What is the midline in a trapezoid? Properties of a trapezoid

In solving planimetric problems, in addition to the sides and angles of a figure, other quantities often take an active part - medians, heights, diagonals, bisectors and others. These include the middle line.
If the original polygon is a trapezoid, what is its midline? This segment is a part of a straight line that intersects the sides of the figure in the middle and is located parallel to the other two sides - the bases.

How to find the midline of a trapezoid through the line of the middle and the base

If the values ​​of the upper and lower bases are known, then the expression will help to calculate the unknown:

a, b – bases, l – midline.

How to find the midline of a trapezoid through an area

If the source data contains the area of ​​the figure, then using this value you can also calculate the length of the line in the middle of the trapezoid. Let's use the formula S = (a+b)/2*h,
S – area,
h – height,
a, b – bases.
But, since l = (a+b)/2, then S = l*h, which means l=S/h.

How to find the midline of a trapezoid through the base and its angles

Given the length of the larger base of the figure, its height, as well as known degree measures of the angles at it, the expression for finding the line of the middle of the trapezoid will have the following form:

l=a – h*(ctgα+ctgβ)/2, while
l is the desired value,
a – larger base,
α, β are the angles at it,
h – height of the figure.

If the value of the smaller base is known (given the same other data), the following relation will help find the midline line:

l=b+h*(ctgα+ctgβ)/2,

l is the desired value,
b – smaller base,
α, β are the angles at it,
h – height of the figure.

Find the midline of a trapezoid using height, diagonals and angles

Let's consider a situation where the problem conditions include the values ​​of the diagonals of the figure, the angles they form when intersecting each other, as well as the height. You can calculate the center line using the following expressions:

l=(d1*d2)/2h*sinγ or l=(d1*d2)/2h*sinφ,

l – midline,
d1, d2 – diagonals,
φ, γ – angles between them,
h – height of the figure.

How to find the midline of a trapezoid For an isosceles figure

If the basic figure is an isosceles trapezoid, the above formulas will have the following form.

• If the values ​​of the trapezoid bases are present, there will be no changes in the expression.

l = (a+b)/2, a, b – bases, l – middle line.

• If the height, base and angles adjacent to it are known, then:

l=a-h*ctgα,
l=b+h*ctgα,

l – midline,
a, b – bases (b< a),
α are the angles at it,
h – height of the figure.

• If the lateral side of the trapezoid and one of the bases are known, then the desired value can be determined by referring to the expression:

l=a-√(c*c-h*h),
l=b+√(c*c-h*h),
l – midline,
a, b – bases (b< a),
h – height of the figure.

• With known values ​​of the height, diagonals (and they are equal to each other) and the angles formed as a result of their intersection, the midline can be found as follows:

l=(d*d)/2h*sinγ or l=(d*d)/2h*sinφ,

l – midline,
d – diagonals,
φ, γ – angles between them,
h – height of the figure.

• The area and height of the figure are known, then:

l=S/h,
S – area,
h – height.

• If the perpendicular height is unknown, it can be determined using the definition of the trigonometric function.

h=c*sinα, therefore
l=S/c*sinα,
l – midline,
S – area,
c – side,
α is the angle at the base.

The midline of a trapezoid, and especially its properties, are very often used in geometry to solve problems and prove certain theorems.

is a quadrilateral with only 2 sides parallel to each other. The parallel sides are called bases (in Figure 1 - AD And B.C.), the other two are lateral (in the figure AB And CD).

Midline of trapezoid is a segment connecting the midpoints of its sides (in Figure 1 - KL).

Properties of the midline of a trapezoid

Proof of the trapezoid midline theorem

Prove that the midline of a trapezoid is equal to half the sum of its bases and is parallel to these bases.

Given a trapezoid ABCD with midline KL. To prove the properties under consideration, it is necessary to draw a straight line through the points B And L. In Figure 2 this is a straight line BQ. And also continue the foundation AD to the intersection with the line BQ.

Consider the resulting triangles L.B.C. And LQD:

1. By definition of the midline KL dot L is the midpoint of the segment CD. It follows that the segments C.L. And LD are equal.
2. ∠BLC = ∠QLD, since these angles are vertical.
3. ∠BCL = ∠LDQ, since these angles lie crosswise on parallel lines AD And B.C. and secant CD.

