Knowing; that the triangles are equal on both sides and the angle between them, we can use a compass and ruler to divide this segment into two equal parts.

If, for example, you need to divide a segment in half A B(Fig. 69), then place the tip of the compass at the points A I B and They describe around them, as if near the centers, two intersecting arcs of equal radius (Fig. 70). Their intersection points WITH And D connected by a straight line, which AB in half: JSC= OB.

To make sure that the segments JSC And OB must be equal, connect the dots C And D with ends A And IN segment (Fig. 71). You will get two triangles ACD And BCD, whose three sides are respectively equal: AC= Sun; AD= BD; CD – common, i.e. belongs to both triangles. This implies the complete equality of these triangles, and therefore the equality of all angles. So, by the way, the angles are equal ACD And BCD. Now comparing the triangles ASO And VSO, we see that they have a side OS – general, A.C.= CB, and the angle between them ASO = ug. VSO. The triangles are equal along two sides and the angle between them; therefore the sides are equal JSC And OB, i.e. point ABOUT there is a midpoint AB.

§ 22. How to construct a triangle using a side and two angles

Finally, consider a problem whose solution leads to the construction of a triangle using a side and two angles:

On the other side of the river (Fig. 72) a milestone is visible A. It is required, without crossing the river, to find out the distance to it from the milestone IN on this shore.

Let's do this. Let's measure from the point IN any distance in a straight line Sun and at the ends of it IN And WITH Let's measure angles 1 and 2 (Fig. 73). If we now measure the distance on a convenient area DE, equal Sun, and build angles at its ends A And b(Fig. 74), equal to angles 1 and 2, then at the point of intersection of their sides we get the third vertex F triangle DEF. It is easy to verify that the triangle DEF equal to a triangle ABC; indeed, if we imagine that the triangle DEF superimposed on ABC so that side DE coincided with its equal side Sun, then ug. A will coincide with angle 1, angle b – with angle 2, and side DF will go to the side VA, and the side E.F. on the side SA. Since two lines can intersect only at one point, then the vertex F should coincide with the top A. So the distance DF equal to the required distance VA.

The problem, as we see, has only one solution. In general, using a side and two angles adjacent to this side, only one triangle can be constructed; There cannot be other triangles with the same side and the same two angles adjacent to it in the same places. All triangles that have one identical side and two identical angles adjacent to it in the same places can be brought into complete coincidence by superposition. This means that this is a sign by which one can establish the complete equality of triangles.

Together with the previously established signs of equality of triangles, we now know the following three:

Triangles:

on three sides;

at the two sides and at the corner between them;

on the side and two sides.

For the sake of brevity, we will further denote these three cases of equality of triangles as follows:

on three sides: SSS;

on two sides and the angle between them: SUS;

along the side and two corners: USU.

Applications

14. To find out the distance to a point A on the other side of the river from the point IN on this bank (Fig. 5), measure some line in a straight line sun, then at point IN construct an angle equal to ABC, on the other side Sun, and at the point WITH- in the same way, an angle equal to DIA Point distance D intersection of the sides of both sides of the angles to the point IN equal to the required distance AB. Why?

Solution: Triangles ABC And BDC equal on one side ( Sun) and two angles (ang. DCB= ug. DIA; ug. DBC= ug. ABC.) Hence, AB= ВD, as sides lying in equal triangles against equal angles.

§ 23. Parallelograms

From triangles we move on to quadrilaterals, i.e., to figures limited by 4 sides. An example of a quadrilateral is a square - a quadrilateral in which all sides are equal and all angles are right (Fig. 76). Another type of quadrilateral, also often found, is a rectangular:

This is the name of any quadrilateral with 4 right angles (Fig. 77 and 78). A square is also a rectangle, but with equal sides.

The peculiarity of a rectangle (and square) is that both pairs of its opposite sides are parallel. In a rectangle ABCD, for example (Fig. 78), AB parallel DC, a AD parallel Sun. This follows from the fact that both opposite sides are perpendicular to the same line, and we know that two perpendiculars to one line are parallel to each other (§ 16).

Another property of every rectangle is that its opposite sides are equal to each other. You can verify this if you connect the opposite vertices of the rectangle with a straight line, that is, draw a diagonal in it. Connecting A With WITH(Drawn 79) we get two triangles ABC And ADC. It is easy to show that these triangles are equal to each other: side AC – total, ug. 1 = angle 2, because these are cross angles with parallel AB And CD for the same reason, angles 3 and 4 are equal. On the same side and two angles, triangles ABC And ACD equal; hence the side AB= side DC, and side AD= side Sun.

