The diagonal of an equilateral trapezoid. What is the average trapezoid line. Types of trapezium. The trapeze is ..

Trapezoid is a special case of a quadrilateral, in which one pair of sides is parallel. The term "trapezoid" comes from the Greek word τράπεζα, meaning "table", "table". In this article, we will look at the trapezium types and its properties. In addition, we will understand how to calculate the individual elements of this geometric figure. For example, the diagonal of an equilateral trapezium, the middle line, the area, etc. The material is described in the style of elementary popular geometry, i.e. in an easily accessible form.

General information

First, let's see what a quadrilateral is. This figure is a special case of a polygon containing four sides and four vertices. Two vertices of a quadrangle that are not adjacent are called opposite vertices. The same can be said about the two non-contiguous sides. The main types of quadrilaterals are a parallelogram, a rectangle, a rhombus, a square, a trapezoid and a deloid.

So, back to the trapezoid. As we have already said, this figure has two sides that are parallel. They are called bases. The other two (non-parallel) are the sides. In the materials of examinations and various tests, it is very often possible to meet the tasks associated with trapezoids, the solution of which often requires the student to have knowledge not provided by the program. The school course of geometry introduces students to the properties of angles and diagonals, as well as the middle line of an isosceles trapezium. But after all, besides this, the mentioned geometric figure has other features. But about them later ...

Types of trapezoid

There are many kinds of this figure. However, two of them are usually considered to be isosceles and rectangular.

1. A rectangular trapezoid is a figure in which one of the lateral sides is perpendicular to the bases. It has two angles always equal to ninety degrees.

2. An isosceles trapezoid is a geometric figure whose sides are equal to each other. This means that the angles of the bases are also equal in pairs.

The main principles of the technique of studying the trapezium properties

The main principle is the use of the so-called problem approach. In fact, there is no need to introduce new properties of this figure into the theoretical geometry course. They can be opened and formulated in the process of solving various problems (better system ones). At the same time, it is very important that the teacher knows what tasks should be set before the schoolchildren at one or another moment of the educational process. Moreover, each trapezium property can be represented as a key task in the system of tasks.

The second principle is the so-called spiral organization of studying the "remarkable" trapezium properties. This implies a return in the learning process to the individual features of the given geometric figure. Thus, students are easier to remember. For example, the property of four points. It can be proved both in the study of similarity and later with the help of vectors. And the equality of the triangles adjacent to the sides of the figure can be proved by applying not only the properties of triangles with equal heights drawn to sides that lie on one line, but also using the formula S = 1/2 (ab * sinα). In addition, one can work out the sine theorem on an inscribed trapezoid or a right triangle on the trapezium described, and so on.

The application of "non-programmatic" features of the geometric figure in the content of the school course is a prudent technology for their teaching. Constant appeal to the studied properties in the passage of other topics allows students to better understand the trapezoid and ensures the success of the solution of the tasks. So, let's start studying this remarkable figure.

Elements and properties of an isosceles trapezoid

As we have already noted, in this geometric figure the sides are equal. She is also known as the right trapezoid. And why is it so remarkable and why did it get such a name? The peculiarity of this figure is that, not only the sides and corners of the bases are equal, but also the diagonals. In addition, the sum of the angles of an isosceles trapezoid is 360 degrees. But that's not all! Of all the known trapezoids, only around an isosceles one can describe a circle. This is due to the fact that the sum of the opposite angles in this figure is 180 degrees, but only under such condition it is possible to describe the circle around the quadrilateral. The next property of the geometrical figure in question is that the distance from the top of the base to the projection of the opposite vertex to the line that contains this base will be equal to the midline.

And now let's figure out how to find the angles of an isosceles trapezoid. Let us consider the solution of this problem, provided that the dimensions of the sides of the figure are known.

Decision

Usually a quadrilateral is usually denoted by letters A, B, C, D, where BS and AD are the bases. In the isosceles trapezium, the sides are equal. We will assume that their size is equal to X, and the sizes of the bases are equal to Y and Z (smaller and larger, respectively). To carry out the calculation it is necessary to draw the height H. from the angle B. As a result, we have a rectangular triangle ABN, where AB is the hypotenuse, and BN and AN are the legs. We calculate the size of the AN: from the larger base we subtract the smaller, and divide the result by 2. We write in the form of the formula: (ZY) / 2 = F. Now, to calculate the acute angle of the triangle, we use the function cos. We get the following notation: cos (β) = X / F. Now calculate the angle: β = arcos (X / F). Further, knowing one corner, we can define the second, for this we make the elementary arithmetic action: 180 - β. All angles are defined.