From these 3 equalities it follows that the previously considered triangles L.B.C. And LQD equal on 1 side and two adjacent angles (see Fig. 3). Hence, ∠LBC = ∠ LQD, BC=DQ and the most important thing - BL=LQ => KL, which is the midline of the trapezoid ABCD, is also the midline of the triangle ABQ. According to the property of the midline of a triangle ABQ we get.

The concept of the midline of the trapezoid

First, let's remember what kind of figure is called a trapezoid.

Definition 1

A trapezoid is a quadrilateral in which two sides are parallel and the other two are not parallel.

In this case, the parallel sides are called the bases of the trapezoid, and the non-parallel sides are called the lateral sides of the trapezoid.

Definition 2

The midline of a trapezoid is a segment connecting the midpoints of the lateral sides of the trapezoid.

Trapezoid midline theorem

Now we introduce the theorem about the midline of a trapezoid and prove it using the vector method.

Theorem 1

The midline of the trapezoid is parallel to the bases and equal to their half-sum.

Proof.

Let us be given a trapezoid $ABCD$ with bases $AD\ and\ BC$. And let $MN$ be the middle line of this trapezoid (Fig. 1).

Figure 1. Midline of trapezoid

Let us prove that $MN||AD\ and\ MN=\frac(AD+BC)(2)$.

Consider the vector $\overrightarrow(MN)$. We next use the polygon rule to add vectors. On the one hand, we get that

On the other side

Let's add the last two equalities and get

Since $M$ and $N$ are the midpoints of the lateral sides of the trapezoid, we will have

We get:

Hence

From the same equality (since $\overrightarrow(BC)$ and $\overrightarrow(AD)$ are codirectional and, therefore, collinear) we obtain that $MN||AD$.

The theorem has been proven.

Examples of problems on the concept of the midline of a trapezoid

Example 1

The lateral sides of the trapezoid are $15\ cm$ and $17\ cm$ respectively. The perimeter of the trapezoid is $52\cm$. Find the length of the midline of the trapezoid.

Solution.

Let us denote the midline of the trapezoid by $n$.

The sum of the sides is equal to

Therefore, since the perimeter is $52\ cm$, the sum of the bases is equal to

So, by Theorem 1, we get

Answer:$10\cm$.

Example 2

The ends of the circle's diameter are $9$ cm and $5$ cm away from its tangent, respectively. Find the diameter of this circle.

Solution.

Let us be given a circle with center at point $O$ and diameter $AB$. Let's draw a tangent $l$ and construct the distances $AD=9\ cm$ and $BC=5\ cm$. Let's draw the radius $OH$ (Fig. 2).

Figure 2.

Since $AD$ and $BC$ are the distances to the tangent, then $AD\bot l$ and $BC\bot l$ and since $OH$ is the radius, then $OH\bot l$, therefore, $OH |\left|AD\right||BC$. From all this we get that $ABCD$ is a trapezoid, and $OH$ is its midline. By Theorem 1, we get

A 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 types of trapezoid and its properties. In addition, we will figure out how to calculate individual elements of this For example, the diagonal of an isosceles trapezoid, the center line, area, etc. The material is presented in the style of elementary popular geometry, i.e. in an easily accessible form.

General information

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

So let's get back to trapezoids. As we have already said, this figure has two parallel sides. They are called bases. The other two (non-parallel) are the lateral sides. In the materials of exams and various tests, you can often find problems related to trapezoids, the solution of which often requires the student to have knowledge not provided for in the program. The school geometry course introduces students to the properties of angles and diagonals, as well as the midline of an isosceles trapezoid. But, in addition to this, the mentioned geometric figure has other features. But more about them a little later...

Types of trapezoid

There are many types of this figure. However, most often it is customary to consider two of them - isosceles and rectangular.

1. A rectangular trapezoid is a figure in which one of the sides is perpendicular to the bases. Her two angles are 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 at the bases are also equal in pairs.