Such quadrilaterals, in which, like rectangles, opposite sides are parallel, are called parallelograms. Damn it. 80 shows an example of a parallelogram: AB parallel DC, A AD parallel BC. Damn.80

A rectangle is one of the parallelograms, namely one in which all the angles are right. It is easy to verify that each parallelogram has the following properties:

OPPOSITE ANGLES PARALLEL GRAMMAR EQUAL; Opposite sides

P a r l l e l o g r a m a v y s.

To verify this, let us draw in a parallelogram ABCD(Fig. 81) straight ВD(diagonal) and compare triangles ABD And VDC. These triangles are equal (case USU): BD– common side; ug. 1 = angle 2, corner 3 = angle 4 (why?). The properties listed earlier follow from this.

A parallelogram with four equal sides is called a rhombus.

Repeat questions

What shape is called a square? Rectangle? – What is called a diagonal? – What figure is called a parallelogram? Diamond? – Indicate the properties of the angles and sides of any parallelogram. – Which rectangle is called a square? – Which parallelogram is called a rectangle? – What are the similarities and differences between a square and a rhombus.

The contours of all images are formed by various lines. The main lines are a straight line, a circle and a series of curves. When drawing the contours of images, geometric constructions and conjugations are used.

When studying the discipline “Descriptive Geometry and Engineering Graphics,” students must learn the rules and sequence of performing geometric constructions and connections.

In this regard the best way acquisition of construction skills are tasks on drawing the contours of complex parts.

Before you begin control task, you need to study the technique of performing geometric constructions and conjugations according to the methodological manual.

1. Division of segments and angles

1.1. Dividing a segment in half

Divide the given segment AB in half.

From the ends of the segment AB, as from the centers, we draw arcs of circles with radius R, the size of which should be slightly larger than half of the segment AB (Fig. 1). These arcs will intersect at points M and N, let's find point C at which straight lines AB and MN intersect. Point C will divide segment AB into two equal parts.

Note. All necessary constructions must and can be carried out only with the help of a compass and a ruler (without divisions).

1.2. Dividing a segment into n equal parts

Divide a given segment into n equal parts.

From the end of the segment - point A, we will draw an auxiliary ray at an arbitrary angle α. (Fig. 2 a) On this ray we will lay 4 equal segments of arbitrary length (Fig. 2b). The end of the last, fourth, segment (point 4) is connected to point B. Next, from all previous points 1...3, we draw segments parallel to segment B4 until they intersect with segment AB at points 1", 2", 3". The points thus obtained divided the segment into equal four segments

1.3. Dividing an angle in half

Divide specified angle YOU in half.

From the vertex of angle A, we draw an arc with an arbitrary radius until it intersects with the sides of the angle at points B and C (Fig. 3 a). Then from points B and C we draw two arcs with a radius greater than half the distance BC until they intersect at point D (Fig. 3 b). By connecting points A and D with a straight line, we obtain the bisector of the angle, which divides the given angle in half (Fig. 3 c)

a) b) c)

2. Dividing a circle into equal parts and constructing regular polygons

2.1. Dividing a circle into three equal parts

From the end of the diameter, for example, point A (Fig. 4), draw an arc of radius R equal to the radius of the given circle. The first and second divisions are obtained - points 1 and 2. The third division, point 3, is located at the opposite end of the same diameter. By connecting points 1,2,3 with chords, you get a regular inscribed triangle.

2.2. Dividing a circle into six equal parts

From the ends of any diameter, for example AB (Fig. 5), arcs of radius R are described. Points A, 1,3,B,4,2 divide the circle into six equal parts. By connecting them with chords, a regular inscribed hexagon is obtained.

Note. Auxiliary arcs should not be drawn completely; it is enough to make notches on the circle.

2.3. Dividing a circle into five equal parts

Two mutually perpendicular diameters AB and CD are drawn (Fig. 6). The OS radius at point O 1 is divided in half.
From point O1, as from the center, draw an arc of radius O1A until it intersects with diameter CD at point E.
The segment AE is equal to the side of a regular inscribed pentagon, and the segment OE is equal to the side of a regular inscribed decagon.
Taking point A as the center, an arc of radius R1 = AE marks points 1 and 4 on the circle. From points 1 and 4, as from centers, arcs of the same radius R1 mark points 3 and 2. Points A, 1, 2, 3, 4 divide the circle into five equal parts.

