The many-sided trapezoid... It can be arbitrary, isosceles or rectangular. And in each case you need to know how to find the area of a trapezoid. Of course, the easiest way is to remember the basic formulas. But sometimes it’s easier to use one that is derived taking into account all the features of a particular geometric figure.
A few words about the trapezoid and its elements
Any quadrilateral whose two sides are parallel can be called a trapezoid. In general, they are not equal and are called bases. The larger one is the lower one, and the other one is the upper one.
The other two sides turn out to be lateral. In an arbitrary trapezoid they have different lengths. If they are equal, then the figure becomes isosceles.
If suddenly the angle between any side and the base turns out to be equal to 90 degrees, then the trapezoid is rectangular.
All these features can help in solving the problem of how to find the area of a trapezoid.
Among the elements of the figure that may be indispensable in solving problems, we can highlight the following:
- height, that is, a segment perpendicular to both bases;
- the middle line, which has at its ends the midpoints of the lateral sides.
What formula can be used to calculate the area if the base and height are known?
This expression is given as a basic one because most often one can recognize these quantities even when they are not given explicitly. So, to understand how to find the area of a trapezoid, you will need to add both bases and divide them by two. Then multiply the resulting value by the height value.
If we denote the bases by the letters a 1 and a 2, the height by n, then the formula for the area will look like this:
S = ((a 1 + a 2)/2)*n.
The formula that calculates the area if its height and center line are given
If you look carefully at the previous formula, it is easy to notice that it clearly contains the value of the midline. Namely, the sum of the bases divided by two. Let the middle line be designated by the letter l, then the formula for the area becomes:
S = l * n.
Ability to find area using diagonals
This method will help if the angle formed by them is known. Suppose that the diagonals are designated by the letters d 1 and d 2, and the angles between them are α and β. Then the formula for how to find the area of a trapezoid will be written as follows:
S = ((d 1 * d 2)/2) * sin α.
You can easily replace α with β in this expression. The result will not change.
How to find out the area if all sides of the figure are known?
There are also situations when exactly the sides of this figure are known. This formula is cumbersome and difficult to remember. But it's possible. Let the sides have the designation: a 1 and a 2, the base a 1 is greater than a 2. Then the area formula will take the following form:
S = ((a 1 + a 2) / 2) * √ (in 1 2 - [(a 1 - a 2) 2 + in 1 2 - in 2 2) / (2 * (a 1 - a 2)) ] 2 ).
Methods for calculating the area of an isosceles trapezoid
The first is due to the fact that a circle can be inscribed in it. And, knowing its radius (it is denoted by the letter r), as well as the angle at the base - γ, you can use the following formula:
S = (4 * r 2) / sin γ.
The last general formula, which is based on knowledge of all sides of the figure, will be significantly simplified due to the fact that the sides have the same meaning:
S = ((a 1 + a 2) / 2) * √ (in 2 - [(a 1 - a 2) 2 / (2 * (a 1 - a 2))] 2 ).
Methods for calculating the area of a rectangular trapezoid
It is clear that any of the above is suitable for any figure. But sometimes it is useful to know about one feature of such a trapezoid. It lies in the fact that the difference between the squares of the lengths of the diagonals is equal to the difference made up of the squares of the bases.
Often the formulas for a trapezoid are forgotten, while the expressions for the areas of a rectangle and triangle are remembered. Then you can use a simple method. Divide the trapezoid into two shapes, if it is rectangular, or three. One will definitely be a rectangle, and the second, or the remaining two, will be triangles. After calculating the areas of these figures, all that remains is to add them up.
This is a fairly simple way to find the area of a rectangular trapezoid.
What if the coordinates of the vertices of the trapezoid are known?
In this case, you will need to use an expression that allows you to determine the distance between points. It can be applied three times: in order to find out both bases and one height. And then just apply the first formula, which is described a little higher.
To illustrate this method, the following example can be given. Given vertices with coordinates A(5; 7), B(8; 7), C(10; 1), D(1; 1). You need to find out the area of the figure.
Before finding the area of the trapezoid, you need to calculate the lengths of the bases from the coordinates. You will need the following formula:
length of the segment = √((difference of the first coordinates of the points) 2 + (difference of the second coordinates of the points) 2 ).