There is also a second solution to this problem. In the beginning, we lower the height H from the angle B. We compute the value of the BN cate- gence. We know that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the legs. We get: BN = √ (X2-F2). Next, we use the trigonometric function tg. As a result we have: β = arctg (BN / F). Acute angle is found. Next, we define the obtuse angle similarly to the first method.

The property of diagonals of an isosceles trapezoid

First, we write down four rules. If the diagonals in an isosceles trapezium are perpendicular, then:

- the height of the figure will be equal to the sum of the bases divided by two;

- its height and middle line are equal;

- the area of the trapezoid will be equal to the square of the height (the middle line, half the sum of the bases);

- the square of the diagonal is equal to half the square of the sum of the bases or to the doubled square of the midline (height).

Now we consider the formulas that determine the diagonal of an equilateral trapezium. This block of information can be divided into four parts:

1. The formula for the length of the diagonal across its sides.

Assume that A is the bottom base, B is the top, C is equal sides, and D is the diagonal. In this case, the length can be determined as follows:

D = √ (C2 + A * B).

2. Formula for the length of the diagonal by the cosine theorem.

Assume that A is the bottom base, B is the top, B is the top side, D is the diagonal, α (at the bottom base) and β (at the upper base) are the corners of the trapezoid. We obtain the following formulas, by means of which we can calculate the length of the diagonal:

- Д = √ (А2 + С2-2А * С * cosα);

- Д = √ (А2 + С2-2А * С * cosβ);

- Д = √ (В2 + С2-2В * С * cosβ);

- Д = √ (В2 + С2-2В * С * cosα).

3. Formula for the length of diagonals of an isosceles trapezoid.

Assume that A is the bottom base, B is the top, D is the diagonal, M is the midline, H is the height, P is the trapezium area, and α and β are the angles between the diagonals. Determine the length of the following formulas:

- D = √ (M2 + H2);

- D = √ (H2 + (A + B) 2/4);

- D = √ (H (A + B) / sinα) = √ (2P / sinα) = √ (2M * H / sinα).

For this case, the equality: sinα = sinβ.

4. Diagonal length formulas through sides and height.

Assume that A is the bottom base, B is the top, C is the side, D is the diagonal, H is the height, and α is the angle with the bottom base.

Determine the length of the following formulas:

- D = √ (H2 + (A-P * ctgα) 2);

- Д = √ (Н2 + (В + Р * ctgα) 2);

- Д = √ (А2 + С2-2А * √ (С2-Н2)).

Elements and properties of a rectangular trapezoid

Let's look at what is interesting about this geometric figure. As we have already said, a rectangular trapezoid has two right angles.

In addition to the classical definition, there are others. For example, a rectangular trapezoid is a trapezoid in which one side is perpendicular to the bases. Or a figure with right angles at the side. In this type of trapezium, the height is equal to the lateral side, which is perpendicular to the bases. The middle line is the segment that connects the middle of the two sides. The property of the element mentioned is that it is parallel to the bases and is equal to half of their sum.

Now let's look at the basic formulas that define this geometric figure. For this we assume that A and B are bases; C (perpendicular to the bases) and D - sides of the rectangular trapezoid, M - middle line, α - acute angle, P - area.

1. The lateral side perpendicular to the bases is equal to the height of the figure (C = H), and is equal to the product of the length of the second side D and the sine of the angle α for a larger base (C = D * sinα). In addition, it is equal to the product of the tangent of the acute angle α and the difference in the bases: C = (A-B) * tgα.

2. The side D (not perpendicular to the bases) is equal to the particular difference A and B and the cosine (α) of the acute angle or the partial height of the figure H and the sine of the acute angle: D = (A-B) / cos α = C / sinα.

3. The side that is perpendicular to the bases is equal to the square root of the difference between the square of D-the second side and the square of the difference in the bases:

C = √ (A2- (A-B) 2).

4. The side D of the rectangular trapezoid is equal to the square root of the sum of the square of the side C and the square of the difference in the bases of the geometric figure: D = √ (C2 + (A-B) 2).

5. The side C is equal to the quotient of dividing the double area by the sum of its bases: C = П / М = 2П / (А + Б).

6. The area is determined by the product M (the middle line of the rectangular trapezoid) to the height or the lateral side perpendicular to the bases: П = М * Н = М * С.

7. The side C is equal to the quotient of dividing the doubled area of the figure by the product of the sine of the acute angle and the sum of its bases: C = П / М * sinα = 2П / ((А + Б) * sinα).

8. The formulas of the lateral side of a rectangular trapezium through its diagonals and the angle between them:

- sinα = sinβ;

- C = (A1 * A2 / (A + B)) * sinα = (A1 * A2 / (A + B)) * sinβ,

Where D1 and D2 are the diagonals of the trapezium; Α and β are the angles between them.

9. Formulas of the lateral side through the angle at the bottom base and other sides: D = (AB) / cosα = C / sinα = H / sinα.