2.4. Dividing a circle into seven equal parts

From the end of the diameter, for example, point A draw an arc of radius R equal to the radius of the circle (Fig. 7). The chord CD is equal to the side of a regular inscribed triangle. Half of the chord CD is, to a sufficient approximation, equal to the side of a regular inscribed heptagon, i.e. divides the circle into seven equal parts.

Rice. 7

Literature

Bogolyubov S.K. Engineering graphics: Textbook for secondary specialized educational institutions. – 3rd ed., rev. And additional - M.: Mechanical Engineering, 2006. – p. 392: ill.
Kuprikov M.Yu. Engineering graphics: textbook for secondary educational institutions - M.: Bustard, 2010 - 495 pp.: ill.
Fedorenko V.A., Shoshin A.I. Handbook of mechanical engineering drawing L.: Mechanical engineering. 1976. 336 p.

TRIANGLES.

§ 28. CONSTRUCTIONS WITH COMPASSES AND RULER.

Until now, when solving construction problems, we have used a compass, a ruler, a drawing triangle and a protractor.

Let us now solve a number of construction problems using only two tools - a compass and a ruler.

Task 1. Divide this segment in half.

Given a segment AB, you need to divide it in half.

Solution. With a radius greater than half of the segment AB, we describe intersecting arcs from points A and B, as from centers (Fig. 161). Through the intersection points of these arcs we draw a straight line CD, which will intersect the segment AB at some point K and divide it in half with this point: AK = KV.

Let's prove it. Let's connect points A and B with points C and D. /\ CAD = /\ SVD, since by construction AC = CB, AD = BD, CD is the common side.

From the equality of these triangles it follows that / ACK = / VSK, i.e. SK is the bisector of the angle at the vertex of the isosceles triangle ASV. And the bisector of the angle at the vertex of an isosceles triangle is also its median, i.e. straight line CD divides segment AB in half.

Task 2. Draw a perpendicular to a given line AB through point O located on this line.

Given a line AB and a point O lying on this line. It is required to draw a perpendicular to line AB passing through point O.

Solution. Let us plot two equal segments OM and ON on the line AB from point O
(drawing 162). From points M and N, as from centers, we will describe two arcs with the same radius, greater than OM. We connect their intersection point K with point O. KO - median in isosceles triangle MKN, therefore, KO_|_A B (§ 18).

Task 3. Draw a perpendicular to a given line AB through a point C located outside this line.

Given a line AB and a point C outside this line, a perpendicular to the line AB passing through point C is required.

Solution. From point C, as from the center, we describe an arc with such a dius that it intersects straight line AB, for example, at points M and N (Fig. 163). From points M and N, as from centers, we will describe arcs with the same radius, greater than half MN. We connect their intersection point E with point C and with points M and N. Triangles CME and CNE are equal on three sides. Means, / 1 = / 2 and CE is the bisector of angle C in the isosceles triangle MCN, and therefore perpendicular to the straight line AB (§ 18).

Knowing; that the triangles are equal on both sides and the angle between them, we can use a compass and ruler to divide this segment into two equal parts.

To make sure that the segments JSC And OB must be equal, connect the dots C And D with ends A And IN segment (Fig. 71). You will get two triangles ACD And BCD, whose three sides are respectively equal: AC= Sun; AD = BD; CD – common, i.e. belongs to both triangles. This implies the complete equality of these triangles, and therefore the equality of all angles. So, by the way, the angles are equal ACD And BCD. Now comparing the triangles ASO And VSO, we see that they have a side OS – general, A.C. = CB, and the angle between them ASO = ug. VSO. The triangles are equal along two sides and the angle between them; therefore the sides are equal JSC And OB, i.e. point ABOUT there is a midpoint AB.

How to construct a triangle using a side and two angles

Finally, consider a problem whose solution leads to the construction of a triangle using a side and two angles:

On the other side of the river (Fig. 72) a milestone is visible A. It is required, without crossing the river, to find out the distance to it from the milestone IN on this shore.

Together with the previously established signs of equality of triangles, we now know the following three:

Triangles:

on three sides;

at the two sides and at the corner between them;

on the side and two sides.

For the sake of brevity, we will further denote these three cases of equality of triangles as follows:

on three sides: SSS;

on two sides and the angle between them: SUS;

along the side and two corners: USU.

Applications

Solution: Triangles ABC And BDC equal on one side ( Sun) and two angles (ang. DCB= ug. DIA; ug. DBC= ug. ABC.) Hence, AB= ВD, as sides lying in equal triangles against equal angles.