The main principles of the methodology for studying the properties of a trapezoid

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

The second principle is the so-called spiral organization of the study of the “remarkable” properties of the trapezoid. This implies a return in the learning process to individual features of a given geometric figure. This makes it easier for students to remember them. For example, the property of four points. It can be proven both when studying similarity and subsequently using vectors. And the equivalence of triangles adjacent to the lateral sides of a figure can be proven by applying not only the properties of triangles with equal heights drawn to the sides that lie on the same straight line, but also using the formula S = 1/2(ab*sinα). In addition, you can work on an inscribed trapezoid or a right triangle on an inscribed trapezoid, etc.

The use of “extracurricular” features of a geometric figure in the content of a school course is a task-based technology for teaching them. Constantly referring to the properties being studied while going through other topics allows students to gain a deeper knowledge of the trapezoid and ensures the success of solving assigned problems. So, let's start studying this wonderful figure.

Elements and properties of an isosceles trapezoid

As we have already noted, this geometric figure has equal sides. It is also known as the correct trapezoid. 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 angles at 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 an isosceles one can be described as a circle. This is due to the fact that the sum of the opposite angles of this figure is equal to 180 degrees, and only under this condition can one describe a circle around a quadrilateral. The next property of the geometric figure under consideration is that the distance from the vertex of the base to the projection of the opposite vertex onto the straight line that contains this base will be equal to the midline.

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

Solution

Typically, a quadrilateral is usually denoted by the letters A, B, C, D, where BS and AD are the bases. In an isosceles trapezoid, 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 angle B. The result is a right triangle ABN, where AB is the hypotenuse, and BN and AN are the legs. We calculate the size of the leg AN: we subtract the smaller one from the larger base, and divide the result by 2. We write it in the form of a formula: (Z-Y)/2 = F. Now, to calculate the acute angle of the triangle, we use the cos function. We get the following entry: cos(β) = X/F. Now we calculate the angle: β=arcos (X/F). Further, knowing one angle, we can determine the second, for this we perform an elementary arithmetic operation: 180 - β. All angles are defined.

There is a second solution to this problem. First, we lower it from the corner to height H. We calculate the value of the leg BN. 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 = arctan (BN/F). An acute angle has been found. Next, we define it similarly to the first method.

Property of diagonals of an isosceles trapezoid

First, let's write down four rules. If the diagonals in an isosceles trapezoid are perpendicular, then:

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

Its height and midline are equal;

The center of the circle is the point at which ;

If the lateral side is divided by the point of tangency into segments H and M, then it is equal to the square root of the product of these segments;

The quadrilateral that is formed by the tangent points, 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 ​​a figure is equal to the product of the bases and the product of half the sum of the bases and its height.

Similar trapezoids

This topic is very convenient for studying the properties of this For example, the diagonals divide a trapezoid into four triangles, and those adjacent to the bases are similar, and those adjacent to the sides are equal in size. This statement can be called a property of the triangles into which the trapezoid is divided by its diagonals. The first part of this statement is proven through the sign of similarity at two angles. To prove the second part, it is better to use the method given below.

Proof of the theorem

We accept that the figure ABSD (AD and BS are the bases of the trapezoid) is divided by diagonals VD and AC. The point of their intersection is O. We get four triangles: AOS - at the lower base, BOS - at the upper base, ABO and SOD at the sides. Triangles SOD and BOS have a common height if the segments BO and OD are their bases. We find that the difference between their areas (P) is equal to the difference between these segments: PBOS/PSOD = BO/OD = K. Therefore, PSOD = PBOS/K. Similarly, triangles BOS and AOB have a common height. We take the segments CO and OA as their bases. We get PBOS/PAOB = CO/OA = K and PAOB = PBOS/K. It follows from this that PSOD = PAOB.

To consolidate the material, students are recommended to find the connection between the areas of the resulting triangles into which the trapezoid is divided by its diagonals by solving the following problem. It is known that triangles BOS and AOD have equal areas; it is necessary to find the area of ​​the trapezoid. Since PSOD = PAOB, it means PABSD = PBOS+PAOD+2*PSOD. From the similarity of triangles BOS and AOD it follows that BO/OD = √(PBOS/PAOD). Therefore, PBOS/PSOD = BO/OD = √(PBOS/PAOD). We get PSOD = √(PBOS*PAOD). Then PABSD = PBOS+PAOD+2*√(PBOS*PAOD) = (√PBOS+√PAOD)2.