The upper base is designated AB, which means its length will be equal to √((8-5) 2 + (7-7) 2 ) = √9 = 3. The lower one is CD = √ ((10-1) 2 + (1-1 ) 2 ) = √81 = 9.
Now you need to draw the height from the top to the base. Let its beginning be at point A. The end of the segment will be on the lower base at the point with coordinates (5; 1), let this be point H. The length of the segment AN will be equal to √((5-5) 2 + (7-1) 2 ) = √36 = 6.
All that remains is to substitute the resulting values into the formula for the area of a trapezoid:
S = ((3 + 9) / 2) * 6 = 36.
The problem was solved without units of measurement, because the scale of the coordinate grid was not specified. It can be either a millimeter or a meter.
Sample problems
No. 1. Condition. The angle between the diagonals of an arbitrary trapezoid is known; it is equal to 30 degrees. The smaller diagonal has a value of 3 dm, and the second is 2 times larger. It is necessary to calculate the area of the trapezoid.
Solution. First you need to find out the length of the second diagonal, because without this it will not be possible to calculate the answer. It is not difficult to calculate, 3 * 2 = 6 (dm).
Now you need to use the appropriate formula for area:
S = ((3 * 6) / 2) * sin 30º = 18/2 * ½ = 4.5 (dm 2). The problem is solved.
Answer: The area of the trapezoid is 4.5 dm2.
No. 2. Condition. In the trapezoid ABCD, the bases are the segments AD and BC. Point E is the middle of the SD side. From it a perpendicular is drawn to the straight line AB, the end of this segment is designated by the letter H. It is known that the lengths AB and EH are equal to 5 and 4 cm, respectively. It is necessary to calculate the area of the trapezoid.
Solution. First you need to make a drawing. Since the value of the perpendicular is less than the side to which it is drawn, the trapezoid will be slightly elongated upward. So EH will be inside the figure.
To clearly see the progress of solving the problem, you will need to perform additional construction. Namely, draw a straight line that will be parallel to side AB. The points of intersection of this line with AD are P, and with the continuation of BC are X. The resulting figure VHRA is a parallelogram. Moreover, its area is equal to the required one. This is due to the fact that the triangles that were obtained during the additional construction are equal. This follows from the equality of the side and two angles adjacent to it, one vertical, the other lying crosswise.
You can find the area of a parallelogram using a formula that contains the product of the side and the height lowered onto it.
Thus, the area of the trapezoid is 5 * 4 = 20 cm 2.
Answer: S = 20 cm 2.
No. 3. Condition. The elements of an isosceles trapezoid have the following values: lower base - 14 cm, upper base - 4 cm, acute angle- 45º. You need to calculate its area.
Solution. Let the smaller base be designated BC. The height drawn from point B will be called VH. Since the angle is 45º, triangle ABH will be rectangular and isosceles. So AN=VN. Moreover, AN is very easy to find. It is equal to half the difference in bases. That is (14 - 4) / 2 = 10 / 2 = 5 (cm).
The bases are known, the heights are calculated. You can use the first formula, which was discussed here for an arbitrary trapezoid.
S = ((14 + 4) / 2) * 5 = 18/2 * 5 = 9 * 5 = 45 (cm 2).
Answer: The required area is 45 cm 2.
No. 4. Condition. There is an arbitrary trapezoid ABCD. Points O and E are taken on its lateral sides, so that OE is parallel to the base of AD. The area of the AOED trapezoid is five times larger than that of the OVSE. Calculate the OE value if the lengths of the bases are known.
Solution. You will need to draw two parallel lines AB: the first through point C, its intersection with OE is point T; the second through E and the point of intersection with AD will be M.
Let the unknown OE=x. The height of the smaller trapezoid OVSE is n 1, the larger AOED is n 2.
Since the areas of these two trapezoids are related as 1 to 5, we can write the following equality:
(x + a 2) * n 1 = 1/5 (x + a 1) * n 2
n 1 / n 2 = (x + a 1) / (5 (x + a 2)).
The heights and sides of the triangles are proportional by construction. Therefore, we can write one more equality:
n 1 / n 2 = (x - a 2) / (a 1 - x).
