How to find the height of a trapezoid: formulas for all occasions. All options on how to find the area of a trapezoid
In mathematics, several types of quadrilaterals are known: square, rectangle, rhombus, parallelogram. Among them is a trapezoid - a kind of convex quadrilateral, in which two sides are parallel, and the other two are not. The parallel opposite sides are called the bases, and the other two are called the sides of the trapezoid. The segment that connects the midpoints of the sides is called the midline. There are several types of trapezoids: isosceles, rectangular, curvilinear. For each type of trapezoid, there are formulas for finding the area.
Trapezium area
To find the area of a trapezoid, you need to know the length of its bases and its height. The height of a trapezoid is a segment perpendicular to the bases. Let the top base be a, the bottom base be b, and the height be h. Then you can calculate the area S by the formula:
S = ½ * (a + b) * h
those. take half the sum of the bases multiplied by the height.
You can also calculate the area of a trapezoid if you know the value of the height and the midline. Let's denote the middle line - m. Then
Let's solve the problem more complicated: we know the lengths of the four sides of the trapezoid - a, b, c, d. Then the area is found by the formula:
If the lengths of the diagonals and the angle between them are known, then the area is sought as follows:
S = ½ * d1 * d2 * sinα
where d with indices 1 and 2 are diagonals. In this formula, the sine of the angle is given in the calculation.
With known base lengths a and b and two angles at the lower base, the area is calculated as follows:
S = ½ * (b2 - a2) * (sin α * sin β / sin(α + β))
Area of an isosceles trapezoid
An isosceles trapezoid is special case trapezoid. Its difference is that such a trapezoid is a convex quadrangle with an axis of symmetry passing through the midpoints of two opposite sides. Her sides are equal.
Find area isosceles trapezium possible in several ways.
- Through the lengths of three sides. In this case, the lengths of the sides will match, therefore they are indicated by one value - c, a and b - the lengths of the bases:
- If the length of the upper base, lateral side and the angle at the lower base are known, then the area is calculated as follows:
S = c * sin α * (a + c * cos α)
where a is the upper base, c is the side.
- If instead of the upper base, the length of the lower base is known - b, the area is calculated by the formula:
S = c * sin α * (b - c * cos α)
- If when two bases and the angle at the lower base are known, the area is calculated using the tangent of the angle:
S = ½ * (b2 - a2) * tg α
- Also, the area is calculated through the diagonals and the angle between them. In this case, the diagonals are equal in length, so each is denoted by the letter d without indices:
S = ½ * d2 * sinα
- Calculate the area of the trapezoid, knowing the length of the lateral side, the midline and the angle at the lower base.
Let the side - c, the middle line - m, the corner - a, then:
S = m * c * sinα
Sometimes a circle can be inscribed in an equilateral trapezoid, the radius of which will be - r.
It is known that a circle can be inscribed in any trapezoid if the sum of the lengths of the bases is equal to the sum of the lengths of its sides. Then the area is found through the radius of the inscribed circle and the angle at the lower base:
S = 4r2 / sinα
The same calculation is made through the diameter D of the inscribed circle (by the way, it coincides with the height of the trapezoid):
Knowing the bases and the angle, the area of an isosceles trapezoid is calculated as follows:
S = a*b/sinα
(this and subsequent formulas are valid only for trapezoids with an inscribed circle).
Through the bases and the radius of the circle, the area is sought as follows:
If only the bases are known, then the area is calculated according to the formula:
through foundations and sideline the area of a trapezoid with an inscribed circle and through the bases and midline - m is calculated as follows:
Square rectangular trapezoid
A trapezoid is called rectangular, in which one of the sides is perpendicular to the bases. In this case, the side length coincides with the height of the trapezoid.
A rectangular trapezoid is a square and a triangle. After finding the area of each of the figures, add up the results and get the total area of the figure.
Also, general formulas for calculating the area of a trapezoid are suitable for calculating the area of a rectangular trapezoid.
- If the lengths of the bases and the height (or perpendicular side) are known, then the area is calculated by the formula:
S = (a + b) * h / 2
As h (height) can be the side with. Then the formula looks like this:
S = (a + b) * c / 2
- Another way to calculate area is to multiply the length of the midline by the height:
or by the length of the lateral perpendicular side:
- The next calculation method is through half the product of the diagonals and the sine of the angle between them:
S = ½ * d1 * d2 * sinα
If the diagonals are perpendicular, then the formula simplifies to:
S = ½ * d1 * d2
- Another way to calculate is through the semi-perimeter (the sum of the lengths of two opposite sides) and the radius of the inscribed circle.