Since the trapezoid with a right angle is a particular case of a trapezoid, the rest of the formulas that define these figures will correspond to a rectangular one.

Inscribed circle properties

If the condition says that a circle is inscribed in a rectangular trapezoid, then you can use the following properties:

- the sum of the bases is equal to the sum of the lateral sides;

- the distances from the top of the rectangular figure to the points of tangency of the inscribed circle are always equal;

- the height of the trapezoid is equal to the lateral side, perpendicular to the bases, and is equal to the diameter of the circle ;

The center of the circle is the point at which the bisectors of the angles intersect;

- if the lateral side is divided by the point of tangency into segments H and M, then the radius of the circle is equal to the square root of the product of these segments;

- a quadrangle that is formed by points of tangency, the vertex of the trapezoid and the center of the inscribed circle is a square whose side is equal to the radius;

- the area of the figure is equal to the product of the bases and the product of the half-sum of the bases to its height.

Similar trapeziums

This topic is very convenient for studying the properties of this geometric figure. For example, the diagonals split the trapezoid into four triangles, the adjacent to the bases being similar, and to the sides being equal. This statement can be called the property of triangles, to which the trapezoid is divided by its diagonals. The first part of this assertion is proved through the similarity criterion at two angles. To prove the second part, it is better to use the method given below.

Proof of the theorem

We assume that the ABSD pattern (AD and BS - the trapezoidal base) is broken by the diagonals of VD and AC. The point of their intersection is O. We obtain four triangles: AOS - at the bottom base, BOS - at the upper base, ABO and SOD at the lateral sides. The triangles of SOD and BFD have a common height in the case when the segments BD and OD are their bases. We get that the difference in their areas (Π) is equal to the difference of these segments: ΠС / / ПСОД = = = / / / Д = = Следовательно. Therefore, the LDPE = NSP / K. Similarly, the triangles BF and AOB have a common height. We take the CO and OA segments as their bases. We get the PBO / PAOB = CO / OA = K and PAOB = PBO / K. From this it follows that the PSCM = PAOB.

To fix the material, students are encouraged to find a connection between the areas of the resulting triangles, to which the trapezium is divided by its diagonals, solving the following problem. It is known that the triangles of the BF and ADN areas are equal, it is necessary to find the area of the trapezoid. Since the LDPE = PAOB, it means that the PABSD = PBO + PAOJD + 2 * PODC. From the similarity of the triangles of BFU and ADN, it follows that BD / DD = √ (PBO / PAOD). Consequently, the BSP / DPPM = BW / DD = √ (PBO / PAOD). We get the LDP = √ (PBO * PAOD). Then the PABSD = PBO + PAOAD + 2 * √ (PAO * PAOD) = (√POPS + √PAOOD) 2.

Similarity properties

Continuing to develop this topic, it is possible to prove other interesting trapezoidal features. Thus, using similarity, we can prove the property of a segment that passes through a point formed by the intersection of the diagonals of this geometric figure, parallel to the bases. To do this, we solve the following problem: it is necessary to find the length of the segment PK that passes through the point O. From the similarity of the triangles ADD and BFD it follows that AO / OC = AD / BS. From the similarity of the triangles AOP and ASB it follows that AO / AC = PO / BS = AD / (BS + AD). From this we obtain that PO = BC * AD / (BS + AD). Similarly, from the similarity of the triangles DKK and DBS it follows that OK = BS * AD / (BS + AD). From this it follows that PO = OK and PK = 2 * BS * AD / (BS + AD). The segment passing through the point of intersection of the diagonals parallel to the bases and connecting the two lateral sides is divided by the intersection point in half. Its length is the average harmonic basis of the figure.

Consider the following trapezoidal quality, which is called the property of four points. The intersection points of the diagonals (O), the intersections of the extension of the lateral sides (E), and also the middle of the bases (T and M) always lie on one line. This is easily proved by the similarity method. The triangles BEC and AED obtained are similar, and in each of them the medians ET and EF divide the angle at the vertex of E into equal parts. Consequently, the points E, T and M lie on one line. In exactly the same way, the points T, 0, and M are located on one straight line. All this follows from the similarity of the triangles BOS and AOD. Hence, we conclude that all four points - E, T, O and M - will lie on one straight line.

Using similar trapezes, you can ask students to find the length of the segment (LF), which breaks the figure into two similar ones. This segment must be parallel to the bases. Since the obtained trapezoids of ALFD and LBSF are similar, then BS / LF = LF / AD. It follows that LF = √ (BS * AD). We get that the segment dividing the trapezoid into two similar ones has a length equal to the average geometric length of the base of the figure.