In the last two entries on the left side there are equal values, which means we can write that (x + a 1) / (5(x + a 2)) is equal to (x - a 2) / (a 1 - x).
A number of transformations are required here. First multiply crosswise. Parentheses will appear to indicate the difference of squares, after applying this formula you will get a short equation.
In it you need to open the brackets and move all the terms with the unknown “x” to left side, and then take the square root.
Answer: x = √ ((a 1 2 + 5 a 2 2) / 6).
To the simple question “How to find the height of a trapezoid?” There are several answers, all because different starting values can be given. Therefore, the formulas will differ.
These formulas can be memorized, but they are not difficult to derive. You just need to apply previously learned theorems.
Notations used in formulas
In all the mathematical notations below, these readings of the letters are correct.
In the source data: all sides
In order to find the height of a trapezoid in the general case, you will need to use the following formula:
n = √(c 2 - (((a - c) 2 + c 2 - d 2)/(2(a - c))) 2). Number 1.
Not the shortest, but also found quite rarely in problems. Usually you can use other data.
The formula that will tell you how to find the height of an isosceles trapezoid in the same situation is much shorter:
n = √(c 2 - (a - c) 2 /4). Number 2.
The problem gives: lateral sides and angles at the lower base
It is assumed that the angle α is adjacent to the side with the designation “c”, respectively, the angle β is to the side d. Then the formula for how to find the height of a trapezoid will be in general form:
n = c * sin α = d * sin β. Number 3.
If the figure is isosceles, then you can use this option:
n = c * sin α= ((a - b) / 2) * tan α. Number 4.
Known: diagonals and angles between them
Typically, these data are accompanied by other known quantities. For example, the bases or the middle line. If the reasons are given, then to answer the question of how to find the height of a trapezoid, the following formula will be useful:
n = (d 1 * d 2 * sin γ) / (a + b) or n = (d 1 * d 2 * sin δ) / (a + b). Number 5.
This is for general view figures. If an isosceles is given, then the notation will change like this:
n = (d 1 2 * sin γ) / (a + b) or n = (d 1 2 * sin δ) / (a + b). Number 6.
When the problem deals with the midline of a trapezoid, the formulas for finding its height become as follows:
n = (d 1 * d 2 * sin γ) / 2m or n = (d 1 * d 2 * sin δ) / 2m. Number 5a.
n = (d 1 2 * sin γ) / 2m or n = (d 1 2 * sin δ) / 2m. Number 6a.
Among the known quantities: area with bases or midline
These are perhaps the shortest and simplest formulas for finding the height of a trapezoid. For an arbitrary figure it will be like this:
n = 2S / (a + b). Number 7.
It’s the same, but with a known middle line:
n = S/m. Number 7a.
Oddly enough, but for an isosceles trapezoid the formulas will look the same.
Tasks
No. 1. To determine the angles at the lower base of the trapezoid.
Condition. Given an isosceles trapezoid whose side is 5 cm. Its bases are 6 and 12 cm. You need to find the sine of an acute angle.
Solution. For convenience, you should enter a designation. Let the lower left vertex be A, all the rest in a clockwise direction: B, C, D. Thus, the lower base will be designated AD, the upper one - BC.
It is necessary to draw heights from vertices B and C. The points that indicate the ends of the heights will be designated H 1 and H 2, respectively. Since all the angles in the figure BCH 1 H 2 are right angles, it is a rectangle. This means that the segment H 1 H 2 is 6 cm.
Now we need to consider two triangles. They are equal because they are rectangular with the same hypotenuses and vertical legs. It follows from this that their smaller legs are equal. Therefore, they can be defined as the quotient of the difference. The latter is obtained by subtracting the upper one from the lower base. It will be divided by 2. That is, 12 - 6 must be divided by 2. AN 1 = N 2 D = 3 (cm).
Now from the Pythagorean theorem you need to find the height of the trapezoid. It is necessary to find the sine of an angle. VN 1 = √(5 2 - 3 2) = 4 (cm).
Using the knowledge of how the sine of an acute angle is found in a triangle with a right angle, we can write the following expression: sin α = ВН 1 / AB = 0.8.
Answer. The required sine is 0.8.
No. 2. To find the height of a trapezoid using a known tangent.
