The concept of the midline of the trapezoid
To begin with, we recall which figure is called a trapezoid.
Definition 1
A trapezoid is a quadrangle 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 not parallel - the sides of the trapezoid.
Definition 2
The middle line of the trapezoid is a segment connecting the midpoints of the sides of the trapezoid.
Trapezium midline theorem
Now we introduce the theorem on the midline of a trapezoid and prove it by the vector method.
Theorem 1
The middle line of the trapezoid is parallel to the bases and equal to their half-sum.
Evidence.
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 the trapezoid
Let us prove that $ MN || AD \\ and \\ MN \u003d \\ frac (AD + BC) (2) $.
Consider the vector $ \\ overrightarrow (MN) $. Next we use the polygon rule to add vectors. On the one hand, we get that
On the other hand
Add up the last two equalities, we obtain
Since $ M $ and $ N $ are the midpoints of the sides of the trapezoid, we will have
We get:
Hence
From the same equality (since $ \\ overrightarrow (BC) $ and $ \\ overrightarrow (AD) $ are co-directed, and therefore collinear), we obtain that $ MN || AD $.
The theorem is proved.
Examples of tasks on the concept of the midline of the 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
Therefore, by Theorem 1, we obtain
Answer: $ 10 \\ cm $.
Example 2
The ends of the diameter of the circle are removed from its tangent by $ 9 $ cm and $ 5 $ cm, respectively. Find the diameter of this circle.
Decision.
Let us be given a circle centered at $ O $ and with diameter $ AB $. Draw the tangent $ l $ and construct the distances $ AD \u003d 9 \\ cm $ and $ BC \u003d 5 \\ cm $. Draw a radius of $ 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 middle line. By Theorem 1, we obtain
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In this article, the next selection of tasks with a trapezoid is made for you. The conditions are somehow related to its middle line. Types of tasks are taken from an open bank of typical tasks. If you wish, you can refresh your theoretical knowledge. The blog has already reviewed the tasks whose conditions are associated with as well. Briefly about the midline:
The midline 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 look at a theoretical example.
Dana trapezoid ABCD. The diagonal AC intersecting with the middle line forms the point K, the diagonal BD the point L. Prove that the segment KL is equal to half the difference of the bases.
Let us first note the fact that the midline of the trapezoid bisects any segment whose ends lie on its bases. This conclusion suggests itself. Imagine a segment connecting two base points, it will break this trapezoid into two others. It turns out that the 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 lines, several equal segments are laid out successively and parallel lines crossing the second line are drawn through their ends, then they will cut off equal segments on the second line.
That is, in this case, K is the middle of AC and L is the middle of BD. Therefore, EK is the midline of the triangle ABC, LF is the midline of the triangle DCB. By the property of the midline of a triangle:
We can now express the segment KL through the bases:
Proven!
This example is given for a reason. In tasks for independent solution there is precisely such a task. Only it does not say that the segment connecting the middle of the diagonals lies on the midline. Consider the tasks:
27819. Find the midline of the trapezoid if its bases are 30 and 16.
Calculated by the formula:
27820. The midline of the trapezoid is 28, and the smaller base is 18. Find the larger base of the trapezoid.
We express a larger basis:
In this way:
27836. A perpendicular, lowered from the top of an obtuse angle to the larger base of an isosceles trapezoid, divides it into parts having lengths of 10 and 4. Find the midline of this trapezoid.
In order to find the middle line you need to know the grounds. The base AB is easy to find: 10 + 4 \u003d 14. Find DC.
Build the second perpendicular DF:
Segments AF, FE, and EB will be 4, 6, and 4, respectively. Why?
In an isosceles trapezoid, perpendiculars lowered to a larger base break it into three segments. Two of them, which are the legs of the 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:
So DC \u003d 6. We calculate:
27839. The base of the trapezoid is 2: 3, and the center line is 5. Find the smaller base.
We introduce the proportionality coefficient x. Then AB \u003d 3x, DC \u003d 2x. We can write:
Therefore, the smaller base is 2 ∙ 2 \u003d 4.
27840. The perimeter of an isosceles trapezoid is 80, its middle line is equal to the side. Find the side of the trapezoid.
Based on the condition, we can write:
If we denote the middle line through the value of x, we get:
The second equation can already be written as:
27841. The middle line of the trapezoid is 7, and one of its bases is larger than the other by 4. Find the larger base of the trapezoid.