Consider the following similarity property. At its base lies a segment that divides the trapezoid into two equal-sized figures. We assume that the trapezoid of ABSD is divided by a section of EH into two similar ones. A height is dropped from the vertex B, which is divided by a segment EH into two parts - B1 and B2. We get: PABSD / 2 = (BS + EH) * B1 / 2 = (AD + EH) * B2 / 2 and PABSD = (BS + AD) * (B1 + B2) / 2. Next, we make up a system whose first equation is (BS + EH) * B1 = (AD + EH) * B2 and the second (BS + EH) * B1 = (BS + AD) * (B1 + B2) / 2. Hence it follows that B2 / B1 = (BS + EH) / (AD + EH) and BS + EH = ((BS + AD) / 2) * (1 + B2 / B1). We obtain that the length of the segment dividing the trapezoid into two equal parts is equal to the mean square root length: √ ((BS2 + AD2) / 2).

Similarity conclusions

Thus, we have proved that:

1. The segment connecting at the trapezium of the middle of the lateral sides is parallel to the arterial and BS and is equal to the arithmetic mean of the BS and AD (the length of the base of the trapezium).

2. The line passing through the point O of the intersection of the diagonals parallel to the AD and BS will be equal to the mean harmonic of the numbers AD and BS (2 * BS * AD / (BS + AD)).

3. The segment dividing the trapezoid into similar ones has the length of the average geometric bases of the BS and AD.

4. The element dividing the figure into two equal parts has the length of the mean square of the numbers AD and BS.

To consolidate the material and realize the connection between the segments examined, the student needs to build them for a specific trapezoid. It can easily display the middle line and the segment that passes through point O - the intersection of the diagonals of the figure - parallel to the bases. But where will the third and fourth be? This answer will lead the student to the discovery of the desired connection between the mean values.

The segment connecting the midpoints of the diagonals of the trapezoid

Consider the following property of this figure. We assume that the segment MN is parallel to the bases and divides the diagonals in half. The points of intersection will be called W and W. This segment will be equal to the base half-difference. Let us analyze this in more detail. The MS is the middle line of the triangle ABC, it is equal to BS / 2. MN is the middle line of the triangle ABD, it is equal to AD / 2. Then we obtain that M, = MN-MN, and consequently, M, = A / 2-BC / 2 = (AD + BC) / 2.

Center of gravity

Let's look at how this element is defined for a given geometric figure. For this, it is necessary to extend the bases in opposite directions. What does it mean? It is necessary to add to the upper base the lower one - to either side, for example, to the right. And the bottom is extended by the length of the upper left. Then connect them with a diagonal. The point of intersection of this segment with the middle line of the figure is the center of gravity of the trapezoid.

Inscribed and described trapeziums

Let's list the features of such figures:

1. A trapezoid can be inscribed in a circle only if it is isosceles.

2. Around the circumference one can describe a trapezoid, provided that the sum of the lengths of their bases is equal to the sum of the lengths of the lateral sides.

Consequences of the inscribed circle:

1. The height of the trapezium described is always equal to two radii.

2. The lateral side of the described trapezium is observed from the center of the circle at a right angle.

The first corollary is obvious, and to prove the second it is required to establish that the angle of the SOD is direct, which, in fact, also does not amount to much difficulty. But knowledge of this property will allow us to apply a right-angled triangle when solving problems.

Now let us concretize these consequences for an isosceles trapezoid, which is inscribed in a circle. We get that the height is the geometric mean of the base of the figure: H = 2R = √ (BS * AD). Working out the basic method of solving problems for trapezoids (the principle of holding two heights), the student must solve the following task. We assume that BT is the height of an isosceles figure of ABSD. It is necessary to find segments AT and TD. Applying the formula described above, this will not be difficult to do.

Now let's figure out how to determine the radius of a circle using the area of the trapezium described. We lower the height from the top B to the base of the blood pressure. Since the circle is inscribed in the trapezoid, then BS + AD = 2AB or AB = (BS + AD) / 2. From the triangle ABN we find sinα = BN / AB = 2 * BN / (BS + AD). PABSD = (BS + AD) * BN / 2, BN = 2R. We get the PABSD = (BS + AD) * R, it follows that R = PABSD / (BS + AD).

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All formulas of the trapezium midline

Now it's time to go to the last element of this geometric figure. Let's see what the middle line of the trapezoid (M) is:

1. Through the bases: M = (A + B) / 2.

2. Through height, base and angles:

• M = A-H * (ctgα + ctgβ) / 2;

• M = B + H * (ctgα + ctgβ) / 2.

3. Through the height, the diagonals and the angle between them. For example, D1 and D2 are diagonals of the trapezoid; Α, β are the angles between them:

M = D1 * D2 * sinα / 2H = D1 * D2 * sinβ / 2H.

4. Through the area and height: M = P / H.