Condition. For an isosceles trapezoid, you need to calculate the height. It is known that its bases are 15 and 28 cm. The tangent of the acute angle is given: 11/13.
Solution. The designation of vertices is the same as in the previous problem. Again you need to draw two heights from the upper corners. By analogy with the solution to the first problem, you need to find AN 1 = N 2 D, which is defined as the difference of 28 and 15 divided by two. After calculations it turns out: 6.5 cm.
Since the tangent is the ratio of two legs, we can write the following equality: tan α = AN 1 / VN 1 . Moreover, this ratio is equal to 11/13 (according to the condition). Since AN 1 is known, the height can be calculated: ВН 1 = (11 * 6.5) / 13. Simple calculations give a result of 5.5 cm.
Answer. The required height is 5.5 cm.
No. 3. To calculate the height using known diagonals.
Condition. It is known about the trapezoid that its diagonals are 13 and 3 cm. You need to find out its height if the sum of the bases is 14 cm.
Solution. Let the designation of the figure be the same as before. Let's assume that AC is the smaller diagonal. From vertex C you need to draw the desired height and designate it CH.
Now you need to do some additional construction. From corner C you need to draw a straight line parallel to the larger diagonal and find the point of its intersection with the continuation of side AD. This will be D 1. The result is a new trapezoid, inside which a triangle ASD 1 is drawn. This is what is needed to further solve the problem.
The desired height will also be in the triangle. Therefore, you can use the formulas studied in another topic. The height of a triangle is defined as the product of the number 2 and the area divided by the side to which it is drawn. And the side turns out to be equal to the sum of the bases of the original trapezoid. This comes from the rule by which the additional construction was made.
In the triangle under consideration, all sides are known. For convenience, we introduce the notation x = 3 cm, y = 13 cm, z = 14 cm.
Now you can calculate the area using Heron's theorem. The semi-perimeter will be equal to p = (x + y + z) / 2 = (3 + 13 + 14) / 2 = 15 (cm). Then the formula for the area after substituting the values will look like this: S = √(15 * (15 - 3) * (15 - 13) * (15 - 14)) = 6 √10 (cm 2).
Answer. The height is 6√10 / 7 cm.
No. 4. To find the height on the sides.
Condition. Given a trapezoid, three sides of which are 10 cm, and the fourth is 24 cm. You need to find out its height.
Solution. Since the figure is isosceles, you will need formula number 2. You just need to substitute all the values into it and count. It will look like this:
n = √(10 2 - (10 - 24) 2 /4) = √51 (cm).
Answer. n = √51 cm.
Geometry is one of the sciences that people encounter in practice almost every day. Among the diversity geometric shapes The trapezoid also deserves special attention. It is a convex figure with four sides, two of which are parallel to each other. The latter are called bases, and the remaining two are called sides. The segment perpendicular to the bases and determining the size of the gap between them will be the height of the trapezoid. How can you calculate its length?
Find the height of an arbitrary trapezoid
Based on the initial data, determining the height of a figure is possible in several ways.
Known area
If the length of the parallel sides is known, and the area of the figure is also indicated, then to determine the desired perpendicular, you can use the following relationship:
S=h*(a+b)/2,
h – the desired value (height),
S – area of the figure,
a and b are sides parallel to each other.
From the above formula it follows that h=2S/(a+b).
The value of the midline is known
If among the initial data, in addition to the area of the trapezoid (S), the length of its midline (l) is also known, then another formula is useful for calculations. First, it’s worth clarifying what the midline is for this type of quadrilateral. The term defines the part of the straight line connecting the midpoints of the lateral sides of the figure.
Based on the trapezoid property l=(a+b)/2,
l – midline,
a, b – base sides of the quadrilateral.
Therefore h=2S/(a+b)=S/l.
4 sides of the figure are known
In this case, the Pythagorean theorem will help. Having lowered the perpendiculars to the larger side-base, use it for the two resulting right-angled triangles. The final expression will look like:
h=√c 2 -(((a-b) 2 +c 2 -d 2)/2(a-b)) 2,
c and d – 2 other sides.
Angles at the base
If you have data on the base angles, use trigonometric functions.
h = c* sinα = d*sinβ,
α and β are the angles at the base of the quadrilateral,
c and d are its sides.