This formula is valid for bases. If we take the lengths of the sides, then one of them will be equal to twice the radius. The formula will look like this:
S = (2r + c) * r
- If a circle is inscribed in a trapezoid, then the area is calculated in the same way:
where m is the length of the midline.
Area of a curvilinear trapezoid
The curvilinear trapezoid is flat figure, bounded by the graph of a non-negative continuous function y = f(x) defined on the segment , the x-axis and the straight lines x = a, x = b. In fact, two of its sides are parallel to each other (bases), the third side is perpendicular to the bases, and the fourth is a curve corresponding to the graph of the function.
The area of a curvilinear trapezoid is sought through the integral using the Newton-Leibniz formula:
How areas are calculated various kinds trapezium. But, in addition to the properties of the sides, trapezoids have the same properties of the angles. Like all existing quadrilaterals, the sum internal corners trapezium is 360 degrees. And the sum of the angles adjacent to the side is 180 degrees.
Geometry is one of the sciences, with the use of which in practice a person encounters almost daily. 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 the bases, and the remaining two are called the 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 \u200b\u200bthe figure is also indicated, then the following relationship can be used to determine the required perpendicular:
S=h*(a+b)/2,
h is the desired value (height),
S is the area of the figure,
a and b are sides parallel to each other.
It follows from the above formula that h=2S/(a+b).
The value of the midline is known
If among the initial data, in addition to the area of \u200b\u200bthe trapezoid (S), the length of its midline (l) is also known, then another formula is useful for calculations. First, it is worth clarifying what the middle line is for this type of quadrilateral. The term defines the part of the straight line connecting the midpoints of the sides of the figure.
Based on the properties of the trapezoid l=(a+b)/2,
l - midline,
a, b are sides-bases of the quadrilateral.
Therefore h=2S/(a+b)=S/l.
4 sides of the figure are known
AT this case the Pythagorean theorem will help. Having lowered the perpendiculars to the large 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 are 2 other sides.
Corners at the base
If you have base angle data, use trigonometric functions.
h = c*sinα = d*sinβ,
α and β are the corners at the base of the quadrilateral,
c and d are its sides.
The diagonals of a figure and the angles they intersect
The length of the diagonal is the length of the segment connecting the opposite vertices of the figure. Let's denote these quantities by symbols d1 and d2, and the angles between them 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 are the diagonals of the trapezoid,
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 - 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 the given trapezoid.
Find the height of an isosceles trapezoid
- As follows from the wording, a distinctive characteristic of an isosceles trapezoid is the equality of its sides. Therefore, to find the height of the figure, use the formula to determine this value in the case when the sides of the trapezoid are known.
So, if c \u003d d, then h \u003d √c 2 - (((a-b) 2 + c 2 -d 2) / 2 (a-b)) 2 \u003d √c 2 - (a-b) 2 / 4,
a, b - side-bases of the quadrilateral,
c = d are its sides.
- In the presence of the magnitude of the angles formed by the two sides (base and side), the height of the trapezoid is determined by the following ratio:
h = c*sinα,
h = с * tgα *cosα = с * tgα * (b - a)/2c = tgα * (b-a)/2,
α is the angle at the base of the figure,
a, b (a< b) – основания фигуры,
c = d are 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 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 break 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. AT school textbooks a math tutor won't see that heading. The description of the reception can only be found in additional teaching aids as an example of solving a problem. I note that tutors in mathematics reveal most of the interesting and useful facts of planimetry to students in the process of performing 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 encountered 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 was half 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 list below is not a methodological support to the topic, it is only a small selection of interesting tasks for 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 challenging tasks(do not be afraid of mekhmat!) with the expectation of the possibility of their 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.
The practice of last year's USE and GIA shows that geometry problems cause difficulties for many students. 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. The same ones can come across to you in KIMs at certification exams or at olympiads. Therefore, treat them carefully.
What you need to know about the trapezoid?
To begin with, let's remember that trapeze a quadrilateral is called, in which two opposite sides, they are also called bases, are parallel, and the other two are not.
In a trapezoid, the height (perpendicular to the base) can also be omitted. 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.
Trapezium area formulas
First, consider the standard formulas for finding the area of a trapezoid. Ways to calculate the area of isosceles and curvilinear trapezoids will be considered below.