Denote the smaller base (DC) as x, then the larger (AB) will be x + 4. We can write
We got that the smaller base is early five, so the larger is 9.
27842. The middle line 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 a larger trapezoid base if we calculate the EO segment. It is the middle line in the triangle ADB, and AB \u003d 2 ∙ EO.
What do we have? It is said that the middle line is 12 and the difference between the EO and OF segments is 2. We can write two equations and solve the system:
It is clear that in this case, a pair of numbers can be selected without calculations, these are 5 and 7. But, nevertheless, we will solve the system:
So ЕО \u003d 12–5 \u003d 7. Thus, the larger base is AB \u003d 2 ∙ EO \u003d 14.
27844. In an isosceles trapezoid, the diagonals are perpendicular. The height of the trapezoid is 12. Find its center line.
Immediately, we note that the height drawn through the intersection of the diagonals in the 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 to calculate the middle line we must find the basis. Here a small dead end arises ... How, knowing the height, in this case, calculate the bases? But no how! There are many such trapezoids with a fixed height and diagonals intersecting at an angle of 90 degrees. How to be
Look at the trapezoid midline formula. After all, we do not need to know the grounds themselves, it is enough to find out their sum (or half-sum). We can do this.
Since the diagonals intersect at right angles, isosceles right triangles form with the height EF:
From the above it follows that FO \u003d DF \u003d FC, and OE \u003d AE \u003d EB. Now we will 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 task, as you understand, for an oral account. But I am sure that the detailed explanation presented is necessary. And so ... If you look at the figure (provided that the angle between the diagonals is observed during construction), the equality FO \u003d DF \u003d FC and OE \u003d AE \u003d EB are immediately evident.
As part of the prototypes, there are also types of tasks with trapezoids. It is built on a sheet in a cell and you need to find the middle line, the side of the cell is usually 1, but there may be a different value.
27848. Find the midline of the trapezoid Abcdif the sides of the square cells are 1.
It's simple, we calculate the bases on the cells and use the formula: (2 + 4) / 2 \u003d 3
If the bases are built at an angle to the cell grid, then there are two ways. For example!
A trapezoid is a special case of a quadrangle in which one pair of sides is parallel. The term "trapeze" is derived from the Greek word τράπεζα, meaning "table", "table". In this article we will consider the types of trapezoid and its properties. In addition, we will figure out how to calculate the individual elements of this example, the diagonal of an isosceles trapezoid, the middle line, area, etc. The material is presented in the style of elementary popular geometry, that is, in an easily accessible form.
General information
First, let's look at what a quadrangle 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. The same can be said of two non-adjacent sides. The main types of quadrangles are a parallelogram, a rectangle, a rhombus, a square, a trapezoid and a deltoid.
So, back to the trapeze. As we have already said, in this figure the two sides are parallel. They are called bases. The other two (non-parallel) are the sides. In the materials of exams and various examinations it is very often possible to meet problems associated with trapezes, the solution of which often requires the student to have knowledge that is 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 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. She has two angles always equal to ninety degrees.
2. An isosceles trapezoid is a geometric figure in which the sides are equal. This means that the angles at the bases are also pairwise equal.
The main principles of the study of trapezoid properties
The main principle is 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 (better than system ones). At the same time, it is very important that the teacher knows what tasks must be set for students at one time or another in the learning process. Moreover, each trapezoid property 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 the individual features of this geometric figure. This makes it easier for students to remember. For example, the property of four points. It can be proved both in the study of similarity, and subsequently using vectors. And the equidistance of triangles adjacent to the lateral sides of the figure can be proved by applying not only the properties of triangles with equal heights drawn to the sides that lie on one straight line, but also using the formula S \u003d 1/2 (ab * sinα). In addition, you can work out on an inscribed trapezoid or a rectangular triangle on a described trapezoid, etc.
The use of “extra-curricular” features of the geometric figure in the content of the school course is the task technology for teaching them. A constant appeal to the studied properties while passing through other topics allows students to learn more about the trapezoid and ensures the success of solving the tasks. So, let's start studying this wonderful figure.