Diagonals of a figure and the angles that intersecting they form
The length of the diagonal is the length of the segment connecting the opposite vertices of the figure. Let us denote these quantities by the symbols d1 and d2, and the angles between them by γ and φ. Then:
h = (d1*d2)/(a+b) sin γ = (d1*d2)/(a+b) sinφ,
h = (d1*d2)/2l sin γ = (d1*d2)/2l sinφ,
a and b are the base sides of the figure,
d1 and d2 – trapezoid diagonals,
γ and φ are the angles between the diagonals.
The height of the figure and the radius of the circle that is inscribed in it
As follows from the definition of this kind of circle, it touches each base at 1 point, which are part of one straight line. Therefore, the distance between them is the diameter – the desired height of the figure. And since the diameter is twice the radius, then:
h = 2 * r,
r is the radius of the circle that is inscribed in this trapezoid.
Find the height of an isosceles trapezoid
- As follows from the formulation, a distinctive characteristic of an isosceles trapezoid is the equality of its lateral sides. Therefore, to find the height of a figure, use the formula for determining this value in the case when the sides of the trapezoid are known.
So, if c = d, then h=√c 2 -(((a-b) 2 +c 2 -d 2)/2(a-b)) 2 = √c 2 -(a-b) 2 /4,
a, b – base sides of the quadrilateral,
c = d – its sides.
- If there are angles formed by two sides (base and side), the height of the trapezoid is determined by the following relationship:
h = c* sinα,
h = с * tgα *cosα = с * tgα * (b – a)/2c = tgα * (b-a)/2,
α – angle at the base of the figure,
a, b (a< b) – основания фигуры,
c = d – its sides.
- If the values of the diagonals of the figure are given, then the expression for finding the height of the figure will change, because d1 = d2:
h = d1 2 /(a+b)*sinγ = d1 2 /(a+b)*sinφ,
h = d1 2 /2*l*sinγ = d1 2 /2*l*sinφ.
There are many ways to find the area of a trapezoid. Usually a math tutor knows several methods of calculating it, let’s look at them in more detail:
1) 
, where AD and BC are the bases, and BH is the height of the trapezoid. Proof: draw the diagonal BD and express the areas of triangles ABD and CDB through the half product of their bases and heights:
, where DP is the external height in
Let us add these equalities term by term and taking into account that the heights BH and DP are equal, we obtain:
Let's put it out of brackets
Q.E.D.
Corollary to 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 equal to 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 through “half the product of the diagonals and the sine of the angle between them” (taken as the angle, add the resulting expressions, take them out of the bracket and factor this bracket using the grouping method to obtain its equality to the expression. Hence
3) Diagonal shift method
This is my name. A math tutor will not come across such a heading in school textbooks. A description of the technique can only be found in additional textbooks as an example of solving a problem. I note that most of the interesting and useful facts mathematics tutors reveal planimetry to students in the process of performing practical work. This is extremely suboptimal, because the student needs to isolate them into separate theorems and call them “big names.” One of these is “diagonal shift”. What are we talking about? 
Let us draw a line parallel to AC through 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. The first equality is important to us now. We have:
Note that the triangle BED, whose area is equal to the area of the trapezoid, has several more remarkable properties:
1) Its area is equal to the area of the trapezoid
2) Its isosceles occurs simultaneously with the isosceles of the trapezoid itself
3) Its upper angle at vertex B is equal to the angle between the diagonals of the 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 encountered the use of this property when preparing a student for Mechanics and Mathematics at Moscow State University using Tkachuk’s textbook, 1973 version (the problem is given at the bottom of the page).
Special techniques for a math tutor.
Sometimes I propose problems using a very tricky way of finding the area of a trapezoid. I classify it as a special technique because in practice the tutor uses them extremely rarely. If you need preparation for the Unified State Exam in mathematics only in Part B, you don’t have to read about them. For others, I'll tell you further. It turns out that the area of a trapezoid is twice the area of a triangle with vertices at the ends of one side and the middle of the other, that is, the triangle ABS in the figure: 
Proof: draw the heights SM and SN in triangles BCS and ADS and express the sum of the areas of these triangles:
Since point S is the middle of CD, then (prove it yourself). Find the sum of the areas of the triangles:
Since this sum turned out to be equal to half the area of the trapezoid, then its second half. Etc.
