home Social science The middle line of the trapezium. Trapeze. Properties, features, area. The middle line of the trapezoid - materials for preparing for the exam in Mathematics. Properties of a trapezoid inscribed in a circle

The middle line of the trapezium. Trapeze. Properties, features, area. The middle line of the trapezoid - materials for preparing for the exam in Mathematics. Properties of a trapezoid inscribed in a circle

The concept of the midline of the trapezoid

First, let's remember what 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, parallel sides are called the bases of the trapezoid, and not parallel - the sides of the trapezoid.

Definition 2

The midline of a trapezoid is a line segment that connects the midpoints of the sides of the trapezoid.

Trapezium midline theorem

We now introduce the theorem on the midline of a trapezoid and prove it by the vector method.

Theorem 1

The median line of the trapezoid is parallel to the bases and equal to half their sum.

Proof.

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

Figure 1. The middle line of the trapezoid

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

Consider the vector $\overrightarrow(MN)$. Next, we use the polygon rule for vector addition. On the one hand, we get that

On the other side

Adding the last two equalities, we get

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

We get:

Hence

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

The theorem has been proven.

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

Example 1

The 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.

Decision.

Denote the midline of the trapezoid by $n$.

The sum of the sides is

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

Hence, by Theorem 1, we obtain

Answer:$10\cm$.

Example 2

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

Decision.

Let us be given a circle with center $O$ and diameter $AB$. Draw the 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 $OH\bot l$, hence $OH |\left|AD\right||BC$. From all this we get that $ABCD$ is a trapezoid, and $OH$ is its midline. By Theorem 1, we get

In this article, another selection of tasks with a trapezoid has been made for you. Conditions are somehow connected with its middle line. Task types are taken from the open bank of typical tasks. If you wish, you can refresh your theoretical knowledge. The blog has already covered tasks whose conditions are associated with, as well as. Briefly about the middle line:

The middle line of the trapezoid connects the midpoints of the sides. It is parallel to the bases and equal to their half-sum.

Before solving problems, let's consider a theoretical example.

Given a trapezoid ABCD. Diagonal AC intersecting with the midline forms a point K, diagonal BD a point L. Prove that the segment KL is equal to half the difference of the bases.

Let's first note the fact that the midline of a trapezoid bisects any segment whose ends lie on its bases. This conclusion suggests itself. Imagine a segment connecting two points of the bases, it will split this trapezoid into two others. It turns out that a segment parallel to the bases of the trapezoid and passing through the middle of the side on the other side will pass through its middle.

It is also based on the Thales theorem:

If on one of the two straight lines several equal segments are sequentially laid aside and parallel lines are drawn through their ends, intersecting the second straight line, then they will cut off equal segments on the second straight line.

That is, in this case, K is the middle of AC and L is the middle of BD. Hence EK is the midline of triangle ABC, LF is the midline of triangle DCB. According to the property of the midline of a triangle:

We can now express the segment KL in terms of bases:

Proven!

This example is not just given. In tasks for independent solution, there is just such a task. Only it does not say that the segment connecting the midpoints of the diagonals lies on the midline. Consider the tasks:

27819. Find the midline of a trapezoid if its bases are 30 and 16.

We calculate by the formula:

27820. The midline of the trapezoid is 28 and the smaller base is 18. Find the larger base of the trapezoid.

Let's express the larger base:

Thus:

27836. A perpendicular dropped from the apex of an obtuse angle to the greater base of an isosceles trapezoid divides it into parts having lengths 10 and 4. Find the midline of this trapezoid.

In order to find the middle line, you need to know the bases. The base AB is easy to find: 10+4=14. Find DC.

Let's construct the second perpendicular DF:

Segments AF, FE and EB will be equal to 4, 6 and 4 respectively. Why?

