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 quadrilateral in which only two sides are parallel is called trapezoid.
The parallel sides of a trapezoid are called its reasons, and those sides that are not parallel are called sides. If the sides are equal, then such a trapezoid is isosceles. The distance between the bases is called the height of the trapezoid.
Middle Line Trapezoid
The midline is a segment connecting the midpoints of the sides of the trapezoid. The midline of the trapezoid is parallel to its bases.
Theorem:
If the straight line crossing the middle of one side is parallel to the bases of the trapezoid, then it bisects the second side of the trapezoid.
Theorem:
The length of the middle line is equal to the arithmetic mean of the lengths of its bases
MN || AB || DCAM = MD; BN=NC
MN midline, AB and CD - bases, AD and BC - lateral sides
MN = (AB + DC)/2
Theorem:
The length of the midline of a trapezoid is equal to the arithmetic mean of the lengths of its bases.
The main task: Prove that the midline of a trapezoid bisects a segment whose ends lie in the middle of the bases of the trapezoid.
Middle Line of the Triangle
The segment connecting the midpoints of two sides of a triangle is called the midline of the triangle. It is parallel to the third side and its length is equal to half the length of the third side.
Theorem: If a line intersecting the midpoint of one side of a triangle is parallel to the other side given triangle, then it divides the third side in half.
AM = MC and BN = NC =>
Applying the midline properties of a triangle and trapezoid
Dividing a segment into a certain number of equal parts.
Task: Divide segment AB into 5 equal parts.
Solution:
Let p be a random ray whose origin is point A and which does not lie on line AB. We sequentially set aside 5 equal segments on p AA 1 = A 1 A 2 = A 2 A 3 = A 3 A 4 = A 4 A 5
We connect A 5 to B and draw such lines through A 4, A 3, A 2 and A 1 that are parallel to A 5 B. They intersect AB respectively at points B 4, B 3, B 2 and B 1. These points divide segment AB into 5 equal parts. Indeed, from the trapezoid BB 3 A 3 A 5 we see that BB 4 = B 4 B 3. In the same way, from the trapezoid B 4 B 2 A 2 A 4 we obtain B 4 B 3 = B 3 B 2
While from the trapezoid B 3 B 1 A 1 A 3, B 3 B 2 = B 2 B 1.
Then from B 2 AA 2 it follows that B 2 B 1 = B 1 A. In conclusion we get:
AB 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 = B 4 B
It is clear that to divide the segment AB into another number of equal parts, we need to project the same number of equal segments onto the ray p. And then continue in the manner described above.
In this article, another selection of problems with trapezoid has been made for you. The conditions are somehow related to its midline. Task types are taken from an open bank of typical tasks. If you wish, you can refresh your theoretical knowledge. The blog has already discussed tasks whose conditions are related to, as well as. Briefly about the middle line:
The midline of the trapezoid connects the midpoints of the lateral sides. It is parallel to the bases and equal to their half-sum.
Before solving problems, let's look at a theoretical example.
Given a 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'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 will pass through the middle of the other side.
This is also based on Thales' theorem:
If several equal segments are laid out in succession on one of two lines and parallel lines are drawn through their ends that intersect the second line, then they will cut off equal segments on the second line.
That is, in in this case K is the middle of AC and L is the middle of BD. Therefore 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 given for a reason. In tasks for independent decision there is just such a task. Only it doesn’t say that the segment connecting the midpoints of the diagonals lies on the midline. Let's consider the tasks:
27819. Find the midline of the trapezoid if its bases are 30 and 16.
We calculate using 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. Perpendicular dropped from the vertex of an obtuse angle to a larger base 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. Let's find DC.
Let's construct the second perpendicular DF:
The segments AF, FE and EB will be equal to 4, 6 and 4 respectively. Why?
In an isosceles trapezoid, perpendiculars lowered to the larger base divide it into three segments. Two of them, which are cut off legs right 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 a rectangle the opposite sides are equal. In this task:
Thus DC=6. We calculate:
27839. The bases of the trapezoid are in the ratio 2:3, and the midline is 5. Find the smaller base.
Let's introduce the proportionality coefficient 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 the value 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 greater than the other. Find the larger base of the trapezoid.
Let us denote the smaller base (DC) as x, then the larger one (AB) will be equal to x+4. We can write it down
We found that the smaller base is early five, which means 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 midline 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 ОF is equal to 2. We can write two equations and solve the system:
It is clear that in this case you can select a pair of numbers without calculations, these are 5 and 7. But, nevertheless, let’s 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.
Let us immediately 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 to calculate the middle line we must find reasons. Here a small dead end arises... How, knowing the height, in this case, calculate the bases? No way! There are many such trapezoids with a fixed height and diagonals intersecting at an angle of 90 degrees. What should I do?
Look at the formula for the midline of a trapezoid. After all, we do not need to know the reasons themselves; it is enough to know their sum (or half-sum). We can do this.
Since the diagonals intersect at right angles, isosceles right triangles are formed with height EF:
From the above it follows that FO=DF=FC, and OE=AE=EB. Now let’s write down what the height is equal to, expressed through the segments DF and AE:
So the middle line is 12.
*In general, this is a problem, as you understand, for mental calculation. But I'm sure the presented detailed explanation 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.
The prototypes also include types of tasks with trapezoids. It is built on a sheet of paper in a square and you need to find the middle line; the side of the cell is usually equal to 1, but it can be a different value.
27848. Find the midline of the trapezoid ABCD, if the sides of square cells are equal to 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!
Lesson objectives:
1) introduce students to the concept of the midline of a trapezoid, consider its properties and prove them;
2) teach how to build the midline of the trapezoid;
3) develop students’ ability to use the definition of the midline of a trapezoid and the properties of the midline of a trapezoid when solving problems;
4) continue to develop students’ ability to speak competently, using the necessary mathematical terms; prove your point of view;
5) develop logical thinking, memory, attention.