In the tutor’s collection of special techniques, I would include the form of calculating the area of an isosceles trapezoid along its sides: where p is the semi-perimeter of the trapezoid. I won't give proof. Otherwise, your math tutor will be left without a job :). Come to class!
Problems on the area of a trapezoid:
Math tutor's note: The list below is not a methodological accompaniment to the topic, it is only a small selection of interesting tasks based on the techniques discussed above.
1) The lower base of an isosceles trapezoid is 13, and the upper 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 the 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 of 6 and 14. Find the height if the lateral side is 4.
9) In a trapezoid, the diagonals are equal to 3 and 5, and the segment connecting the midpoints of the bases is equal to 2. Find the area of the trapezoid (Mekhmat MSU, 1970).
I didn't choose the best complex tasks(don’t be afraid of the mechanics and mathematics department!) with the expectation that they can be solved independently. Decide for your health! If you need preparation for the Unified State Exam in mathematics, then without the participation of the formula for the area of a trapezoid in this process, serious problems may arise 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.
Mathematics tutor in Moscow, preparation for the Unified State Exam in Strogino.
The practice of last year's Unified State Exam and State Examination shows that geometry problems cause difficulties for many schoolchildren. You can easily cope with them if you memorize all the necessary formulas and practice solving problems.
In this article you will see formulas for finding the area of a trapezoid, as well as examples of problems with solutions. You may come across the same ones in KIMs during certification exams or at Olympiads. Therefore, treat them carefully.
What you need to know about the trapezoid?
To begin with, let us remember that trapezoid is called a quadrilateral in which two opposite sides, also called bases, are parallel, and the other two are not.
In a trapezoid, the height (perpendicular to the base) can also be lowered. The middle line is drawn - this is a straight line that is parallel to the bases and equal to half of their sum. As well as diagonals that can intersect, forming acute and obtuse angles. Or, in some cases, at a right angle. In addition, if the trapezoid is isosceles, a circle can be inscribed in it. And describe a circle around it.
Trapezoid area formulas
First, let's look at the standard formulas for finding the area of a trapezoid. We will consider ways to calculate the area of isosceles and curvilinear trapezoids below.
So, imagine that you have a trapezoid with bases a and b, in which height h is lowered to the larger base. Calculating the area of a figure in this case is as easy as shelling pears. You just need to divide the sum of the lengths of the bases by two and multiply the result by the height: S = 1/2(a + b)*h.
Let's take another case: suppose in a trapezoid, in addition to the height, there is a middle line m. We know the formula for finding the length of the middle line: m = 1/2(a + b). Therefore, we can rightfully simplify the formula for the area of a trapezoid to the following type: S = m*h. In other words, to find the area of a trapezoid, you need to multiply the center line by the height.
Let's consider another option: the trapezoid contains diagonals d 1 and d 2, which do not intersect at right angles α. To calculate the area of such a trapezoid, you need to divide the product of the diagonals by two and multiply the result by the sin of the angle between them: S= 1/2d 1 d 2 *sinα.
Now consider the formula for finding the area of a trapezoid if nothing is known about it except the lengths of all its sides: a, b, c and d. This is a cumbersome and complex formula, but it will be useful for you to remember it just in case: S = 1/2(a + b) * √c 2 – ((1/2(b – a)) * ((b – a) 2 + c 2 – d 2)) 2.
By the way, the above examples are also true for the case when you need the formula for the area of a rectangular trapezoid. This is a trapezoid, the side of which adjoins the bases at a right angle.
Isosceles trapezoid
A trapezoid whose sides are equal is called isosceles. We will consider several options for the formula for the area of an isosceles trapezoid.
First option: for the case when a circle with radius r is inscribed inside an isosceles trapezoid, and the side and larger base form an acute angle α. A circle can be inscribed in a trapezoid provided that the sum of the lengths of its bases is equal to the sum of the lengths of the sides.
The area of an isosceles trapezoid is calculated as follows: multiply the square of the radius of the inscribed circle by four and divide it all by sinα: S = 4r 2 /sinα. Another area formula is a special case for the option when the angle between the large base and the side is 30 0: S = 8r2.