Properties of similarity

Continuing to develop this topic, we can prove other interesting features of trapezoids. Thus, using similarity, one can prove the property of a segment that passes through the point formed by the intersection of the diagonals of this geometric figure, parallel to the bases. To do this, let's solve the following problem: we need to find the length of the segment RK that passes through point O. From the similarity of triangles AOD and BOS it follows that AO/OS = AD/BS. From the similarity of triangles AOP and ASB it follows that AO/AC=RO/BS=AD/(BS+AD). From here we get that RO=BS*BP/(BS+BP). Similarly, from the similarity of triangles DOC and DBS, it follows that OK = BS*AD/(BS+AD). From here we get that RO=OK and RK=2*BS*AD/(BS+AD). A segment passing through the point of intersection of the diagonals, parallel to the bases and connecting two lateral sides, is divided in half by the point of intersection. Its length is the harmonic mean of the figure's bases.

Consider the following property of a trapezoid, which is called the property of four points. The intersection points of the diagonals (O), the intersection of the continuation of the sides (E), as well as the midpoints of the bases (T and F) always lie on the same line. This can be easily proven by the similarity method. The resulting triangles BES and AED are similar, and in each of them the medians ET and EJ divide the vertex angle E into equal parts. Therefore, points E, T and F lie on the same straight line. In the same way, points T, O, and Zh are located on the same straight line. All this follows from the similarity of triangles BOS and AOD. From here we conclude that all four points - E, T, O and F - will lie on the same straight line.

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

Consider the following similarity property. It is based on a segment that divides the trapezoid into two equal figures. We assume that the trapezoid ABSD is divided by the segment EH into two similar ones. From vertex B a height is omitted, which is divided by segment EN into two parts - B1 and B2. We get: PABSD/2 = (BS+EN)*B1/2 = (AD+EN)*B2/2 and PABSD = (BS+AD)*(B1+B2)/2. Next, we compose a system whose first equation is (BS+EN)*B1 = (AD+EN)*B2 and the second (BS+EN)*B1 = (BS+AD)*(B1+B2)/2. It follows that B2/B1 = (BS+EN)/(AD+EN) and BS+EN = ((BS+AD)/2)*(1+B2/B1). We find that the length of the segment dividing the trapezoid into two equal ones is equal to the root mean square of the lengths of the bases: √((BS2+AD2)/2).

Similarity findings

Thus, we have proven that:

1. The segment connecting the midpoints of the lateral sides of a trapezoid is parallel to AD and BS and is equal to the arithmetic mean of BS and AD (the length of the base of the trapezoid).

2. The line passing through the point O of the intersection of the diagonals parallel to AD and BS will be equal to the harmonic mean 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 geometric mean of the bases BS and AD.

4. An element dividing a figure into two equal ones has the length of the root mean square of the numbers AD and BS.

To consolidate the material and understand the connection between the considered segments, the student needs to construct them for a specific trapezoid. He 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 located? This answer will lead the student to the discovery of the desired relationship between average values.

A segment connecting the midpoints of the diagonals of a trapezoid

Consider the following property of this figure. We assume that the segment MH is parallel to the bases and bisects the diagonals. Let's call the intersection points Ш and Ш. This segment will be equal to half the difference of the bases. Let's look at this in more detail. MS is the middle line of the ABS triangle, it is equal to BS/2. MSH is the middle line of triangle ABD, it is equal to AD/2. Then we get that ShShch = MSh-MSh, therefore, ShShch = AD/2-BS/2 = (AD+VS)/2.

Center of gravity

Let's look at how this element is determined for a given geometric figure. To do this, it is necessary to extend the bases in opposite directions. What does it mean? You need to add the lower base to the upper base - in any direction, for example, to the right. And we extend the lower one by the length of the upper one to the left. Next, we connect them diagonally. The point of intersection of this segment with the midline of the figure is the center of gravity of the trapezoid.

Inscribed and circumscribed trapezoids

Let's list the features of such figures:

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

2. A trapezoid can be described around a circle, provided that the sum of the lengths of their bases is equal to the sum of the lengths of the sides.

Corollaries of the incircle:

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

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

The first corollary is obvious, but to prove the second it is necessary to establish that the angle SOD is right, which, in fact, is also not difficult. But knowledge of this property will allow you to use a right triangle when solving problems.