So, imagine that you have a trapezoid with bases a and b, in which the height h is lowered to the larger base. Calculating the area of a figure in this case is easy. You just need to divide by two the sum of the lengths of the bases and multiply what happens by the height: S = 1/2(a + b)*h.
Let's take another case: suppose that in addition to the height, the trapezoid has a median line m. We know the formula for finding the length of the midline: m = 1/2(a + b). Therefore, we can rightfully simplify the formula for the area of a trapezoid to the following kind: S = m * h. In other words, to find the area of a trapezoid, you need to multiply the midline by the height.
Let's consider one more option: diagonals d 1 and d 2 are drawn in a trapezoid, which intersect not at a right angle α. To calculate the area of such a trapezoid, you need to halve the product of the diagonals and multiply what you get 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 complicated formula, but it will be useful for you to remember it just in case: S \u003d 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 variants of the formula for the area of an isosceles trapezoid.
The first option: for the case when a circle with radius r is inscribed inside an isosceles trapezoid, and the lateral side and the 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.
The 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 convert the trapezoid area formula already familiar to you into this form: S = h2.
The formula for the area of a curvilinear trapezoid
Let's start by understanding: what is a curvilinear trapezoid. 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 \u003d f (x) - at the top, the x axis - 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 interval. And the area of the curvilinear trapezoid corresponds to the increment of the antiderivative on a given segment.
Task examples
To make all these formulas better in your head, here are some examples of problems for finding the area of a trapezoid. It would be best if you first try to solve the problems yourself, and only then check the answer you received with the ready-made solution.
Task #1: Given a trapezoid. Its larger base is 11 cm, the smaller one is 4 cm. The trapezium has diagonals, one 12 cm long, the other 9 cm long.
Solution: Build a trapezoid AMRS. Draw line RX through vertex P so that it is parallel to diagonal MC and intersects line AC at point X. You get triangle APX.
We will consider two figures obtained as a result of these manipulations: the triangle APX and the parallelogram CMPX.
Thanks to the parallelogram, we learn that PX = MC = 12 cm and CX = MP = 4 cm. Where can we calculate the side AX of the triangle ARCH: AX \u003d AC + CX \u003d 11 + 4 \u003d 15 cm.
We can also prove that the triangle ARCH is right-angled (to do this, apply the Pythagorean theorem - AX 2 \u003d AP 2 + PX 2). And calculate its area: S APX \u003d 1/2 (AP * PX) \u003d 1/2 (9 * 12) \u003d 54 cm 2.
Next, you need to prove that the triangles AMP and PCX are equal in area. The basis will be the equality of the sides MP 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 assert that S AMPC \u003d S APX \u003d 54 cm 2.
Task #2: Given a trapezoid KRMS. Points O and E are located on its lateral sides, while OE and KS are parallel. It is also known that the areas of the trapezoid ORME and OXE are in the ratio 1:5. PM = a and KS = b. You need to find an OE.
Solution: Draw a line through point M parallel to RK, 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 of KS.
Let's introduce one more notation - OE = x. As well as 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 OXE are related as 1:5, which gives us the right to draw up the following equation: (x + a) * h 1 \u003d 1/5 (b + x) * h 2. Let's transform and get: h 1 / h 2 \u003d 1/5 * ((b + x) / (x + a)).
Since the triangles TME and AEC are similar, we have h 1 / h 2 = (x - a) / (b - x). Combine both entries and get: (x - a) / (b - x) \u003d 1/5 * ((b + x) / (x + a)) ↔ 5 (x - a) (x + a) \u003d (b + x) (b - x) ↔ 5 (x 2 - a 2) \u003d (b 2 - x 2) ↔ 6x 2 \u003d b 2 + 5a 2 ↔ x \u003d √ (5a 2 + b 2) / 6.
Thus, OE \u003d x \u003d √ (5a 2 + b 2) / 6.
Conclusion
Geometry is not the easiest of the sciences, but you can certainly cope with exam tasks. It just takes a little patience in preparation. And, of course, remember all the necessary formulas.
We tried to collect in one place all the formulas for calculating the area of a trapezoid so that you can use them when you prepare for exams and repeat the material.
Be sure to tell your classmates and friends about this article in in social networks. Let good grades there will be more for the USE and GIA!
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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 heights BHA and CFD are right, and the angles at the lateral sides BAH and CDF are equal, since the trapezoid ABCD is isosceles, so 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.
- Coming from right triangle ABH by 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.