Diagonal an equilateral trapezoid.

-Line - is a special case of a quadrilateral that has one pair of parallel sides is.The term "Keystone" is derived from the Greek word τράπεζα, meaning "table", "table".In this article we consider the types of trapeze and its properties.Also, we look at how to calculate the individual elements of the geometric figure.For example, the diagonal of an equilateral trapezoid, the middle line, area, and others. The material is presented in the style of the popular elementary geometry, t. E. In an easily accessible form.


First, let's understand what the quadrangle.This figure is a special case of a polygon having four sides and four vertices.Two vertices of the quadrangle that are not adjacent are called opposite.The same can be said of the two non-adjacent sides.The main types of quadrangles - a parallelogram, rectangle, diamond, square, trapezoid and deltoid.

So back to the trapeze.As we have said, this figure the two sides are parallel.They are called bases.The other two (non-parallel) - sides.The materials of the various examinations and examinations very often you can find the tasks associated with trapezoids whose solution often requires the student's knowledge, is not provided by the program.The school geometry course introduces students to the properties of angles and diagonals and midline of an isosceles trapezoid.But other than that referred to a geometric figure has other features.But about them later ...


Types There are many types of this figure.However, most agreed to consider two of them - isosceles and rectangular.

1. Rectangular Trapezoid - a figure in which one of the sides perpendicular to the base.She has two angles are always ninety degrees.

2. isosceles trapezoid - a geometrical figure whose sides are equal.And that means, and the angles at the base pairs as equal.

main principles of methods for studying the properties of a trapezoid

to the basic principles include the use of so-called task approach.In fact, there is no need to enter into a theoretical course Geometry of new properties of this figure.They can be open or in the process of formulating the various tasks (better system).It is very important that the teacher know what tasks you need to put in front of students in a given moment of the educational process.Furthermore, each property trapeze can be represented as a key task in the task.

The second principle is the so-called spiral organization of the study "remarkable" property trapeze.This implies a return to the process of learning to the individual features of the geometric figure.Thus, it is easier for students to memorize them.For example, four feature points.It can be proved as in the study of similarity, and subsequently using the vectors.And of equal triangles adjacent to the sides of the figure, it is possible to prove, using not only the properties of triangles with equal heights, carried out to the sides, which lie on a straight line, but also by the formula S = 1/2 (ab * sinα).In addition, it is possible to work out the law of sines inscribed on the trapeze or a right triangle described on the trapeze, and so on. D.

use of "extracurricular" features a geometric figure in the content of school course - tasking is the technology of their teaching.Constant reference to study the properties of the passage of the other allows students to learn the trapeze deeper and provides the solution of the tasks.So, we proceed to the study of this remarkable figure.

elements and properties of an isosceles trapezoid

As we have noted, in this geometrical figure the sides are equal.Yet it is known as a right trapezoid.And what is she so remarkable and why got its name?The special features of this figure relates that she not only equal sides and angles at the bases, but also diagonally.In addition, the angles of an isosceles trapezoid is equal to 360 degrees.But that's not all!Of all the isosceles trapezoids only around a circle can be described.This is due to the fact that the sum of opposite angles in the figure is 180 degrees, but only when this condition can be described by a circle around the quad.The following properties of geometric figures is considered that the distance from the top of the base opposite to the projection of the vertex on a straight line which contains this base will be equal to the midline.

Now let's look at how to find the corners of an isosceles trapezoid.Consider the case of solutions to this problem provided that the known dimensions of the sides of the figure.


usually rectangle is denoted by the letters A, B, C, D, where BC and AD - a foundation.The isosceles trapezium sides are equal.We assume that X is equal to their size, and the size of the base is Y, and Z (smaller and larger, respectively).To carry out the calculation of the angle necessary to hold in height H. The result is a right-angled triangle ABN, where AB - the hypotenuse, and the BN and AN - legs are.We calculate the size of the leg AN: With base takes less and the result is divided by 2. We write as a formula: (ZY) / 2 = F. Now, for the calculation of the acute angle of the triangle we use the function cos.We get the following entry: cos (β) = X / F.Now we calculate the angle: β = arcos (X / F).Further, knowing one corner, we can determine the second, for it is elementary arithmetic operation: 180 - β.All angles are defined.

There is a second solution to this problem.At the beginning we omit from corner to calculate the value of the height H. leg BN.We know that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.We get: BN = √ (X2 F2).Next, we use the trigonometric function tg.The result is: β = arctg (BN / F).Acute angle found.Next, we define an obtuse angle similar to the first method.

property diagonals of an isosceles trapezoid

write the first four rules.If the diagonal in an isosceles trapezoid perpendicular, then:

- the height of the figure is the sum of the bases, divided by two;

- its height and the middle line are equal;

- area of ​​a trapezoid is equal to the square of the height (the middle line, half the sum of the bases);

- diagonal square is half the sum of the square of bases or twice the square of the mean line (height).