In an isosceles trapezoid, the perpendiculars dropped to the larger base divide it into three segments. Two of them, which are the legs of cut off right-angled triangles, are equal to each other. The third segment is equal to the smaller base, since when constructing the indicated heights, a rectangle is formed, and in the rectangle, the opposite sides are equal. In this task:

Thus DC=6. We calculate:

27839. The bases of the trapezoid are in ratio 2:3, and the midline is 5. Find the smaller base.

Let's introduce the coefficient of proportionality x. Then AB=3x, DC=2x. We can write:

Therefore, the smaller base is 2∙2=4.

27840. The perimeter of an isosceles trapezoid is 80, its midline is equal to the lateral side. Find the side of the trapezoid.

Based on the condition, we can write:

If we denote the middle line through x, we get:

The second equation can already be written as:

27841. The midline of the trapezoid is 7, and one of its bases is 4 more than the other. Find the larger base of the trapezoid.

Let's denote the smaller base (DC) as x, then the larger one (AB) will be equal to x + 4. We can record

We got that the smaller base is early than five, which means that the larger one is equal to 9.

27842. The midline of the trapezoid is 12. One of the diagonals divides it into two segments, the difference of which is 2. Find the larger base of the trapezoid.

We can easily find the larger base of the trapezoid if we calculate the segment EO. It is the middle line in triangle ADB, and AB=2∙EO.

What do we have? It is said that the middle line is equal to 12 and the difference between the segments EO and OF is equal to 2. We can write down two equations and solve the system:

It is clear that in this case it is possible to select a pair of numbers without calculations, these are 5 and 7. But, nevertheless, we will solve the system:

So EO=12–5=7. Thus, the larger base is equal to AB=2∙EO=14.

27844. In an isosceles trapezoid, the diagonals are perpendicular. The height of the trapezoid is 12. Find its midline.

Immediately, we note that the height drawn through the intersection point of the diagonals in an isosceles trapezoid lies on the axis of symmetry and divides the trapezoid into two equal rectangular trapezoids, that is, the bases of this height are divided in half.

It would seem that in order to calculate the average line, we must find the grounds. Here a small dead end arises ... How, knowing the height, in this case, calculate the bases? And no how! Many such trapezoids with a fixed height and diagonals intersecting at an angle of 90 degrees can be built. How to be?

Look at the formula for the midline of a trapezoid. After all, we do not need to know the bases themselves, it is enough to know their sum (or half-sum). This we can do.

Since the diagonals intersect at right angles, isosceles right triangles are formed with height EF:

It follows from the above that FO=DF=FC, and OE=AE=EB. Now let's write down what the height expressed through the segments DF and AE is equal to:

So the middle line is 12.

* In general, this is a problem, as you understand, for an oral account. But I'm sure the detailed explanation provided is necessary. And so ... If you look at the figure (provided that the angle between the diagonals is observed during construction), the equality FO=DF=FC, and OE=AE=EB immediately catches your eye.

As part of the prototypes, there are also types of tasks with trapezoids. It was built on a sheet in a cell and it is required to find the middle line, the side of the cell is usually 1, but there may be another value.

27848. Find the midline of the trapezoid ABCD if the sides of the square cells are 1.

It's simple, we calculate the bases by cells and use the formula: (2 + 4) / 2 = 3

If the bases are built at an angle to the cell grid, then there are two ways. For example!

In this article, we will try to reflect the properties of the trapezoid as fully as possible. In particular, we will talk about the general signs and properties of a trapezoid, as well as about the properties of an inscribed trapezoid and about 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 considered properties will help you sort things out in your head and better remember the material.

Trapeze and all-all-all

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

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

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

About the various properties associated with all these elements and their combinations, we will now talk.

Properties of the diagonals of a trapezoid

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

Properties of the midline of a trapezoid

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

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.
If you draw any segment (height, for example) through both bases of the trapezoid, the middle line will divide it into two equal parts.

Property of the bisector of a trapezoid

Pick any angle of the trapezoid and draw a bisector. Take, for example, the angle KAE of our trapezoid ACME. Having completed the construction on your own, you can easily see 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.