Parallelograms

This is the name of any quadrilateral with 4 right angles (Fig. 77 and 78). A square is also a rectangle, but with equal sides.

Such quadrilaterals, in which, like rectangles, opposite sides are parallel, are called parallelograms. Damn it. 80 shows an example of a parallelogram: AB parallel DC, A AD parallel BC. Damn.80

A rectangle is one of the parallelograms, namely one in which all the angles are right. It is easy to verify that each parallelogram has the following properties:

OPPOSITE ANGLES PARALLEL GRAMMAR EQUAL; Opposite sides

P a r l l e l o g r a m a v y s.

A parallelogram with four equal sides is called a rhombus.

Repeat questions

Applications

15. A square is drawn like this: having set aside one side, draw perpendiculars to it at the ends, put the same lengths on them and connect the ends with a straight line (Drawing 82). How can you be sure that the fourth side of a drawn quadrilateral is equal to the other three and that all its angles are right angles?

Solution. If the formation was carried out in such a way that to the side AB at points A And IN perpendiculars were drawn on which were laid: AC = AB And DВ= AB, then it remains to prove that the angles WITH And D straight and what CD equals AB. To do this, let's draw (Fig. 83) a diagonal A.D. Ugh. CAD = A.D.B. as corresponding (for which parallel ones?); AC= D.B., and therefore triangles CAD And BAD equal (based on SUS). From this we deduce that CD = AB and ug. C = right angle IN. How to prove that the fourth angle CDB is it also straight?

16. How to draw a rectangle? Why can a drawn figure be called a rectangle? (Show that all angles of the drawn figure are right).

The solution is similar to the solution to the previous problem.

17. Prove that both diagonals of the rectangle are equal.

The solution (Fig. 84) follows from the equality of triangles ABC And ABD(based on SUS).

18. Prove that the diagonals of a parallelogram bisect each other.

Solution: Comparing (Fig. 85) triangles AVO And DCO, we make sure that they are equal (based on USU). From here JSC= OS, 0V= OD.

19. The length of the common perpendicular between two parallel lines is called the distance between them. Prove that the distance between parallels is the same everywhere.

Hint: What shape is formed by parallel lines with two perpendiculars between them?

IV. MEASUREMENT OF AREA

Square measures. Palette

In figures, it is often necessary to measure not only the length of the lines and the angles between them, but also the size of the area that they cover - that is, their area. In what units is area measured? A certain length (meter, centimeter) is taken as a measure of length, and a certain angle (1°) is taken as a measure of angles; a certain area is taken as a measure of area, namely, the area of a square with a side of 1 meter, 1 cm, etc. Such a square is called “square meter”, “square centimeter”, etc. To measure an area means to find out how many square units of measure there are in it.

If the area being measured is not large (fits on a sheet of paper), it can be measured as follows. Transparent paper is cut into centimeter squares and placed on the figure being measured. Then it is not difficult to directly count how many square centimeters are contained within the boundaries of the figure. In this case, incomplete squares near the border are taken (by eye) for half a square, a quarter square, etc., or mentally connect them several at a time into whole squares. Transparent paper graphed in this way is called a pallet. This method is often used to measure the areas of irregular areas on a plan.

But it is not always possible or convenient to impose a network of squares on the measured figure. It is not possible, for example, to measure floor area or land plot. In such cases, instead of directly measuring the area, they resort to an unpleasant method, which consists in measuring only the length of some linear figures and performing certain actions on the resulting numbers. Later we will show how this is done.

Repeat questions

What measures are used to determine the area of figures? – What is a palette and how is it used?

Area of a rectangle

Suppose you need to determine the area of some rectangle, for example, ABDC(drawing 86). Measured with a linear unit, e.g. meter, the length of this section. Let's assume that the meter is laid out 5 times in length. Let's divide the area into transverse strips one meter wide, as shown in Fig. 87. Obviously, there will be 5 such stripes. Next, let’s measure the width of the area with a meter; let it be equal to 3 meters. We will divide the area into longitudinal strips 1 meter wide, as shown in Fig. 88; of course, there will be 3 of them. Each of the five transverse strips will be cut into 3 square meters, and the entire plot will be divided into 5 x 3 = 15 squares with a side of 1 meter: we learned that the plot contains 15 square meters. meters. But we could get the same number 15 without graphing the area, but only by multiplying its length by its width. So, to find out how much square meters in a rectangle, you need to measure its length, its width and multiply both numbers.

