Trapeze is a quadrilateral with two parallel sides, which are the bases, and two non-parallel sides, which are the sides.
There are also names such as isosceles or isosceles.
It is a trapezoid with right angles on the lateral side.
Trapeze elements
a, b bases of a trapezoid(a parallel to b ),
m, n — sides trapeze,
d 1 , d 2 — diagonals trapeze,
h- height trapezoid (a segment connecting the bases and at the same time perpendicular to them),
MN- middle line(a segment connecting the midpoints of the sides).
Trapezium area
- Through half the sum of the bases a, b and the height h : S = \frac(a + b)(2)\cdot h
- Through the midline MN and height h : S = MN\cdot h
- Through the diagonals d 1 , d 2 and the angle (\sin \varphi ) between them: S = \frac(d_(1) d_(2) \sin \varphi)(2)
Trapezoid Properties
Median line of the trapezoid
middle line is parallel to the bases, equal to their half-sum, and divides each segment with ends located on straight lines that contain the bases (for example, the height of the figure) in half:
MN || a, MN || b, MN = \frac(a + b)(2)
The sum of the angles of a trapezoid
The sum of the angles of a trapezoid, adjacent to each side, is equal to 180^(\circ) :
\alpha + \beta = 180^(\circ)
\gamma + \delta =180^(\circ)
Equal area triangles of a trapezoid
Equal-sized, that is, having equal areas, are the segments of the diagonals and the triangles AOB and DOC formed by the sides.
Similarity of formed trapezoid triangles
similar triangles are AOD and COB, which are formed by their bases and diagonal segments.
\triangle AOD \sim \triangle COB
similarity coefficient k is found by the formula:
k = \frac(AD)(BC)
Moreover, the ratio of the areas of these triangles is equal to k^(2) .
The ratio of the lengths of segments and bases
Each segment connecting the bases and passing through the point of intersection of the diagonals of the trapezoid is divided by this point in relation to:
\frac(OX)(OY) = \frac(BC)(AD)
This will also be true for the height with the diagonals themselves.
Trapeze is special case a quadrilateral with one pair of sides parallel. The term "trapezoid" comes from the Greek word τράπεζα, meaning "table", "table". In this article we will consider the types of trapezium and its properties. In addition, we will figure out how to calculate the individual elements of this example, the diagonal of an isosceles trapezoid, the midline, area, etc. The material is presented in the style of elementary popular geometry, that is, in an easily accessible form.
General information
First, let's understand what a quadrilateral is. This figure is a special case of a polygon containing four sides and four vertices. Two vertices of a quadrilateral that are not adjacent are called opposite. The same can be said about two non-adjacent sides. The main types of quadrilaterals are parallelogram, rectangle, rhombus, square, trapezoid and deltoid.
So, back to the trapeze. As we have already said, this figure has two sides that are parallel. They are called bases. The other two (non-parallel) are the sides. In exam materials and various control works very often you can find tasks related to trapezoids, the solution of which often requires the student to have knowledge that is not provided for by the program. The school geometry course introduces students to the properties of angles and diagonals, as well as the midline of an isosceles trapezoid. But after all, in addition to this, the mentioned geometric figure has other features. But more on them later...
Types of trapezoid
There are many types of this figure. However, most often it is customary to consider two of them - isosceles and rectangular.
1. A rectangular trapezoid is a figure in which one of the sides is perpendicular to the bases. It has two angles that are always ninety degrees.
2. An isosceles trapezoid is a geometric figure whose sides are equal to each other. This means that the angles at the bases are also pairwise equal.
The main principles of the methodology for studying the properties of a trapezoid
The main principle is the use of the so-called task approach. In fact, there is no need to introduce new properties of this figure into the theoretical course of geometry. They can be discovered and formulated in the process of solving various problems (better than systemic ones). At the same time, it is very important that the teacher knows what tasks need to be set for students at one time or another. educational process. Moreover, each property of the trapezoid can be represented as a key task in the task system.