Now consider the formula determining the diagonal of an equilateral trapezoid.This piece of information can be divided into four parts:

1. Formula length diagonally across her.

accepted that A - lower base, B - upper C - equal sides, D - diagonal.In this case, the length can be determined as follows:

D = √ (C 2 + A * B).

2. Formula for the length of the diagonal of the law of cosines.

accepted that A - lower base, B - upper C - equal sides, D - diagonal, α (at the lower base) and β (the upper base) - the corners of a trapezoid.We get the following formula, with which you can calculate the length of the diagonal:

- D = √ (A2 + S2-2A * On * cosα);

- D = √ (A2 + S2-2A * C * cosβ);

- D = √ (B2 + S2-2V * C * cosβ);

- D = √ (B2 + S2-2V * C * cosα).

3. Formula lengths of the diagonals of an isosceles trapezoid.

accepted that A - lower base, B - upper, D - diagonal, M - middle line, H - height, P - the area of ​​a trapezoid, α and β - the angle between the diagonals.Determine the length of the following formulas:

- D = √ (M2 + H2);

- D = √ (H2 + (A + B) 2/4);

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

Adhoc equality: sinα = sinβ.

4. Formula diagonally across the length and height of the part.

accepted that A - lower base, B - upper C - sides, D - diagonal, H - height, α - angle of the lower base.

Determine the length of the following formulas:

- D = √ (H2 + (A-P * ctgα) 2);

- D = √ (H2 + (B + P * ctgα) 2);

- D = √ (A2 + S2-2A * √ (C 2 H 2)).

elements and properties of rectangular trapezoid

Let's see what this is interesting geometric shapes.As we have said, we have a rectangular trapezoid two right angles.

Besides the classical definition, there are others.For example, a rectangular trapezium - a trapezoid, one side of which is perpendicular to the substrates.Or shapes having at the side angles.In this type of trapezoids height is the side that is perpendicular to the base.The middle line - a segment connecting the midpoints of the two sides.The property of said element is that it is parallel to the bases, and is equal to half of their sum.

Now let's consider the basic formulas that define the geometric shapes.To do this we assume that the A and B - base;C (perpendicular to the base) and D - the part of the rectangular trapezoid, M - middle line, α - an acute angle, P - Square.

1. The side, perpendicular to the base, a figure equal to the height (C = N), and is equal to the length of the second side A and the sine of the angle α at a higher basis (C = A * sinα).Moreover, it is equal to the product of the tangent of the acute angle α and the difference in bases: C = (A-B) * tgα.

2. The side of the D (not perpendicular to the base) equal to the quotient of the difference of A and B and the cosine (α) an acute angle or a private figure height H and sinus acute angle: A = (A-B) / cos α = C / sinα.

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

C = √ (q2 (AB) 2).

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

5. The side of C is equal to the quotient of the sum of double the area of ​​its grounds: C = P / M = 2n / (A + B).

6. Area defined by the product M (middle line of a rectangular trapezoid) to the height or the side, perpendicular to the base: P = M * N = M * C.

7. Party C is equal to the quotient of twice the area of ​​the figure in the work of the sinus acute angle and the sum of its bases: C = P / M * sinα = 2n / ((A + B) * sinα).

8. Formula side of the rectangular trapezoid across its diagonal and the angle between them:

- sinα = sinβ;

- C = (D1 * D2 / (A + B)) * sinα = (D1 * D2 / (A + B)) * sinβ,

where D1 and D2 - diagonal trapezoid;α and β - the angle between them.

9. Formula side through a corner at the lower base and the other parties: D = (A-B) / cosα = C / sinα = N / sinα.

Since trapezoid with a right angle is a special case of the trapezoid, the other formulas that determine these figures will meet and rectangular.

Properties inscribed circle

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

- the amount is the sum of the bases sides;

- the distance from the top of a rectangular shape to the points of contact of the inscribed circle is always equal;

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

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

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

- quadrangle, which formed the points of contact, the apex of the trapezoid and the center of the inscribed circle - a square whose side is equal to the radius;

- area of ​​the figure is equal to the product of half-sum basis and grounds for its height.

Similar trapeze

This topic is very useful for studying the properties of geometric figures.For example, diagonally split trapeze into four triangles, and adjacent to the bases are similar, and to the sides - by equal.This statement may be called a property of triangles, which are broken trapeze its diagonals.The first part of this statement is proved by an indication of similarity in the two corners.To prove the second part is better to use the method below.