Trapezoid angle properties

Whichever of the two pairs of angles adjacent to the side you choose, the sum of the angles in a pair is always 180 0: α + β = 180 0 and γ + δ = 180 0 .
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 TX segment is easy to calculate based on the difference in the lengths of the bases, divided in half: TX \u003d (AE - KM) / 2.
If parallel lines are drawn through the sides of the angle of a trapezoid, they will divide the sides of the angle into proportional segments.

Properties of an isosceles (isosceles) trapezoid

In an isosceles trapezoid, the angles at any of the bases are equal.
Now build a trapezoid again to make it easier to imagine what it is about. Look carefully at the base of AE - the vertex of the opposite base of 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 midline of an isosceles trapezoid are equal.
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.
Only near an isosceles trapezoid can a circle be described, since the sum of the opposite angles of a quadrilateral 180 0 is a prerequisite for this.
The property of an isosceles trapezoid follows from the previous paragraph - if a circle can be described near a trapezoid, it is isosceles.
From the features of an isosceles trapezoid, the property of the height of a trapezoid follows: if its diagonals intersect at a right angle, then the length of the height is equal to half the sum of the bases: h = (a + b)/2.
Draw the line TX again 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.
This time lower to the larger base (let's call it a) the height from the opposite vertex of the trapezoid. You will get two cuts. 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's dwell on this issue in more detail. In particular, where is the center of the circle in relation to the trapezoid. Here, too, it is recommended not to be too lazy to pick up a pencil and draw what will be discussed below. So you will understand faster, and remember better.

The location of the center of the circle is determined by the angle of inclination of the diagonal of the trapezoid to its side. For example, a diagonal may emerge from the top of a trapezoid at right angles to the side. In this case, the larger base intersects the center of the circumscribed circle exactly in the middle (R = ½AE).
The diagonal and the side can also meet at an acute angle - then the center of the circle is inside the trapezoid.
The center of the circumscribed circle may be outside the trapezoid, beyond its large base, if there is an obtuse angle between the diagonal of the trapezoid and the lateral side.
The angle formed by the diagonal and the large base of the trapezoid ACME (inscribed angle) is half of the central angle that corresponds to it: MAE = ½MY.
Briefly about two ways to find the radius of the circumscribed circle. Method one: look carefully at your drawing - what do you see? You will easily notice that the diagonal splits the trapezoid into two triangles. The radius can be found through the ratio of the side of the triangle to the sine of the opposite angle, multiplied by two. For example, R \u003d AE / 2 * sinAME. Similarly, the formula can be written for any of the sides of both triangles.
Method two: we find the radius of the circumscribed circle through the area of the triangle formed by the diagonal, side and base of the trapezoid: R \u003d AM * ME * AE / 4 * S AME.

Properties of a trapezoid circumscribed about a circle

You can inscribe a circle in a trapezoid if one condition is met. More about it below. And together this combination of figures has a number of interesting properties.

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.
For a trapezoid ACME, circumscribed 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.
From this property of the bases of a trapezoid, the converse statement follows: a circle can be inscribed in that trapezoid, the sum of the bases of which is equal to the sum of the sides.
The tangent point of a circle with radius r inscribed in a trapezoid divides the lateral side into two segments, let's call them a and b. The radius of a circle can be calculated using the formula: r = √ab.
And one more property. In order not to get confused, draw this example yourself. We have the good old ACME trapezoid, circumscribed around a circle. Diagonals are drawn in it, intersecting at the point O. The triangles AOK and EOM formed by the segments of the diagonals and the sides are rectangular.
The heights of these triangles, lowered to the hypotenuses (i.e., the sides of the trapezoid), coincide with the radii of the inscribed circle. And the height of the trapezoid is the same as the diameter of the inscribed circle.

Properties of a rectangular trapezoid

A trapezoid is called rectangular, one of the corners of which is right. And its properties stem from this circumstance.

A rectangular trapezoid has one of the sides perpendicular to the bases.
The height and side of the trapezoid adjacent to the 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.
For a rectangular trapezoid, the general properties of the trapezoid diagonals already described above are relevant.