The second principle is the so-called spiral organization of the study of the "remarkable" properties of the trapezium. This implies a return in the learning process to individual features of a given geometric figure. Thus, it is easier for students to memorize them. For example, the property of four points. It can be proved both in the study of similarity and subsequently with the help of vectors. And the equal area of triangles adjacent to the sides of the figure can be proved by applying not only the properties of triangles with equal heights drawn to the sides that lie on the same straight line, but also using the formula S= 1/2(ab*sinα). In addition, you can work out on an inscribed trapezoid or a right triangle on a circumscribed trapezoid, etc.
The use of "out-of-program" features of a geometric figure in the content school course is a task technology of their teaching. The constant appeal to the studied properties when passing through other topics allows students to gain a deeper knowledge of the trapezoid and ensures the success of solving the tasks. So, let's start studying this wonderful figure.
Elements and properties of an isosceles trapezoid
As we have already noted, the sides of this geometric figure are equal. It is also known as the right trapezoid. Why is it so remarkable and why did it get such a name? The features of this figure include the fact that not only the sides and corners at the bases are equal, but also the diagonals. Also, the sum of the angles of an isosceles trapezoid is 360 degrees. But that's not all! Of all known trapezoids, only around an isosceles one can a circle be described. This is due to the fact that the sum of the opposite angles of this figure is 180 degrees, and only under this condition can a circle be described around the quadrilateral. The next property of the geometric figure under consideration is that the distance from the base vertex to the projection of the opposite vertex onto the straight line that contains this base will be equal to the midline.
Now let's figure out how to find the angles of an isosceles trapezoid. Consider a solution to this problem, provided that the dimensions of the sides of the figure are known.
Solution
Usually, a quadrilateral is usually denoted by the letters A, B, C, D, where BS and AD are the bases. In an isosceles trapezoid, the sides are equal. We will assume that their size is X, and the sizes of the bases are Y and Z (smaller and larger, respectively). To carry out the calculation, it is necessary to draw a height H from angle B. The result is a right-angled triangle ABN, where AB is the hypotenuse, and BN and AN are the legs. We calculate the size of the leg AN: we subtract the smaller one from the larger base, and divide the result by 2. We write it in the form of a formula: (Z-Y) / 2 \u003d F. Now, to calculate the acute angle of the triangle, we will use cos function. We get the following record: cos(β) = Х/F. Now we calculate the angle: β=arcos (Х/F). Further, knowing one angle, we can determine the second, for this we perform an elementary arithmetic operation: 180 - β. All angles are defined.
There is also a second solution to this problem. At the beginning, we lower the height H from the corner B. We calculate the value of the BN leg. We know that the square of the hypotenuse right triangle is equal to the sum squares of legs. We get: BN \u003d √ (X2-F2). Next, we use trigonometric function tg. As a result, we have: β = arctg (BN / F). Sharp corner found. Next, we determine in the same way as the first method.
Property of the diagonals of an isosceles trapezoid
Let's write down four rules first. If the diagonals in an isosceles trapezoid are perpendicular, then:
The height of the figure will be equal to the sum of the bases divided by two;
Its height and median line are equal;
The center of the circle is the point where the ;
If the lateral side is divided by the point of contact into segments H and M, then it is equal to square root products of these segments;
The quadrilateral, which was formed by the tangent points, the vertex of the trapezoid and the center of the inscribed circle, is a square whose side is equal to the radius;
The area of a figure is equal to the product of the bases and the product of half the sum of the bases and its height.
Similar trapeziums
This topic is very convenient for studying the properties of this one. For example, the diagonals divide the trapezoid into four triangles, and those adjacent to the bases are similar, and those adjacent to the sides are equal. This statement can be called a property of the triangles into which the trapezoid is divided by its diagonals. The first part of this assertion is proved through the criterion of similarity in two angles. To prove the second part, it is better to use the method given below.