Elements and properties of an isosceles trapezoid
As we have already noted, in this geometric figure the sides are equal. It is also known as the correct trapezoid. But why is it so remarkable and why did it get such a name? The features of this figure include the fact that it has equal sides and angles at the bases, but also diagonals. In addition, the sum of the angles of an isosceles trapezoid is 360 degrees. But that is not all! Of all the known trapezoids, only around the isosceles one can describe the circle. This is due to the fact that the sum of the opposite angles of this figure is 180 degrees, and only under this condition can we describe the circle around the quadrangle. The next property of the considered geometric figure is that the distance from the top of the base to the projection of the opposite peak on the 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. Consider a solution to this problem, provided that the dimensions of the sides of the figure are known.
Decision
Usually, a quadrangle 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 assume that their size is X, and the sizes of the bases are Y and Z (smaller and larger, respectively). To carry out the calculation, it is necessary to draw a height N from angle B. As a result, we obtain a right-angled triangle ABN, where AB is the hypotenuse, and BN and AN are legs. We calculate the size of the leg AN: subtract the smaller from the larger base, and divide the result by 2. We write in the form of the formula: (Z-Y) / 2 \u003d F. Now we use the cos function to calculate the acute angle of the triangle. We get the following record: cos (β) \u003d X / F. Now we calculate the angle: β \u003d arcos (X / F). Further, knowing one angle, we can determine the second, for this we produce an elementary arithmetic action: 180 - β. All angles are defined.
There is a second solution to this problem. At the beginning we lower N. from the angle B to the height. We calculate the value of the BN leg. We know that the square of the hypotenuse of a right triangle is the sum of the squares of the legs. We get: BN \u003d √ (X2-F2). Next, we use the trigonometric function tg. As a result, we have: β \u003d arctan (BN / F). An acute angle is found. Next, we determine similarly to the first method.
The property of diagonal isosceles trapezoid
First we 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 they intersect;
If the 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 quadrangle formed by the points of tangency, the top of the trapezoid and the center of the inscribed circle is a square whose side is equal to the radius;
The area of \u200b\u200bthe figure is equal to the product of the bases and the product of half the sum of the bases to its height.
Similar trapezoid
This topic is very convenient for studying the properties of this. For example, diagonals divide the trapezoid into four triangles, and adjacent to the bases are similar, and to the sides - are equal. This statement can be called a property of triangles into which the trapezoid is divided by its diagonals. The first part of this statement is proved through the sign of similarity in two angles. To prove the second part, it is better to use the method below.
Proof of the theorem
We accept that the ABSD figure (HELL and BS - the basis of the trapezoid) is divided by the diagonals of the VD and AS. 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. The triangles SOD and BOS have a common height if the segments BO and OD are their bases. We get that the difference in their areas (P) is equal to the difference of these segments: CBE / PSOD \u003d BO / OD \u003d K. Therefore, PSOD \u003d CBE / K. Similarly, the BFB and AOB triangles have a common height. We take for their base segments of CO and OA. We get CBE / PAOB \u003d CO / OA \u003d K and PAOB \u003d CBE / K. It follows that PSOD \u003d PAOB.
To consolidate the material, students are advised to find a connection between the areas of the resulting triangles into which the trapezoid is divided by its diagonals, having solved the following problem. It is known that the area of \u200b\u200bthe triangles BOS and AOD is equal, it is necessary to find the area of \u200b\u200bthe trapezoid. Since PSOD \u003d PAOB, it means that PABSD \u003d CBE + PAOD + 2 * PSOD. From the similarity of the triangles BOS and AOD, it follows that BO / OD \u003d √ (TBC / PAOD). Consequently, CBE / PSOD \u003d BO / OD \u003d √ (CBE / PAOD). We get PSOD \u003d √ (CBOS * PAOD). Then PABSD \u003d TBC + PAOD + 2 * √ (TBC * PAOD) \u003d (√ CBED + √ PAOD) 2.
Similarity Properties
Continuing to develop this topic, one can prove other interesting features of the trapezoid. So, 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 will solve the following problem: it is necessary to find the length of the segment of the RK that passes through the point O. From the similarity of the triangles AOD and BOS it follows that AO / OS \u003d AD / BS. From the similarity of the triangles AOR and ASB, it follows that AO / AC \u003d PO / BS \u003d AD / (BS + AD). From this we obtain that PO \u003d BS * AD / (BS + AD). Similarly, from the similarity of triangles DOK and DBS, it follows that OK \u003d BS * HELL / (BS + HELL). From this we obtain that PO \u003d OK and PK \u003d 2 * BS * HELL / (BS + HELL). A line passing through the diagonal intersection point parallel to the bases and connecting the two sides is divided in half by the intersection point. Its length is the harmonic mean of the base of the figure.
