AND . Now we can begin to consider the question of how to find the area of a trapezoid. This task in everyday life it occurs very rarely, but sometimes it turns out to be necessary, for example, to find the area of \u200b\u200ba room in the form of a trapezoid, which is increasingly used in the construction of modern apartments, or in renovation design projects.
Trapeze is geometric figure, formed by four intersecting segments, two of which are parallel to each other and are called the bases of a trapezoid. The other two segments are called the sides of the trapezoid. In addition, we will need another definition later on. This is the midline of the trapezoid, which is a segment connecting the midpoints of the sides and the height of the trapezoid, which is equal to the distance between the bases.
Like triangles, a trapezoid has particular types in the form of an isosceles (isosceles) trapezoid, in which the lengths of the sides are the same and rectangular trapezoid, in which one of the sides forms a right angle with the bases.
Trapezoids have some interesting properties:
- The midline of a trapezoid is half the sum of the bases and parallel to them.
- Isosceles trapeziums have equal sides and angles that they form with the bases.
- The midpoints of the diagonals of a trapezoid and the point of intersection of its diagonals are on the same straight line.
- If the sum of the sides of a trapezoid is equal to the sum of the bases, then a circle can be inscribed in it
- If the sum of the angles formed by the sides of a trapezoid at any of its bases is 90, then the length of the segment connecting the midpoints of the bases is equal to their half-difference.
- An isosceles trapezoid can be described by a circle. And vice versa. If a trapezoid is inscribed in a circle, then it is isosceles.
- The segment passing through the midpoints of the bases of an isosceles trapezoid will be perpendicular to its bases and represents the axis of symmetry.
How to find the area of a trapezoid.
The area of a trapezoid will be half the sum of its bases multiplied by its height. In the form of a formula, this is written as an expression:
where S is the area of the trapezoid, a,b is the length of each of the bases of the trapezoid, h is the height of the trapezoid.
You can understand and remember this formula as follows. As follows from the figure below, a trapezoid using the midline can be converted into a rectangle, the length of which will be equal to half the sum of the bases.
You can also decompose any trapezoid into simpler shapes: a rectangle and one or two triangles, and if it’s easier for you, then find the area of the trapezoid as the sum of the areas of its constituent figures.
There is one more simple formula to calculate its area. According to it, the area of the trapezoid is equal to the product of its midline and the height of the trapezoid and is written as: S = m * h, where S is the area, m is the length of the midline, h is the height of the trapezoid. This formula is more suitable for math problems than for everyday problems, since in real conditions you will not know the length of the midline without preliminary calculations. And you will only know the lengths of the bases and sides.
In this case, the area of the trapezoid can be found using the formula:
S \u003d ((a + b) / 2) * √c 2 - ((b-a) 2 + c 2 -d 2 / 2 (b-a)) 2
where S is the area, a,b are the bases, c,d are the sides of the trapezoid.
There are several more ways to find the area of a trapezoid. But, they are about as inconvenient as the last formula, which means it makes no sense to dwell on them. Therefore, we recommend that you use the first formula from the article and wish you always get accurate results.
Trapeze is called a quadrilateral only two sides are parallel to each other.
They are called the bases of the figure, the rest - the sides. A parallelogram is considered a special case of a figure. There is also a curvilinear trapezoid, which includes a function graph. The formulas for the area of a trapezoid include almost all of its elements, and the best solution selected depending on the given values.
The main roles in the trapezoid are assigned to height and midline. middle line- this is a line connecting the midpoints of the sides. Height the trapezoid is drawn at a right angle from the top corner to the base.
The area of a trapezoid through the height is equal to the product of half the sum of the lengths of the bases, multiplied by the height:
If the median line is known according to the conditions, then this formula is greatly simplified, since it is equal to half the sum of the lengths of the bases:
If, according to the conditions, the lengths of all sides are given, then we can consider an example of calculating the area of a trapezoid through these data: 
Suppose a trapezoid is given with bases a = 3 cm, b = 7 cm and sides c = 5 cm, d = 4 cm. Find the area of the figure: 
Area of an isosceles trapezoid
A separate case is an isosceles or, as it is also called, an isosceles trapezoid.
A special case is also finding the area of an isosceles (isosceles) trapezoid. Formula derived different ways- through the diagonals, through the angles adjacent to the base and the radius of the inscribed circle.
If the length of the diagonals is specified by the conditions and the angle between them is known, you can use the following formula:
Remember that the diagonals of an isosceles trapezoid are equal to each other!