Proofs of some properties of a trapezoid

Equality of angles at the base of an isosceles trapezoid:

You probably already guessed that here we again need the ACME trapezoid - draw an isosceles trapezoid. Draw a 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 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 trapezium ACME is isosceles:

To begin with, let's draw a straight line МХ – МХ || KE. We get a parallelogram KMHE (base - MX || KE and KM || EX).

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

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

It turned out that the triangles AKE and EMA are equal to each other, because AM \u003d KE and AE is the common side of the two triangles. And also MAE \u003d MXE. We can conclude that AK = ME, and hence it follows that the trapezoid AKME is isosceles.

Task to repeat

The bases of the trapezoid ACME are 9 cm and 21 cm, the side of the KA, equal to 8 cm, forms an angle of 150 0 with a 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. Which means they add up to 1800. Therefore, KAN = 30 0 (based on the property of the angles of the trapezoid).

Consider now the rectangular ∆ANK (I think this point is obvious to readers without further proof). From it we find the height of the trapezoid KH - in a triangle it is a leg, which lies opposite the angle of 30 0. Therefore, KN \u003d ½AB \u003d 4 cm.

The area of the trapezoid is found by the formula: S AKME \u003d (KM + AE) * KN / 2 \u003d (9 + 21) * 4/2 \u003d 60 cm 2.

Afterword

If you carefully and thoughtfully studied this article, were not too lazy to draw trapezoids for all the above 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 saw that the difference is huge.

Now you have a detailed summary of all the general properties of a trapezoid. As well as specific properties and features 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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middle line figures in planimetry - a segment connecting the midpoints of the two sides of a given figure. The concept is used for the following figures: triangle, quadrilateral, trapezoid.

Middle line of the triangle

Properties

the midline of a triangle is parallel to the base and equal to half of it.
the middle line cuts off a triangle similar and homothetic to the original one with a factor of 1/2; its area is equal to one-fourth the area of the original triangle.
three middle lines divide the original triangle into four equal triangles. The central of these triangles is called the complementary or medial triangle.

signs

If a segment in a triangle passes through the midpoint of one of its sides, intersects the second and is parallel to the third, then this segment is the midline.
The area and, accordingly, the volume of the triangle cut off by the middle line is equal to 1/4 of the area and, accordingly, the volume of the entire given triangle.

Middle line of the quadrilateral

Middle line of the quadrilateral A line segment that joins the midpoints of opposite sides of a quadrilateral.

Properties

The first line connects 2 opposite sides. The second connects 2 other opposite sides. The third one connects the centers of the two diagonals (not in all quadrilaterals the diagonals are bisected by the intersection point).

If in a convex quadrilateral the midline forms equal angles with the diagonals of the quadrilateral, then the diagonals are equal.
The length of the midline of a quadrilateral is less than or equal to half the sum of the other two sides if these sides are parallel, and only in this case.
The midpoints of the sides of an arbitrary quadrilateral are the vertices of a parallelogram. Its area is equal to half the area of the quadrilateral, and its center lies at the point of intersection of the median lines. This parallelogram is called the Varignon parallelogram;
The last point means the following: In a convex quadrilateral, four middle lines of the second kind. Middle lines of the second kind- four segments inside the quadrangle passing through the midpoints of its adjacent sides parallel to the diagonals. Four middle lines of the second kind convex quadrilateral cut it into four triangles and one central quadrilateral. This central quadrilateral is the Varignon parallelogram.
The point of intersection of the midlines of the quadrilateral is their common midpoint and bisects the segment connecting the midpoints of the diagonals. In addition, it is the centroid of the vertices of the quadrilateral.
In an arbitrary quadrilateral, the midline vector is equal to half the sum of the base vectors.

Median line of the trapezoid

Median line of the trapezoid- a segment connecting the midpoints of the sides of this trapezoid. The segment connecting the midpoints of the bases of the trapezoid is called the second midline of the trapezoid.

It is calculated by the formula: E F = A D + B C 2 (\displaystyle EF=(\frac (AD+BC)(2))), where AD and BC- the base of the trapezoid.