In the case considered, the unit of length - the meter - was placed on both sides of the rectangle an integer number of times. Detailed mathematics textbooks prove that the rule now established is also true when the sides of the rectangle do not contain an integer number of units of length. In all cases:

Area of rectangular area

the product of the length by the width,

or, as they say, in geometry, – its

“base” on “height”.

If the length of the base of a rectangle is indicated by the letter A, and the length of the height is the letter b, then its area S equal to

S = a? b,

or just S = ab, because the multiplication sign is not placed between the letters.

It is easy to understand that to determine the area of a square, you need to multiply the length of its side by itself, that is, “raise it by the square.” In other words:

The area of a square is equal to the square side. If the side length of a square A, then its area S equal to

S= a? a = a 2.

Knowing this, it is possible to establish the relationship between the various square units. For example, a square meter contains square decimeters 10 X 10, i.e. 100, and square centimeters 100 X 100, i.e. 10,000, because the linear centimeter is placed to the side square decimeter 10 times, and a square meter is 100 times.

To measure land plots it is used special measure– hectare, containing 10,000 square meters. A square plot with a side of 100 meters has an area of 1 hectare; a rectangular plot with a base of 200 meters and a height of 150 meters has an area of 200 x 150, i.e. 30,000 square meters. m or 3 hectares. Large areas - such as counties and districts - are measured

SQUARE KILOMETERS.

The abbreviated designation for square measures is:

square meter………………………………. sq. m or m2

square decimeter…………………………. sq. dm or dm2

square centimeter………………………… sq. cm or cm2

square millimeter……………………….. sq. mm or mm2

hectare…………………………………….. ha

Repeat questions

How is the area of a rectangle calculated? Square? - How many sq. cm to sq. m? How many sq. mm in sq. m? – What is a hectare? – How many hectares in a square? km? What is the abbreviation for square measures?

Applications

20. It is required to paint the interior of the room shown in the drawing. 6. Dimensions are indicated in meters. How much materials and labor will be needed for this, if it is known that to paint one square meter? meters of wooden floors with putty of cracks and branches over previously painted, for two, required (according to the Urgent Regulations):

Malyarov………………………………….. 0.044

Drying oils, kilograms…………………….… 0.18

Light ocher, kg…………………………… 0;099

Putties, kg…………………………………0.00225

Pumice, kg………………………………….. 0.0009.

Solution: Is the floor area 8? 12 = 96 sq. m.

The consumption of materials and labor is as follows

Malyarov........ 0.044? 96 = 4.2

Drying oils......0.18? 96= 17 kg

Ocher......... 0.099? 96 – 9.9 kg

Putties......0.00225? 96 = 0.22 kg

Pumice.........0.0009? 96 = 0.09 kg.

21. Make a statement of the consumption of labor and materials for wallpapering the previous room. tasks. For pasting walls simple wallpaper with curbs is required (according to the Local Regulations) per sq. meter:

Painters or upholsterers………………………… 0.044

Wallpaper (44 cm wide) pieces……………………… 0.264

Curb (according to calculation)

Starch grams………………………………. 90.

Solution - according to the sample indicated in the previous problem. Let us only note that when calculating the required amount of wallpaper, in practice, the openings of the walls are not subtracted from their area (since when fitting figures in adjacent panels, some of the wallpaper is lost).

Area of a triangle

Let's first consider how the area of a right triangle is calculated. Suppose we need to determine the area of a triangle ABC(Fig. 89), in which the angle IN– straight. Let's take you through the peaks A And WITH straight lines parallel to opposite sides. We get (Fig. 90) a rectangle ABCD(why is this figure a rectangle?), which is divided by a diagonal AC into two equal triangles (why?). The area of this rectangle is ah; the area of our triangle is half the area of the rectangle, i.e. equal to 1/2 ah. So, the area of each right triangle equal to half the product of its sides enclosing a right angle.

Suppose now you need to determine the area of an oblique (i.e., not rectangular) triangle - for example. ABC(drawing 91). We draw a perpendicular through one of its vertices to the opposite side; such a perpendicular is called the height of this triangle, and the side to which it is drawn is the base of the triangle. Let us denote the height by h, and the segments into which it divides the base are p And q. Area of a right triangle ABD, as we already know, is equal to 1/2 ph; square VDC = 1/2 qh. Square S triangle ABC equal to the sum of these areas: S= 1/2 ph + 1/2 qh = 1/2 h (r+ q). But r+ q = a; hence S = 1/2 ah.