That is, knowing one of their bases, side and angle, you can easily calculate the area.
Area of a curvilinear trapezoid
A separate case is curvilinear trapezoid. It is located on the coordinate axis and is limited to a graph of a continuous positive function.
Its base is located on the X axis and is limited to two points: 
Integrals help calculate the area of a curvilinear trapezoid.
The formula is written like this:
Consider an example of calculating the area of a curvilinear trapezoid. Formula Requires certain knowledge to work with definite integrals. First, let's analyze the value of the definite integral: 
Here F(a) is the value of the antiderivative function f(x) at point a , F(b) is the value of the same function f(x) at point b . 
Now let's solve the problem. The figure shows a curvilinear trapezoid bounded by a function. Function 
We need to find the area of the selected figure, which is a curvilinear trapezoid, bounded on top by a graph, on the right is a straight line x = (-8), on the left is a straight line x = (-10) and the axis OX is below.
We will calculate the area of this figure using the formula: 
We are given a function by the conditions of the problem. Using it, we will find the values of the antiderivative at each of our points:
Now
Answer: the area of a given curvilinear trapezoid is 4.
There is nothing difficult in calculating this value. Only the utmost care in calculations is important.
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.
V classical form the trapezium 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 of 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.
Download (view) the material of the article in *pdf format
That's all. Good luck to you!
Sincerely, Alexander.
There are many ways to find the area of a trapezoid. Usually a math tutor knows several methods for calculating it, let's dwell on them in more detail:
1) 
, where AD and BC are the bases, and BH is the height of the trapezoid. Proof: draw a diagonal BD and express the areas of triangles ABD and CDB in terms of the half product of their bases and height:
, where DP is the external height in
We add these equalities term by term and, given that the heights of BH and DP are equal, we get:
Let's take it out of the bracket
Q.E.D.
Consequence from the formula for the area of a trapezoid:
Since the half sum of the bases is equal to MN - the midline of the trapezoid, then
2) Application of the general formula for the area of a quadrilateral.
The area of a quadrilateral is half the product of the diagonals multiplied by the sine of the angle between them
To prove it, it is enough to divide the trapezoid into 4 triangles, express the area of each in terms of “half the product of the diagonals and the sine of the angle between them” (it is taken as the angle, add the resulting expressions, put it out of the bracket and decompose this bracket into factors using the grouping method to get its equality to the expression. From here
3) Diagonal shift method
This is my title. In school textbooks, a math tutor will not find such a heading. The description of the reception can only be found in additional teaching aids as an example of solving a problem. I note that most of the interesting and useful facts planimetry math tutors open to students in the process of doing practical work. This is extremely suboptimal, because the student needs to separate them into separate theorems and call them "big names". One of these is “diagonal shift”. About what in question?
Let us draw a straight line parallel to AC through the vertex B until it intersects with the lower base at point E. In this case, the quadrilateral EBCA will be a parallelogram (by definition) and therefore BC=EA and EB=AC. We are now concerned with the first equality. We have:
Note that triangle BED, whose area is equal to the area of a trapezoid, has several other remarkable properties:
1) Its area is equal to the area of a trapezoid
2) Its isosceles occurs simultaneously with the isosceles of the trapezoid itself
3) Its upper corner at vertex B equal to the angle between the diagonals of a trapezoid (which is very often used in problems) 
4) Its median BK is equal to the distance QS between the midpoints of the bases of the trapezoid. I recently came across the use of this property when preparing a student for the Mekhmat of Moscow State University using Tkachuk's textbook, version of 1973 (the task is given at the bottom of the page).
Mathematics tutor specials.
Sometimes I propose tasks in a very tricky way of finding the square of a trapezoid. I attribute it to special moves, because in practice the tutor rarely uses them. If you need to prepare for the exam in mathematics only in part B, you can not read about them. For others, I'll tell you more. It turns out the area of the trapezoid is twice more area a triangle with vertices at the ends of one side and the middle of the other, that is, the ABS triangle in the figure: 
Proof: draw heights SM and SN in triangles BCS and ADS and express the sum of the areas of these triangles:
Since the point S is the midpoint of CD, then (prove it yourself). Let's find the sum of the areas of triangles:
Since this amount turned out to be equal to half the area of the trapezoid, then - its second half. Ch.t.d.
I would include the form of calculating the area of an isosceles trapezoid along its sides into the treasury of special moves of a tutor: where p is the half-perimeter of the trapezoid. I will not give proof. Otherwise, your math tutor will be out of work :). Come to class!
