points roulette- V different sides from the center, but always along line AB (neighbor’s fence line). The closer points A and B are to the center, the farther from it the guide points: C and D, and the more more accurate measurements. In the figure, this distance is taken to be about a quarter of the pedometer radius = 260mm.
Ill.4. Constructing a right angle using a pedometer and tape measure
- This scheme of actions is no less relevant when constructing any rectangle, in particular the contour of a rectangular foundation. You will receive it perfect. Its diagonals, of course, need to be checked, but isn't the effort reduced? – Compared to when the diagonals, corners and sides of the foundation contour are moved back and forth until the corners meet..
Actually, we decided geometric problem on the ground. To make your actions more confident on the site, practice on paper - using a regular compass. Which is basically no different.
In construction problems we will consider the construction geometric figure which can be done using a ruler and compass.
Using a ruler you can:
arbitrary straight line;
an arbitrary straight line passing through a given point;
a straight line passing through two given points.
Using a compass, you can describe a circle of a given radius from a given center.
Using a compass you can plot a segment on a given line from a given point.
Let's consider the main construction tasks.
Task 1. Construct a triangle with given sides a, b, c (Fig. 1).
Solution. Using a ruler, draw an arbitrary straight line and take on it arbitrary point B. Using a compass opening equal to a, we describe a circle with center B and radius a. Let C be the point of its intersection with the line. With a compass opening equal to c, we describe a circle from center B, and with a compass opening equal to b, we describe a circle from center C. Let A be the intersection point of these circles. Triangle ABC has sides equal to a, b, c.
Comment. In order for three straight segments to serve as sides of a triangle, it is necessary that the largest of them be less than the sum of the other two (and< b + с).
Task 2.
Solution. This angle with vertex A and the ray OM are shown in Figure 2.
Let us draw an arbitrary circle with its center at vertex A of the given angle. Let B and C be the points of intersection of the circle with the sides of the angle (Fig. 3, a). With radius AB we draw a circle with the center at point O - the starting point of this ray (Fig. 3, b). Let us denote the point of intersection of this circle with this ray as C 1 . Let us describe a circle with center C 1 and radius BC. Point B 1 of the intersection of two circles lies on the side of the desired angle. This follows from the equality Δ ABC = Δ OB 1 C 1 (the third sign of equality of triangles).
Task 3. Construct the bisector of this angle (Fig. 4).
Solution. From vertex A of a given angle, as from the center, we draw a circle of arbitrary radius. Let B and C be the points of its intersection with the sides of the angle. From points B and C we describe circles with the same radius. Let D be their intersection point, different from A. Ray AD bisects angle A. This follows from the equality Δ ABD = Δ ACD (the third criterion for the equality of triangles).
Task 4. Draw a perpendicular bisector to this segment (Fig. 5).
Solution. Using an arbitrary but identical compass opening (larger than 1/2 AB), we describe two arcs with centers at points A and B, which will intersect each other at some points C and D. The straight line CD will be the desired perpendicular. Indeed, as can be seen from the construction, each of the points C and D is equally distant from A and B; therefore, these points must lie on the perpendicular bisector to segment AB.
Task 5. Divide this segment in half. It is solved in the same way as problem 4 (see Fig. 5).
Task 6. Through a given point draw a line perpendicular to the given line.
Solution. There are two possible cases:
1) given point O lies on a given straight line a (Fig. 6).
From point O we draw arbitrary radius a circle intersecting straight line a at points A and B. Draw circles from points A and B with the same radius. Let O 1 be the point of their intersection, different from O. We obtain OO 1 ⊥ AB. In fact, points O and O 1 are equidistant from the ends of the segment AB and, therefore, lie on the perpendicular bisector to this segment.
When building or developing home design projects, it is often necessary to build an angle equal to an existing one. Templates and school knowledge of geometry come to the rescue.
Instructions
- An angle is formed by two straight lines emanating from one point. This point will be called the vertex of the angle, and the lines will be the sides of the angle.
- Use three letters to represent corners: one at the top, two at the sides. The angle is named starting with the letter that stands on one side, then the letter that stands at the apex is named, and then the letter on the other side. Use other ways to indicate angles if you prefer otherwise. Sometimes only one letter is named, which is at the top. Can you mark the angles? Greek letters, for example, α, β, γ.
- There are situations when it is necessary to draw an angle so that it is equal to an already given angle. If it is not possible to use a protractor when constructing a drawing, you can only get by with a ruler and compass. Let's say on a straight line marked in the drawing with the letters MN, you need to construct an angle at point K, so that it is equal to angle B. That is, from point K it is necessary to draw a straight line forming an angle with the line MN, which will be equal to angle B.
- First, mark a point on each side of a given angle, for example, points A and C, then connect points C and A with a straight line. Get triangle ABC.
- Now construct the same triangle on line MN so that its vertex B is on the line at point K. Use the rule for constructing a triangle on three sides. Lay off the segment KL from point K. It must be equal to the segment BC. Get the L point.
- From point K, draw a circle with a radius equal to segment BA. From L, draw a circle with radius CA. Connect the resulting point (P) of intersection of two circles with K. Obtain a triangle KPL, which will be equal to triangle ABC. This way you will get angle K. It will be equal to angle B. To make this construction more convenient and faster, set aside from vertex B equal segments, using one compass opening, without moving the legs, describe a circle with the same radius from point K.