Consider the following quality of the 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 middle of the bases (T and G) always lie on the same line. This is easily proved by the method of similarity. The resulting triangles BES and AED are similar, and in each of them the medians ET and EJ divide the angle at the vertex E into equal parts. Consequently, the points E, T and M lie on the same line. In the same way, the points T, O, and G are located on one straight line. All this follows from the similarity of the triangles BOS and AOD. From this we conclude that all four points - E, T, O and W - will lie on one straight line.
Using similar trapezoids, students can be asked to find the length of the segment (LF), which divides the figure into two similar ones. This segment should be parallel to the bases. Since the resulting trapezoid ALPD and LBSF are similar, then BS / LF \u003d LF / AD. It follows that LF \u003d √ (BS * HELL). We get that the segment dividing the trapezoid into two similar ones has a length equal to the geometric mean of the lengths of the base of the figure.
Consider the following similarity property. It is based on a segment that divides the trapezoid into two equal figures. We accept that the trapezoid ABSD is divided by the EN segment into two similar ones. From the top B, the height is omitted, which is divided by the segment EH into two parts - B1 and B2. We get: PABSD / 2 \u003d (BS + ЕН) * В1 / 2 \u003d (HELL + ЕН) * В2 / 2 and PABSD \u003d (BS + AD) * (В1 + В2) / 2. Next, we compose a system whose first equation is (BS + EH) * B1 \u003d (HELL + EN) * B2 and the second (BS + EH) * B1 \u003d (BS + HELL) * (B1 + B2) / 2. It follows that B2 / B1 \u003d (BS + EH) / (AD + EH) and BS + EH \u003d ((BS + AD) / 2) * (1 + B2 / B1). We get that the length of the segment dividing the trapezoid into two equal ones is equal to the root mean square of the base lengths: √ ((BS2 + AD2) / 2).
Similarity conclusions
Thus, we have proved that:
1. The segment connecting the middle of the sides at the trapezoid is parallel to HELL and BS and is equal to the arithmetic mean of BS and HELL (trapezoid base length).
2. The line passing through the point O of the intersection of the diagonals parallel to HELL and BS will be equal to the harmonic mean of the numbers HELL and BS (2 * BS * HELL / (BS + HELL)).
3. The segment dividing the trapezoid into similar ones has the length of the geometric mean of the BS and HELL bases.
4. The element dividing the figure into two equal, has the length of the mean square numbers AD and BS.
To consolidate the material and understand the connection between the considered segments, the student needs to build 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? This answer will lead the student to discover the desired relationship between the averages.
The line connecting the midpoints of the diagonals of the trapezoid
Consider the following property of this figure. We accept that the segment MN is parallel to the bases and divides the diagonals in half. The intersection points are called W and U. This segment will be equal to the half-difference of the bases. We will analyze this in more detail. MS - the middle line of the triangle of the ABS, it is equal to BS / 2. MSC - the middle line of the triangle ABD, it is equal to AD / 2. Then we get that ШШ \u003d МЩ-МШ, therefore, ШЩ \u003d HELL / 2-BS / 2 \u003d (HELL + BC) / 2.
Center of gravity
Let's look at how this element is determined for a given geometric figure. To do this, extend the bases in opposite directions. What does it mean? It is necessary to add the lower base to the upper base - to either side, for example, to the right. And the bottom is extended by the length of the top left. Next, connect them with a diagonal. The intersection point of this segment with the midline of the figure is the center of gravity of the trapezoid.
Trapezoid inscribed and described
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.
Consequences of the inscribed circle:
1. The height of the described trapezoid is always equal to two radii.
2. The lateral side of the described trapezoid is observed from the center of the circle at a right angle.
The first corollary is obvious, but in order to prove the second, it is required to establish that the angle of the SOD is straight, which, in fact, will not be difficult. But knowledge of this property will allow us to use a right triangle when solving problems.
Now we specify 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 \u003d 2R \u003d √ (BS * HELL). Working out the main method for solving problems for trapezoids (the principle of holding two heights), the student must solve the following task. We accept that BT is the height of the isosceles figure of the ABS. It is necessary to find segments of AT and TD. Applying the formula described above, this will not be difficult.