Proof of the theorem
We accept that the figure ABSD (AD and BS - the bases of the trapezoid) is divided by the diagonals VD and AC. Their intersection point is O. We get four triangles: AOS - at the lower base, BOS - at the upper base, ABO and SOD at the sides. Triangles SOD and BOS have a common height if the segments BO and OD are their bases. We get that the difference between their areas (P) is equal to the difference between these segments: PBOS / PSOD = BO / OD = K. Therefore, PSOD = PBOS / K. Similarly, the BOS and AOB triangles have a common height. We take the segments CO and OA as their bases. We get PBOS / PAOB \u003d CO / OA \u003d K and PAOB \u003d PBOS / K. It follows from this that PSOD = PAOB.
To consolidate the material, students are advised to find a connection between the areas of the resulting triangles, into which the trapezoid is divided by its diagonals, by solving the following problem. It is known that the areas of triangles BOS and AOD are equal, it is necessary to find the area of the trapezoid. Since PSOD \u003d PAOB, it means that PABSD \u003d PBOS + PAOD + 2 * PSOD. From the similarity of the triangles BOS and AOD it follows that BO / OD = √ (PBOS / PAOD). Therefore, PBOS/PSOD = BO/OD = √(PBOS/PAOD). We get PSOD = √ (PBOS * PAOD). Then PABSD = PBOS+PAOD+2*√(PBOS*PAOD) = (√PBOS+√PAOD)2.
similarity properties
Continuing to develop this topic, we can prove other interesting features trapezium. So, using similarity, you can prove the property of a segment that passes through a point formed by the intersection of the diagonals of this geometric figure, parallel to the bases. To do this, we solve the following problem: it is necessary to find the length of the segment RK, which passes through the point O. From the similarity of triangles AOD and BOS, it follows that AO/OS=AD/BS. From the similarity of triangles AOP and ASB, it follows that AO / AS \u003d RO / BS \u003d AD / (BS + AD). From here we get that RO \u003d BS * AD / (BS + AD). Similarly, from the similarity of the triangles DOK and DBS, it follows that OK \u003d BS * AD / (BS + AD). From here we get that RO=OK and RK=2*BS*AD/(BS+AD). The segment passing through the point of intersection of the diagonals, parallel to the bases and connecting the two sides, is divided by the point of intersection in half. Its length is the harmonic mean of the bases of the figure.
Consider the following property of a trapezoid, which is called the property of four points. The intersection points of the diagonals (O), the intersections of the continuation of the sides (E), as well as the midpoints of the bases (T and W) always lie on the same line. This is easily proved by the similarity method. The resulting triangles BES and AED are similar, and in each of them the medians ET and EZH divide the angle at the vertex E into equal parts. Therefore, the points E, T and W lie on the same straight line. In the same way, the points T, O, and G are located on the same straight line. All this follows from the similarity of the triangles BOS and AOD. From this we conclude that all four points - E, T, O and W - will lie on one straight line.
Using similar trapezoids, students can be asked to find the length of the segment (LF) that divides the figure into two similar ones. This segment should be parallel to the bases. Since the resulting trapezoids ALFD and LBSF are similar, then BS/LF=LF/AD. It follows that LF=√(BS*BP). We get that the segment that divides the trapezoid into two similar ones has a length equal to the geometric mean of the lengths of the bases of the figure.
Consider the following similarity property. It is based on a segment that divides the trapezoid into two equal-sized figures. We accept that the trapezoid ABSD is divided by the segment EN into two similar ones. From the vertex B, the height is omitted, which is divided by the segment EH into two parts - B1 and B2. We get: PABSD / 2 \u003d (BS + EH) * B1 / 2 \u003d (AD + EH) * B2 / 2 and PABSD \u003d (BS + HELL) * (B1 + B2) / 2. Next, we compose a system whose first equation is (BS + EH) * B1 \u003d (AD + EH) * B2 and the second (BS + EH) * B1 \u003d (BS + HELL) * (B1 + B2) / 2. It follows that B2/ B1 = (BS+EN)/(AD+EN) and BS+EN = ((BS+AD)/2)*(1+B2/ B1). We get that the length of the segment dividing the trapezoid into two equal ones is equal to the mean square of the lengths of the bases: √ ((BS2 + AD2) / 2).