During the classes
1. Homework is checked during the lesson. The homework was oral, remember:
a) definition of a trapezoid; types of trapezoids;
b) determining the midline of the triangle;
c) property of the midline of a triangle;
d) sign of the middle line of the triangle.
2. Studying new material.
a) The board shows a trapezoid ABCD.
b) The teacher asks you to remember the definition of a trapezoid. Each desk has a hint diagram to help you remember the basic concepts in the topic “Trapezoid” (see Appendix 1). Appendix 1 is issued to each desk.
Students draw the trapezoid ABCD in their notebooks.
c) The teacher asks you to remember in which topic the concept of a midline was encountered (“Midline of a triangle”). Students recall the definition of the midline of a triangle and its properties.
e) Write down the definition of the midline of the trapezoid, drawing it in a notebook.
Middle line A trapezoid is a segment connecting the midpoints of its sides.
The property of the midline of a trapezoid remains unproven at this stage, so the next stage of the lesson involves working on proving the property of the midline of a trapezoid.
Theorem. The midline of the trapezoid is parallel to its bases and equal to their half-sum.
Given: ABCD – trapezoid,
MN – middle line ABCD
Prove, What:
1. BC || MN || A.D.
2. MN = (AD + BC).
We can write down some corollaries that follow from the conditions of the theorem:
AM = MB, CN = ND, BC || A.D.
It is impossible to prove what is required based on the listed properties alone. The system of questions and exercises should lead students to the desire to connect the midline of a trapezoid with the midline of some triangle, the properties of which they already know. If there are no proposals, then you can ask the question: how to construct a triangle for which the segment MN would be the midline?
Let us write down an additional construction for one of the cases.
Let us draw a straight line BN intersecting the continuation of side AD at point K.
Additional elements appear - triangles: ABD, BNM, DNK, BCN. If we prove that BN = NK, then this will mean that MN is the midline of ABD, and then we can use the property of the midline of a triangle and prove the necessary.
Proof:
1. Consider BNC and DNK, they contain:
a) CNB =DNK (property of vertical angles);
b) BCN = NDK (property of internal cross-lying angles);
c) CN = ND (by corollary to the conditions of the theorem).
This means BNC =DNK (by the side and two adjacent angles).
Q.E.D.
The proof can be done orally in class, and can be reconstructed and written down in a notebook at home (at the teacher’s discretion).
It is necessary to say about other possible ways of proving this theorem:
1. Draw one of the diagonals of the trapezoid and use the sign and property of the triangle’s midline.
2. Carry out CF || BA and consider the parallelogram ABCF and DCF.
3. Carry out EF || BA and consider the equality of FND and ENC.
g) At this stage it is specified homework: paragraph 84, textbook ed. Atanasyan L.S. (proof of the property of the midline of a trapezoid using a vector method), write it down in your notebook.
h) We solve problems using the definition and properties of the midline of a trapezoid using ready-made drawings (see Appendix 2). Appendix 2 is given to each student, and the solution to the problems is written out on the same sheet in a short form.
Area of a trapezoid. Greetings! In this publication we will look at this formula. Why is she exactly like this and how to understand her. If there is understanding, then you don’t need to teach it. If you just want to look at this formula and urgently, then you can immediately scroll down the page))
Now in detail and in order.
A trapezoid is a quadrilateral, two sides of this quadrilateral are parallel, the other two are not. Those that are not parallel are the bases 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 classic form The trapezoid is depicted as follows: the larger base is at the bottom, and the smaller base is at the top. But no one forbids depicting her and vice versa. Here are the sketches:
Next important concept.
The midline of a trapezoid is a segment that connects the midpoints of the sides. The middle line is parallel to the bases of the trapezoid and equal to their half-sum.
Now let's delve deeper. Why is this so?
Consider a trapezoid with 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 until they intersect with the bases:
*Letter designations for vertices and other points are not included intentionally to avoid unnecessary designations.
Look, triangles 1 and 2 are equal according to the second sign of equality of triangles, triangles 3 and 4 are the same. From the equality of triangles follows the equality of the elements, namely the legs (they are indicated in blue and red, respectively).
Now attention! If we mentally “cut off” the blue and red segments from the lower base, then we will be left with a segment (this is the side of the rectangle) equal to the middle line. Next, if we “glue” the cut blue and red segments to the upper base of the trapezoid, then we will also get a segment (this is also the side of the rectangle) equal to the midline of the trapezoid.
Got it? It turns out that the sum of the bases will be equal to the two middle lines of the trapezoid:
View another explanation
Let's do the following - construct a straight line passing through the lower base of the trapezoid and a straight line that will pass through points A and B:
We get triangles 1 and 2, they are equal along the side and adjacent angles (the second sign of equality of triangles). This means that the resulting segment (in the sketch it is indicated in blue) is equal to the upper base of the trapezoid.
Now consider the triangle:
*The midline of this trapezoid and the midline of the triangle coincide.
It is known that a triangle is equal to half of the base parallel to it, that is:
Okay, we figured it out. Now about the area of the trapezoid.
Trapezoid area formula:
They say: the area of a 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 center line and the height:
You've probably already noticed that this is obvious. Geometrically, this can be expressed this way: if we mentally cut off triangles 2 and 4 from the trapezoid and place them on triangles 1 and 3, respectively:
Then we get a rectangle in area equal to area our trapezoid. The area of this rectangle will be equal to the product of the center line and the height, that is, we can write:
But the point here is not in writing, of course, but in understanding.
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That's all. Good luck to you!
Sincerely, Alexander.
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