Now let us specify these consequences for an isosceles trapezoid inscribed in a circle. We find that the height is the geometric mean of the bases of the figure: H=2R=√(BS*AD). While practicing the basic technique for solving problems for trapezoids (the principle of drawing two heights), the student must solve the following task. We assume that BT is the height of the isosceles figure ABSD. It is necessary to find the segments AT and TD. Using 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 circumscribed trapezoid. We lower the height from vertex B to the base AD. Since the circle is inscribed in a trapezoid, then BS+AD = 2AB or AB = (BS+AD)/2. From triangle ABN we find sinα = BN/AB = 2*BN/(BS+AD). PABSD = (BS+BP)*BN/2, BN=2R. We get PABSD = (BS+BP)*R, it follows that R = PABSD/(BS+BP).

All formulas for the midline of a trapezoid

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

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

2. Through height, base and corners:

M = A-H*(ctgα+ctgβ)/2;

M = B+N*(ctgα+ctgβ)/2.

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

M = D1*D2*sinα/2N = D1*D2*sinβ/2N.

4. Through area and height: M = P/N.

In this article we will try to reflect the properties of a trapezoid as fully as possible. In particular, we will talk about the general characteristics and properties of a trapezoid, as well as the properties of an inscribed trapezoid and a circle inscribed in a trapezoid. We will also touch on the properties of an isosceles and rectangular trapezoid.

An example of solving a problem using the properties discussed will help you sort it into places in your head and better remember the material.

Trapeze and all-all-all

To begin with, let us briefly recall what a trapezoid is and what other concepts are associated with it.

So, a trapezoid is a quadrilateral figure, two of whose sides are parallel to each other (these are the bases). And the two are not parallel - these are the sides.

In a trapezoid, the height can be lowered - perpendicular to the bases. The center line and diagonals are drawn. It is also possible to draw a bisector from any angle of the trapezoid.

We will now talk about the various properties associated with all these elements and their combinations.

Properties of trapezoid diagonals

To make it clearer, while you are reading, sketch out the trapezoid ACME on a piece of paper and draw diagonals in it.

1. If you find the midpoints of each of the diagonals (let's call these points X and T) and connect them, you get a segment. One of the properties of the diagonals of a trapezoid is that the segment HT lies on the midline. And its length can be obtained by dividing the difference of the bases by two: ХТ = (a – b)/2.
2. Before us is the same trapezoid ACME. The diagonals intersect at point O. Let's look at the triangles AOE and MOK, formed by segments of the diagonals together with the bases of the trapezoid. These triangles are similar. The similarity coefficient k of triangles is expressed through the ratio of the bases of the trapezoid: k = AE/KM.
   The ratio of the areas of triangles AOE and MOK is described by the coefficient k 2 .
3. The same trapezoid, the same diagonals intersecting at point O. Only this time we will consider the triangles that the segments of the diagonals formed together with the sides of the trapezoid. The areas of triangles AKO and EMO are equal in size - their areas are the same.
4. Another property of a trapezoid involves the construction of diagonals. So, if you continue the sides of AK and ME in the direction of the smaller base, then sooner or later they will intersect at a certain point. Next, draw a straight line through the middle of the bases of the trapezoid. It intersects the bases at points X and T.
   If we now extend the line XT, then it will connect together the point of intersection of the diagonals of the trapezoid O, the point at which the extensions of the sides and the middle of the bases X and T intersect.
5. Through the point of intersection of the diagonals we will draw a segment that will connect the bases of the trapezoid (T lies on the smaller base KM, X on the larger AE). The intersection point of the diagonals divides this segment in the following ratio: TO/OX = KM/AE.
6. Now, through the point of intersection of the diagonals, we will draw a segment parallel to the bases of the trapezoid (a and b). The intersection point will divide it into two equal parts. You can find the length of the segment using the formula 2ab/(a + b).

Properties of the midline of a trapezoid

Draw the middle line in the trapezoid parallel to its bases.

1. The length of the midline of a trapezoid can be calculated by adding the lengths of the bases and dividing them in half: m = (a + b)/2.
2. If you draw any segment (height, for example) through both bases of the trapezoid, the middle line will divide it into two equal parts.