Similarity inferences
Thus, we have proven that:
1. The segment connecting the midpoints of the sides of the trapezoid is parallel to AD and BS and is equal to the arithmetic mean of BS and AD (the length of the base of the trapezoid).
2. The line passing through the point O of the intersection of the diagonals parallel to AD and BS will be equal to the harmonic mean of the numbers AD and BS (2 * BS * AD / (BS + AD)).
3. The segment that divides the trapezoid into similar ones has the length of the geometric mean of the bases BS and AD.
4. An element that divides a figure into two equal ones has the length of the mean square numbers AD and BS.
To consolidate the material and understand the connection between the considered segments, the student needs to build them for a specific trapezoid. He can easily display the midline and the segment that passes through the point O - the intersection of the diagonals of the figure - parallel to the bases. But where will be the third and fourth? This answer will lead the student to the discovery of the desired relationship between the averages.
A line segment that joins the midpoints of the diagonals of a trapezoid
Consider the following property of this figure. We accept that the segment MH is parallel to the bases and bisects the diagonals. Let's call the intersection points W and W. This segment will be equal to the half-difference of the bases. Let's analyze this in more detail. MSH - the middle line of the triangle ABS, it is equal to BS / 2. MS - the middle line of the triangle ABD, it is equal to AD / 2. Then we get that ShShch = MShch-MSh, therefore, Sshch = AD / 2-BS / 2 = (AD + VS) / 2.
Center of gravity
Let's look at how this element is determined for a given geometric figure. To do this, it is necessary to extend the bases in opposite directions. What does it mean? It is necessary to add the lower base to the upper base - to any of the sides, for example, to the right. And the bottom is extended by the length of the top to the left. Next, we connect them with a diagonal. The point of intersection of this segment with the middle line of the figure is the center of gravity of the trapezoid.
Inscribed and circumscribed trapezoids
Let's list the features of such figures:
1. A trapezoid can only be inscribed in a circle if it is isosceles.
2. A trapezoid can be described around a circle, provided that the sum of the lengths of their bases is equal to the sum of the lengths of the sides.
Consequences of the inscribed circle:
1. The height of the described trapezoid is always equal to two radii.
2. The lateral side of the described trapezoid is observed from the center of the circle at a right angle.
The first corollary is obvious, and to prove the second one it is required to establish that the SOD angle is right, which, in fact, will also not be difficult. But knowledge of this property will allow us to use a right-angled triangle in solving problems.
Now we specify these consequences for an isosceles trapezoid, which is inscribed in a circle. We get that the height is the geometric mean of the bases of the figure: H=2R=√(BS*AD). Practicing the main technique for solving problems for trapezoids (the principle of drawing two heights), the student must solve the following task. We accept that BT is the height of the isosceles figure ABSD. It is necessary to find segments AT and TD. Using the formula described above, this will not be difficult to do.
Now let's figure out how to determine the radius of a circle using the area of the circumscribed trapezoid. We lower the height from top B to the base AD. Since the circle is inscribed in a trapezoid, then BS + AD \u003d 2AB or AB \u003d (BS + AD) / 2. From the triangle ABN we find sinα = BN / AB = 2 * BN / (BS + AD). PABSD \u003d (BS + AD) * BN / 2, BN \u003d 2R. We get PABSD \u003d (BS + HELL) * R, it follows that R \u003d PABSD / (BS + HELL).