Tasks for the area of the trapezoid:
Math tutor's note: The following list is not a methodological guide to the topic, it is only small selection interesting problems on the above methods.
1) The lower base of an isosceles trapezoid is 13, and the upper one is 5. Find the area of the trapezoid if its diagonal is perpendicular to the side.
2) Find the area of a trapezoid if its bases are 2cm and 5cm and its sides are 2cm and 3cm.
3) In an isosceles trapezoid, the larger base is 11, the side is 5, and the diagonal is Find the area of the trapezoid.
4) The diagonal of an isosceles trapezoid is 5, and the midline is 4. Find the area.
5) In an isosceles trapezoid, the bases are 12 and 20, and the diagonals are mutually perpendicular. Calculate the area of a trapezoid
6) The diagonal of an isosceles trapezoid makes an angle with its lower base. Find the area of a trapezoid if its height is 6 cm.
7) The area of the trapezoid is 20, and one of its sides is 4 cm. Find the distance to it from the middle of the opposite side.
8) The diagonal of an isosceles trapezoid divides it into triangles with areas 6 and 14. Find the height if the side is 4.
9) In a trapezoid, the diagonals are 3 and 5, and the segment connecting the midpoints of the bases is 2. Find the area of the trapezoid (Mekhmat of Moscow State University, 1970).
I chose not the most difficult tasks (do not be afraid of the Mekhmat!) with the expectation that they could independent decision. Decide on health! If you need to prepare for the exam in mathematics, then without participating in this process, the trapezoid area formulas may arise serious problems even with problem B6 and even more so with C4. Do not start the topic and in case of any difficulties, ask for help. A math tutor is always happy to help you.
Kolpakov A.N.
Math tutor in Moscow, preparation for the exam in Strogino.
A trapezoid is a convex quadrilateral in which two opposite sides are parallel and the other two are non-parallel. If all opposite sides of a quadrilateral are pairwise parallel, then it is a parallelogram.
You will need
- - all sides of the trapezoid (AB, BC, CD, DA).
Instruction
- Non-parallel sides trapeze are called lateral sides, and parallel ones are called bases. The line between the bases, perpendicular to them - the height trapeze. If the sides trapeze equal, it is called isosceles. First consider the solution for trapeze, which is not isosceles.
- Draw line BE from point B to lower base AD parallel to side trapeze CD. Since BE and CD are parallel and drawn between parallel bases trapeze BC and DA, then BCDE is a parallelogram, and its opposite sides BE and CD are equal. BE=CD.
- Consider triangle ABE. Calculate side AE. AE=AD-ED. Foundations trapeze BC and AD are known, and in parallelogram BCDE opposite sides ED and BC are equal. ED=BC, so AE=AD-BC.
- Now find out the area of triangle ABE using Heron's formula by calculating the semi-perimeter. S=root(p*(p-AB)*(p-BE)*(p-AE)). In this formula, p is the semiperimeter of triangle ABE. p=1/2*(AB+BE+AE). To calculate the area, you know all the necessary data: AB, BE=CD, AE=AD-BC.
- Next, write down the area of triangle ABE in a different way - it is equal to half the product of the height of the triangle BH and the side AE to which it is drawn. S=1/2*BH*AE.
- Express from this formula height triangle, which is also the height trapeze. BH=2*S/AE. Calculate it.
- Swipe from vertex C height CF.
- Examine the HBCF figure. HBCF is a rectangle because two of its sides are heights and the other two are bases trapeze, that is, the angles are right and the opposite sides are parallel. This means that BC=HF.
- Look at right triangles ABH and FCD. The angles at the heights BHA and CFD are right, and the angles at the lateral sides BAH and CDF are equal, since the trapezoid ABCD is isosceles, which means that the triangles are similar. Since heights BH and CF are equal or the sides of an isosceles trapeze AB and CD are congruent, then similar triangles are also congruent. Hence, their sides AH and FD are also equal.
- Find AH. AH+FD=AD-HF. Since from a parallelogram HF=BC, and from triangles AH=FD, then AH=(AD-BC)*1/2.
- Next, from the right triangle ABH, using the Pythagorean theorem, calculate height B.H. The square of the hypotenuse AB is equal to the sum of the squares of the legs AH and BH. BH=root(AB*AB-AH*AH).
If the trapezoid is isosceles, the solution can be done differently. Consider triangle ABH. It is rectangular because one of the corners, BHA, is straight.