This reasoning cannot be directly applied to a triangle with an obtuse angle (Fig. 92), because the perpendicular CD does not meet the base AB, and its continuation. In this case, we have to think differently. Let us denote the segment AD through p, BD- through, q, so the basis A triangle is equal p – q. Area of our triangle ABC is equal to the difference in the areas of two triangles ADC – BDC = 1/2 ph – 1/2 qh = 1/2 h (p – q) = 1/2 ah.

So, in all cases, the area of a triangle is equal to half the product of any of its bases and the corresponding height.

It follows that triangles with equal bases and altitudes have equal areas, or, as they say,

equals.

Equal-sized figures are generally those that have equal areas, at least the figures themselves were not equal (that is, they did not coincide when superimposed).

Repeat questions

What is the height of a triangle called? The base of the triangle? – How many heights can be drawn in one triangle? – Draw a triangle with an obtuse angle and draw all the heights in it. – How is the area of a triangle calculated? How to express this rule in a formula? – What figures are called equal in size?

Applications

22. The vegetable garden has the shape of a triangle with a base of 13.4 m and a height of 37.2 m... How many seeds (by weight) are required to plant it with cabbage, if per sq. m is 0.5 grams of seeds?

Solution: Is the area of the vegetable garden 13.4? 37.2 = 498 sq. m.

You will need 250 g of seeds.

23. The parallelogram is divided by diagonals into 4 triangular parts. Which one has the most large area?

Solution. All 4 triangles are equal in size, since they have equal bases and heights.

Area of a parallelogram

The rule for calculating the area of a parallelogram is established very simply if you divide it by a diagonal into two triangles. For example, the area of a parallelogram ABCD(Fig. 93) is equal to double the mercy of each of the two equal triangles, into which it is divided by the diagonal AC. Marking the base of the triangle ADC through A, and the height through h, we get the area S parallelogram

Perpendicular h is called the “parallelogram height”, and the side A, to which it is drawn - “the base of the parallelogram”. Therefore, the rule now established can be stated as follows:

The area of the parallelogram is equal to the product of any new height.

Repeat questions

What is the base and height of a parallelogram? How is the area of a parallelogram calculated? – Express this rule in a formula. – How many times is the area of a parallelogram greater than the area of a triangle that has the same base and height? – Given equal heights and bases, which figure has the largest area: a rectangle or a parallelogram?

Application

24. A square with a side of 12.4 cm is equal in size to a parallelogram with a height of 8.8 cm. Find the base of the parallelogram.

Solution. The area of this square, and therefore the parallelogram, is 12.42 = 154 square meters. cm. The required base is 154: 8.8 = 18 cm.

Area of trapezoid

In addition to parallelograms, let's consider another type of quadrilaterals - namely those that have only one pair of parallel sides (Fig. 94). Such figures are called trapezoids. The parallel sides of a trapezoid are called its bases, and the non-parallel sides are called its sides.

Crap. 94 Damn. 95

Let us establish a rule for calculating the area of a trapezoid. Suppose we need to calculate the area of a trapezoid ABCD(Fig. 95), the length of the bases of which a And b. Let's draw a diagonal AC, which cuts a trapezoid into two triangles ACD And ABC. We know that

area ACD = 1/2 ah

area ABC = 1/2 bh.

area ABCD= 1/2 ah+ 1/2 bh= 1/2 (a+ b) h.

Since the distance h between the bases of a trapezoid is called its height, then the rule for calculating the area of a trapezoid can be stated as follows:

The area of a trapezoid is equal to half the sum multiplied by and in you with about t at.

Repeat questions

What shape is called a trapezoid? What are the bases of a trapezoid, its sides and height called? – How is the area of a trapezoid calculated?

Applications

25. A section of the street has the shape of a trapezoid with bases of 180 m and 170 m and a height of 8.5 m. How many wooden blocks will be required to lay it, if per sq. m. m there are 48 checkers?

Solution. The area of the plot is 8.5 H = (180 + 170)/ 2 = 1490 sq. m. Number of checkers = 72,000.

26. The roof slope has the shape of a trapezoid, the bases of which are 23.6 m and 19.8 m, and the height is 8.2 m. How much material and labor will be required to cover it, if per sq. m required:

Iron sheets...... 1.23

Roofing nails kg.... 0.032

Drying oils kg........0.036

Roofers...... 0.45.

Solution: Is the area of the slope equal to 8.2? (23.6 + 19.8)/ 2 = 178 sq. m. It remains to multiply all the numbers on the tablet by 178.