All formulas of the midline of a trapezoid
Now it's time to move on to the last element of this geometric figure. Let's figure out what the middle line of the trapezoid (M) is equal to:
1. Through the bases: M \u003d (A + B) / 2.
2. Through height, base and angles:
M \u003d A-H * (ctgα + ctgβ) / 2;
M \u003d B + H * (ctgα + ctgβ) / 2.
3. Through height, diagonals and the angle between them. For example, D1 and D2 are the diagonals of a trapezoid; α, β - angles between them:
M = D1*D2*sinα/2H = D1*D2*sinβ/2H.
4. Through the area and height: M = P / N.
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In the materials of various tests and exams, very often there are tasks for the trapezoid, the solution of which requires knowledge of its properties.
Let us find out what interesting and useful properties a trapezoid has for solving problems.
After studying the properties of the midline of a trapezoid, we can formulate and prove property of a segment connecting the midpoints of the diagonals of a trapezoid. The segment connecting the midpoints of the diagonals of a trapezoid is equal to the half-difference of the bases.
MO is the midline of triangle ABC and is equal to 1/2BC (Fig. 1).
MQ is the midline of triangle ABD and is equal to 1/2AD.
Then OQ = MQ – MO, hence OQ = 1/2AD – 1/2BC = 1/2(AD – BC).
When solving many problems on a trapezoid, one of the main tricks is to hold two heights in it.
Consider the following task.
Let BT be the height of an isosceles trapezoid ABCD with bases BC and AD, where BC = a, AD = b. Find the lengths of the segments AT and TD.
Solution.
Problem solving is not difficult (Fig. 2), but it allows you to get property of the height of an isosceles trapezoid drawn from the vertex of an obtuse angle: the height of an isosceles trapezoid, drawn from the apex of an obtuse angle, divides the greater base into two segments, the smaller of which is half the difference of the bases, and the larger is half the sum of the bases.
When studying the properties of a trapezoid, you need to pay attention to such a property as similarity. So, for example, the diagonals of a trapezoid divide it into four triangles, and the triangles adjacent to the bases are similar, and the triangles adjacent to the sides are equal. This statement can be called property of the triangles into which a trapezoid is divided by its diagonals. Moreover, the first part of the assertion is proved very easily through the sign of similarity of triangles in two angles. Let's prove the second part of the statement.
Triangles BOC and COD have the same height (Fig. 3), if we take the segments BO and OD as their bases. Then S BOC /S COD = BO/OD = k. Therefore, S COD = 1/k · S BOC .
Similarly, triangles BOC and AOB have a common height if we take segments CO and OA as their bases. Then S BOC /S AOB = CO/OA = k and S A O B = 1/k · S BOC .
It follows from these two propositions that S COD = S A O B.
We will not dwell on the stated statement, but find the relation between the areas of the triangles into which a trapezoid is divided by its diagonals. To do this, we will solve the following problem.
Let the point O be the point of intersection of the diagonals of the trapezoid ABCD with the bases BC and AD. It is known that the areas of triangles BOC and AOD are equal to S 1 and S 2 , respectively. Find the area of the trapezoid.
Since S COD \u003d S A O B, then S ABC D \u003d S 1 + S 2 + 2S COD.
From the similarity of triangles BOC and AOD, it follows that BO / OD \u003d √ (S₁ / S 2).
Therefore, S₁/S COD = BO/OD = √(S₁/S 2), and hence S COD = √(S 1 S 2).
Then S ABC D = S 1 + S 2 + 2√(S 1 S 2) = (√S 1 + √S 2) 2 .
Using similarity, one can also prove property of a line segment passing through the point of intersection of the diagonals of a trapezoid parallel to the bases.
Consider task:
Let the point O be the point of intersection of the diagonals of the trapezoid ABCD with the bases BC and AD. BC=a, AD=b. Find the length of the segment PK passing through the point of intersection of the diagonals of the trapezoid parallel to the bases. What segments does PK divide into by the point O (Fig. 4)?