The proof

accepted that the figure ABSD (AD and BC - the basis of the trapezoid) is broken diagonals HP and AC.The point of intersection - O. We get four triangles: AOC - at the lower base, BOS - at the upper base, ABO and SOD at the sides.Triangles SOD and biofeedback have a common height in that case, if the segments CD and OD are their bases.We find that the difference in their areas (P) is equal to the difference between these segments: PBOS / PSOD = BO / ML = K. Hence PSOD PBOS = / K.Similarly, the triangles AOB and biofeedback have a common height.We accept their base segments SB and OA.We get the PBOS / PAOB = CO / OA = K and PAOB PBOS = / K.It follows that PSOD = PAOB.

To consolidate the material is recommended for students to find a connection between the areas of triangles obtained, which is broken trapeze its diagonals, deciding the next task.It is known that triangles BOS and ADP areas are equal, you must find the area of ​​a trapezoid.Since PSOD = PAOB, then PABSD PBOS + = PAOD + 2 * PSOD.From the similarity of triangles BOS and ADP shows that BO / OD = √ (PBOS / PAOD).Consequently, PBOS / PSOD = BO / OD = √ (PBOS / PAOD).We get the PSOD = √ (* PBOS PAOD).Then PABSD PBOS + = PAOD + 2 * √ (PAOD PBOS *) = (+ √PBOS √PAOD) 2.

Properties similarity

Continuing to develop this theme, you can prove the other interesting features of the trapezoids.Thus, using the similarity can prove property section which passes through the point formed by the intersection of the diagonals of this geometrical figure, parallel to the base.To do this will solve the following problem: you need to find the length of the segment of the RK, which passes through the point O. From the similarity of triangles ADP and biofeedback follows that AO / OS = BP / BS.From the similarity of triangles ADP and ASB follows that AB / AC = PO / BS = AD / (BS + BP).This implies that PO = BS * BP / (BS + BP).Similarly, from the similarity of triangles MLC and DBS follows that OK = BS * BP / (BS + BP).This implies that PO = OK and RK = 2 * BS * BP / (BS + BP).The segment passing through the point of intersection of the diagonals, parallel to the base and connecting the two sides of the divided point of intersection of two.Its length - is the harmonic mean of the bases of the figure.

Consider the following quality trapezoid, which is called the property of the four points.The points of intersection of the diagonals (D), the intersections continue sides (E) and the middle base (T and G) always lie on the same line.This is easily proved by similarity.These triangles BES and AED are similar, and in each of them, and the median ET HEDGEHOG divide the apex angle E in equal parts.Consequently, the point E, T and F are collinear.Similarly, on the same line are arranged in terms of T, O, and G. This follows from the similarity of triangles BOS and ADP.Hence, we conclude that all four points - E, T, O and F - will lie on a straight line.

Using similar trapezoids, can be offered to students to find the length of the segment (LF), which divides into two similar figure.This segment must be parallel to the bases.Since obtained trapeze ALFD and LBSF similar, the BS / LF = LF / AD.This implies that the LF = √ (BS * BP).We find that the segment breaking like a trapezoid into two, has a length equal to the geometric mean length of the base figure.

Consider the following property of similarity.It is based on the segment, which divides the trapezoid into two equal size pieces.We accept that Keystone ABSD segment is divided into two like EN.From the top of the B lowered the height of that segment is divided into two parts EN - B1 and B2.We obtain PABSD / 2 = (BS EN +) * B1 / 2 = (AD + EN) * B2 / 2 = PABSD (BS + BP) * (B1 + B2) / 2.Next compose the system, the first equation is (BS EN +) * B1 = (AD + EN) * B2 and the second (BS EN +) * B1 = (BS + BP) * (B1 + B2) / 2.It follows that B2 / B1 = (BS EH +) / (AD + EH) and BS EN + = ((BS + BP) / 2) * (1 + B2 / B1).We find that the length of the segment, dividing the trapezoid into two equal size, equal to the average quadratic length of the base: √ ((BS2 + w2) / 2).

Conclusions similarity

Thus, we have proved that:

1. The line segment joining in the middle of the trapezoid sides, parallel to AD and BC and is equal to the average BC and AD (the length of the base of the trapezoid).

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

3. Cut, breaking on the trapeze like, has a length of the geometric mean of the bases BC and AD.

4. The element that divides the figure into two equal size, has a length of average square numbers of AD and BC.

To consolidate the material and understanding of linkages between the segments of the student is necessary to build them for a particular trapeze.What does it mean?