Trapezoid Bisector Property

Select any angle of the trapezoid and draw a bisector. Let's take, for example, the angle KAE of our trapezoid ACME. Having completed the construction yourself, you can easily verify that the bisector cuts off from the base (or its continuation on a straight line outside the figure itself) a segment of the same length as the side.

Properties of trapezoid angles

1. Whichever of the two pairs of angles adjacent to the side you choose, the sum of the angles in the pair is always 180 0: α + β = 180 0 and γ + δ = 180 0.
2. Let's connect the midpoints of the bases of the trapezoid with a segment TX. Now let's look at the angles at the bases of the trapezoid. If the sum of the angles for any of them is 90 0, the length of the segment TX can be easily calculated based on the difference in the lengths of the bases, divided in half: TX = (AE – KM)/2.
3. If parallel lines are drawn through the sides of a trapezoid angle, they will divide the sides of the angle into proportional segments.

Properties of an isosceles (equilateral) trapezoid

1. In an isosceles trapezoid, the angles at any base are equal.
2. Now build a trapezoid again to make it easier to imagine what we're talking about. Look carefully at the base AE - the vertex of the opposite base M is projected to a certain point on the line that contains AE. The distance from vertex A to the projection point of vertex M and the middle line of the isosceles trapezoid are equal.
3. A few words about the property of the diagonals of an isosceles trapezoid - their lengths are equal. And also the angles of inclination of these diagonals to the base of the trapezoid are the same.
4. Only around an isosceles trapezoid can a circle be described, since the sum of the opposite angles of a quadrilateral is 180 0 - a prerequisite for this.
5. The property of an isosceles trapezoid follows from the previous paragraph - if a circle can be described near the trapezoid, it is isosceles.
6. From the features of an isosceles trapezoid follows the property of the height of a trapezoid: if its diagonals intersect at right angles, then the length of the height is equal to half the sum of the bases: h = (a + b)/2.
7. Again, draw the segment TX through the midpoints of the bases of the trapezoid - in an isosceles trapezoid it is perpendicular to the bases. And at the same time TX is the axis of symmetry of an isosceles trapezoid.
8. This time, lower the height from the opposite vertex of the trapezoid onto the larger base (let's call it a). You will get two segments. The length of one can be found if the lengths of the bases are added and divided in half: (a + b)/2. We get the second one when we subtract the smaller one from the larger base and divide the resulting difference by two: (a – b)/2.

Properties of a trapezoid inscribed in a circle

Since we are already talking about a trapezoid inscribed in a circle, let us dwell on this issue in more detail. In particular, on where the center of the circle is in relation to the trapezoid. Here, too, it is recommended that you take the time to pick up a pencil and draw what will be discussed below. This way you will understand faster and remember better.

1. The location of the center of the circle is determined by the angle of inclination of the trapezoid's diagonal to its side. For example, a diagonal may extend from the top of a trapezoid at right angles to the side. In this case, the larger base intersects the center of the circumcircle exactly in the middle (R = ½AE).
2. The diagonal and the side can also meet at an acute angle - then the center of the circle is inside the trapezoid.
3. The center of the circumscribed circle may be outside the trapezoid, beyond its larger base, if there is an obtuse angle between the diagonal of the trapezoid and the side.
4. The angle formed by the diagonal and the large base of the trapezoid ACME (inscribed angle) is half the central angle that corresponds to it: MAE = ½MOE.
5. Briefly about two ways to find the radius of a circumscribed circle. Method one: look carefully at your drawing - what do you see? You can easily notice that the diagonal splits the trapezoid into two triangles. The radius can be found by the ratio of the side of the triangle to the sine of the opposite angle, multiplied by two. For example, R = AE/2*sinAME. In a similar way, the formula can be written for any of the sides of both triangles.
6. Method two: find the radius of the circumscribed circle through the area of ​​the triangle formed by the diagonal, side and base of the trapezoid: R = AM*ME*AE/4*S AME.

Properties of a trapezoid circumscribed about a circle

You can fit a circle into a trapezoid if one condition is met. Read more about it below. And together this combination of figures has a number of interesting properties.

