What is the length of the midline. Trapezium area
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The segment of the straight line connecting the midpoints of the sides of the trapezoid is called the midline of the trapezoid. How to find the middle line of the trapezoid and how it relates to other elements of this figure, we will describe below.
Midline theorem
Let's draw a trapezoid in which AD is the larger base, BC is the smaller base, EF is the middle line. Let's extend the base AD beyond point D. Draw the line BF and continue it until it intersects with the continuation of the base AD at point O. Consider the triangles ∆BCF and ∆DFO. Angles ∟BCF = ∟DFO as vertical. CF = DF, ∟BCF = ∟FDO, because VS // AO. Therefore, triangles ∆BCF = ∆DFO. Hence the sides BF = FO.
Now consider ∆ABO and ∆EBF. ∟ABO is common to both triangles. BE/AB = ½ by convention, BF/BO = ½ because ∆BCF = ∆DFO. Therefore, triangles ABO and EFB are similar. Hence the ratio of the sides EF / AO = ½, as well as the ratio of the other sides.
We find EF = ½ AO. The drawing shows that AO = AD + DO. DO = BC as sides equal triangles, so AO = AD + BC. Hence EF = ½ AO = ½ (AD + BC). Those. length middle line trapezium is half the sum of the bases.
Is the midline of a trapezoid always equal to half the sum of the bases?
Suppose there is a special case where EF ≠ ½ (AD + BC). Then BC ≠ DO, hence ∆BCF ≠ ∆DCF. But this is impossible, since they have two equal angles and sides between them. Therefore, the theorem is true under all conditions.
The problem of the middle line
Suppose, in our trapezoid ABCD AD // BC, ∟A=90°, ∟С = 135°, AB = 2 cm, diagonal AC is perpendicular to the side. Find the midline of the trapezoid EF.
If ∟A = 90°, then ∟B = 90°, so ∆ABC is rectangular.
∟BCA = ∟BCD - ∟ACD. ∟ACD = 90° by convention, therefore ∟BCA = ∟BCD - ∟ACD = 135° - 90° = 45°. 
If in right triangle∆ABC one angle is 45°, which means that the legs in it are equal: AB = BC = 2 cm.
Hypotenuse AC \u003d √ (AB² + BC²) \u003d √8 cm.
Consider ∆ACD. ∟ACD = 90° by convention. ∟CAD = ∟BCA = 45° as the angles formed by the secant of the parallel bases of the trapezoid. Therefore, the legs AC = CD = √8.
Hypotenuse AD = √(AC² + CD²) = √(8 + 8) = √16 = 4 cm.
The median line of the trapezoid EF = ½(AD + BC) = ½(2 + 4) = 3 cm.
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The area of the trapezoid. Greetings! In this publication, we will consider this formula. Why is it the way it is and how can you understand it? If there is an understanding, then you do not need to learn it. If you just want to see this formula and what is urgent, 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 trapezium. 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 the classical form, the trapezoid is depicted as follows - the larger base is at the bottom, respectively, the smaller one is at the top. But no one forbids depicting it and vice versa. Here are the sketches:
The next important concept.
The median line of a trapezoid is a segment that connects the midpoints of the sides. The median line is parallel to the bases of the trapezoid and is equal to their half-sum.
Now let's delve deeper. Why exactly?
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 of vertices and other points are not entered 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 respectively in blue and red).
Now attention! If we mentally “cut off” the blue and red segments from the lower base, then we will have a segment (this is the side of the rectangle) equal to the midline. Further, if we “glue” the cut off 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 medians of the trapezoid:
See another explanation
Let's do the following - build 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 in side and adjacent angles (the second sign of equality of triangles). This means that the resulting segment (in the sketch it is marked in blue) is equal to the upper base of the trapezoid.
Now consider a triangle:
*The median line of this trapezoid and the median line of the triangle coincide.
It is known that the triangle is equal to half of the base parallel to it, that is:
Okay, got it. Now about the area of the trapezoid.
Trapezium 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 midline and height:
You probably already noticed that this is obvious. Geometrically, this can be expressed as follows: if we mentally cut off 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 our trapezoid. The area of this rectangle will be equal to the product of the midline and 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.