Second option: this time we take an isosceles trapezoid, in which in addition the diagonals d 1 and d 2 are drawn, as well as the height h. If the diagonals of a trapezoid are mutually perpendicular, the height is half the sum of the bases: h = 1/2(a + b). Knowing this, it is easy to transform the formula for the area of a trapezoid that is already familiar to you into this form: S = h 2.
Formula for the area of a curved trapezoid
Let's start by figuring out what a curved trapezoid is. Imagine a coordinate axis and a graph of a continuous and non-negative function f that does not change sign within a given segment on the x-axis. A curvilinear trapezoid is formed by the graph of the function y = f(x) - at the top, the x axis is at the bottom (segment), and on the sides - straight lines drawn between points a and b and the graph of the function.
It is impossible to calculate the area of such a non-standard figure using the above methods. Here you need to apply mathematical analysis and use the integral. Namely: the Newton-Leibniz formula - S = ∫ b a f(x)dx = F(x)│ b a = F(b) – F(a). In this formula, F is the antiderivative of our function on the selected segment. And the area of a curvilinear trapezoid corresponds to the increment of the antiderivative on a given segment.
Sample problems
To make all these formulas easier to understand in your head, here are some examples of problems for finding the area of a trapezoid. It will be best if you first try to solve the problems yourself, and only then compare the answer you receive with the ready-made solution.
Task #1: Given a trapezoid. Its larger base is 11 cm, the smaller one is 4 cm. The trapezoid has diagonals, one 12 cm long, the second 9 cm.
Solution: Construct a trapezoid AMRS. Draw a straight line РХ through vertex P so that it is parallel to the diagonal MC and intersects the straight line AC at point X. You will get a triangle APХ.
We will consider two figures obtained as a result of these manipulations: triangle APX and parallelogram CMRX.
Thanks to the parallelogram, we learn that PX = MC = 12 cm and CX = MR = 4 cm. From where we can calculate the side AX of the triangle ARX: AX = AC + CX = 11 + 4 = 15 cm.
We can also prove that the triangle APX is right-angled (to do this, apply the Pythagorean theorem - AX 2 = AP 2 + PX 2). And calculate its area: S APX = 1/2(AP * PX) = 1/2(9 * 12) = 54 cm 2.
Next you will need to prove that triangles AMP and PCX are equal in area. The basis will be the equality of the parties MR and CX (already proven above). And also the heights that you lower on these sides - they are equal to the height of the AMRS trapezoid.
All this will allow you to say that S AMPC = S APX = 54 cm 2.
Task #2: The trapezoid KRMS is given. On its lateral sides there are points O and E, while OE and KS are parallel. It is also known that the areas of trapezoids ORME and OKSE are in the ratio 1:5. RM = a and KS = b. You need to find OE.
Solution: Draw a line parallel to RK through point M, and designate the point of its intersection with OE as T. A is the point of intersection of a line drawn through point E parallel to RK with the base KS.
Let's introduce one more notation - OE = x. And also the height h 1 for the triangle TME and the height h 2 for the triangle AEC (you can independently prove the similarity of these triangles).
We will assume that b > a. The areas of the trapezoids ORME and OKSE are in the ratio 1:5, which gives us the right to create the following equation: (x + a) * h 1 = 1/5(b + x) * h 2. Let's transform and get: h 1 / h 2 = 1/5 * ((b + x)/(x + a)).
Since the triangles TME and AEC are similar, we have h 1 / h 2 = (x – a)/(b – x). Let's combine both entries and get: (x – a)/(b – x) = 1/5 * ((b + x)/(x + a)) ↔ 5(x – a)(x + a) = (b + x)(b – x) ↔ 5(x 2 – a 2) = (b 2 – x 2) ↔ 6x 2 = b 2 + 5a 2 ↔ x = √(5a 2 + b 2)/6.
Thus, OE = x = √(5a 2 + b 2)/6.
Conclusion
Geometry is not the easiest of sciences, but you can certainly cope with the exam questions. It is enough to show a little perseverance in preparation. And, of course, remember all the necessary formulas.
We tried to collect all the formulas for calculating the area of a trapezoid in one place so that you can use them when you prepare for exams and revise the material.
Be sure to tell your classmates and friends about this article. social networks. Let there be more good grades for the Unified State Examination and State Examinations!
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