1. If a circle is inscribed in a trapezoid, the length of its midline can be easily found by adding the lengths of the sides and dividing the resulting sum in half: m = (c + d)/2.
2. For the trapezoid ACME, described about a circle, the sum of the lengths of the bases is equal to the sum of the lengths of the sides: AK + ME = KM + AE.
3. From this property of the bases of a trapezoid, the converse statement follows: a circle can be inscribed in a trapezoid whose sum of bases is equal to the sum of its sides.
4. The tangent point of a circle with radius r inscribed in a trapezoid divides the side into two segments, let's call them a and b. The radius of a circle can be calculated using the formula: r = √ab.
5. And one more property. To avoid confusion, draw this example yourself too. We have the good old trapezoid ACME, described around a circle. It contains diagonals that intersect at point O. The triangles AOK and EOM formed by the segments of the diagonals and the lateral sides are rectangular.
   The heights of these triangles, lowered to the hypotenuses (i.e., the lateral sides of the trapezoid), coincide with the radii of the inscribed circle. And the height of the trapezoid coincides with the diameter of the inscribed circle.

Properties of a rectangular trapezoid

A trapezoid is called rectangular if one of its angles is right. And its properties stem from this circumstance.

1. A rectangular trapezoid has one of its sides perpendicular to its base.
2. The height and side of a trapezoid adjacent to a right angle are equal. This allows you to calculate the area of ​​a rectangular trapezoid (general formula S = (a + b) * h/2) not only through the height, but also through the side adjacent to the right angle.
3. For a rectangular trapezoid, the general properties of the diagonals of a trapezoid already described above are relevant.

Evidence of some properties of the trapezoid

Equality of angles at the base of an isosceles trapezoid:

• You probably already guessed that here we will need the AKME trapezoid again - draw an isosceles trapezoid. Draw a straight line MT from vertex M, parallel to the side of AK (MT || AK).

The resulting quadrilateral AKMT is a parallelogram (AK || MT, KM || AT). Since ME = KA = MT, ∆ MTE is isosceles and MET = MTE.

AK || MT, therefore MTE = KAE, MET = MTE = KAE.

Where does AKM = 180 0 - MET = 180 0 - KAE = KME.

Q.E.D.

Now, based on the property of an isosceles trapezoid (equality of diagonals), we prove that trapezoid ACME is isosceles:

• First, let’s draw a straight line MX – MX || KE. We obtain a parallelogram KMHE (base – MX || KE and KM || EX).

∆AMX is isosceles, since AM = KE = MX, and MAX = MEA.

MH || KE, KEA = MXE, therefore MAE = MXE.

It turned out that the triangles AKE and EMA are equal to each other, since AM = KE and AE are the common side of the two triangles. And also MAE = MXE. We can conclude that AK = ME, and from this it follows that the trapezoid AKME is isosceles.

Review task

The bases of the trapezoid ACME are 9 cm and 21 cm, the side side KA, equal to 8 cm, forms an angle of 150 0 with the smaller base. You need to find the area of ​​the trapezoid.

Solution: From vertex K we lower the height to the larger base of the trapezoid. And let's start looking at the angles of the trapezoid.

Angles AEM and KAN are one-sided. This means that in total they give 180 0. Therefore, KAN = 30 0 (based on the property of trapezoidal angles).

Let us now consider the rectangular ∆ANC (I believe this point is obvious to readers without additional evidence). From it we will find the height of the trapezoid KH - in a triangle it is a leg that lies opposite the angle of 30 0. Therefore, KH = ½AB = 4 cm.

We find the area of ​​the trapezoid using the formula: S ACME = (KM + AE) * KN/2 = (9 + 21) * 4/2 = 60 cm 2.

Afterword

If you carefully and thoughtfully studied this article, were not too lazy to draw trapezoids for all the given properties with a pencil in your hands and analyze them in practice, you should have mastered the material well.

Of course, there is a lot of information here, varied and sometimes even confusing: it is not so difficult to confuse the properties of the described trapezoid with the properties of the inscribed one. But you yourself have seen that the difference is huge.

Now you have a detailed outline of all the general properties of a trapezoid. As well as specific properties and characteristics of isosceles and rectangular trapezoids. It is very convenient to use to prepare for tests and exams. Try it yourself and share the link with your friends!

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