Now let's figure out how to determine the radius of a circle using the area of \u200b\u200bthe described trapezoid. Lower the height from the top B to the base of HELL. Since the circle is inscribed in the trapezoid, then BS + AD \u003d 2AB or AB \u003d (BS + AD) / 2. From the triangle ABN we find sinα \u003d BN / AB \u003d 2 * BN / (BS + HELL). PABSD \u003d (BS + HELL) * BN / 2, BN \u003d 2R. We get PABSD \u003d (BS + HELL) * R, it follows that R \u003d PABSD / (BS + HELL).
All trapezium midline formulas
Now it's time to move on to the last element of this geometric shape. Let's figure out what the middle line of the trapezoid (M) is equal to:
1. Through the grounds: M \u003d (A + B) / 2.
2. Through height, base and angles:
M \u003d A-H * (ctgα + ctgβ) / 2;
M \u003d B + H * (ctgα + ctgβ) / 2.
3. Through the height, diagonals and the angle between them. For example, D1 and D2 are the diagonals of the trapezoid; α, β are the angles between them:
M \u003d D1 * D2 * sinα / 2H \u003d D1 * D2 * sinβ / 2H.
4. Through the area and height: M \u003d P / N.
Area of \u200b\u200bthe trapezoid. Greetings! In this publication we will consider the specified formula. Why is she like that and how to understand her. If there is understanding, then you do not need to teach it. If you just want to see this formula and what’s urgent, you can immediately scroll down the page))
Now in detail and in order.
A trapezoid is a quadrangle, the two sides of this quadrangle are parallel, the other two are not. Those that are not parallel are the foundations of the trapezoid. The other two are called sides.
If the sides are equal, then the trapezoid is called isosceles. If one of the sides is perpendicular to the bases, then such a trapezoid is called rectangular.
In a classic form, the trapezoid is depicted as follows - the larger base is at the bottom, respectively, the smaller at the top. But no one forbids portraying her and vice versa. Here are the thumbnails:
The next important concept.
The midline of the trapezoid is a segment that connects the midpoints of the sides. The middle line is parallel to the bases of the trapezoid and is equal to their half-sum.
Now let's go deeper. Why so?
Consider the trapezoid with the bases a and b and with the middle line l , and perform some additional constructions: draw straight lines through the bases, and perpendiculars through the ends of the midline to the intersection with the bases:
* Alphabetic designations of vertices and other points are not intentionally entered to avoid unnecessary designations.
Look, triangles 1 and 2 are equal in the second sign of equality of triangles, triangles 3 and 4 are the same. The equality of the triangles implies the equality of the elements, namely the legs (they are indicated by blue and red, respectively).
Now attention! If we mentally “cut” the blue and red line from the lower base, then we will have a line (this is the side of the rectangle) equal to the midline. Further, if we “glue” the cut blue and red line to the upper base of the trapezoid, then we also get a line (this is also the side of the rectangle) equal to the midline of the trapezoid.
Did you catch it? It turns out that the sum of the bases will be equal to the two middle lines of the trapezoid:
See another explanation
Let's do the following - we will build a straight line passing through the lower base of the trapezoid and a straight line that passes through points A and B:
We get triangles 1 and 2, they are equal on the side and the corners adjacent to it (the second sign of the equality of triangles). This means that the resulting segment (indicated in blue on the sketch) is equal to the upper base of the trapezoid.
Now consider the triangle:
* The middle line of this trapezoid and the middle line of the triangle coincide.
It is known that a triangle is equal to half of the base parallel to it, that is:
Ok, sorted it out. Now about the area of \u200b\u200bthe trapezoid.
Trapezoid area formula:
They say: the area of \u200b\u200bthe trapezoid is equal to the product of half the sum of its bases and height.
That is, it turns out that it is equal to the product of the midline and height:
You probably already noticed that this is obvious. Geometrically, this can be expressed as follows: if we mentally cut triangles 2 and 4 from the trapezoid and put them on triangles 1 and 3, respectively:
Then we get a rectangle in area equal to the area of \u200b\u200bour trapezoid. The area of \u200b\u200bthis rectangle will be equal to the product of the midline and height, that is, we can write:
But the point here is not in the record, of course, but in understanding.
Download (view) the material of the article in * pdf format
That's all. Success to you!
Sincerely, Alexander.