From the similarity of triangles AOD and BOC it follows that АO/OC = AD/BC = b/a.
From the similarity of triangles AOP and ACB it follows that AO/AC = PO/BC = b/(a + b).
Hence PO = BC b / (a + b) = ab/(a + b).
Similarly, from the similarity of triangles DOK and DBC, it follows that OK = ab/(a + b).
Hence PO = OK and PK = 2ab/(a + b).
So, the proven property can be formulated as follows: a segment parallel to the bases of the trapezoid, passing through the intersection point of the diagonals and connecting two points on the sides, is divided by the intersection point of the diagonals in half. Its length is the harmonic mean of the bases of the trapezoid.
Following property of four points: in a trapezoid, the intersection point of the diagonals, the intersection point of the continuation of the sides, the midpoints of the bases of the trapezoid lie on the same line.
Triangles BSC and ASD are similar (Fig. 5) and in each of them the medians ST and SG divide the vertex angle S into equal parts. Therefore, the points S, T and G lie on the same line.
Similarly, points T, O and G are located on the same line. This follows from the similarity of triangles BOC and AOD.
Hence, all four points S, T, O and G lie on the same line.
You can also find the length of the segment dividing the trapezoid into two similar ones.
If the trapezoids ALFD and LBCF are similar (Fig. 6), then a/LF = LF/b.
Hence LF = √(ab).
Thus, the segment dividing the trapezoid into two similar trapezoids has a length equal to the geometric mean of the lengths of the bases. 
Let's prove property of a line segment that divides a trapezoid into two equal parts.
Let the area of the trapezoid be S (Fig. 7). h 1 and h 2 are parts of the height, and x is the length of the desired segment.
Then S/2 = h 1 (a + x)/2 = h 2 (b + x)/2 and
S \u003d (h 1 + h 2) (a + b) / 2.
Let's make a system
(h 1 (a + x) = h 2 (b + x)
(h 1 (a + x) = (h 1 + h 2) (a + b)/2.
Solving this system, we get x \u003d √ (1/2 (a 2 + b 2)).
In this way, the length of the segment dividing the trapezoid into two equal ones is √ ((a 2 + b 2) / 2)(root mean square lengths of the bases).
So, for the trapezoid ABCD with bases AD and BC (BC = a, AD = b) we proved that the segment:
1) MN, connecting the midpoints of the sides of the trapezoid, is parallel to the bases and equal to their half-sum (average arithmetic numbers a and b);
2) PK passing through the point of intersection of the diagonals of the trapezoid parallel to the bases is equal to
2ab/(a + b) (harmonic mean of numbers a and b);
3) LF, dividing the trapezoid into two similar trapezoids, has a length equal to the average geometric numbers a and b, √(ab);
4) EH dividing the trapezoid into two equal ones has length √((a 2 + b 2)/2) (root mean square of numbers a and b).
Sign and property of an inscribed and circumscribed trapezoid.
Property of an inscribed trapezoid: A trapezoid can be inscribed in a circle if and only if it is isosceles.
Properties of the described trapezoid. A trapezoid can be described about a circle if and only if the sum of the lengths of the bases is equal to the sum of the lengths of the sides.
Useful consequences of the fact that a circle is inscribed in a trapezoid:
1. The height of the circumscribed trapezoid is equal to two radii of the inscribed circle.
2. The lateral side of the circumscribed trapezoid is visible from the center of the inscribed circle at a right angle.
The first is obvious. To prove the second corollary, it is necessary to establish that the angle COD is a right one, which is also not difficult. But the knowledge of this consequence allows us to use a right-angled triangle in solving problems.
We concretize consequences for the isosceles circumscribed trapezoid:
The height of an isosceles circumscribed trapezoid is the geometric mean of the bases of the trapezoid
h = 2r = √(ab).
The considered properties will allow a deeper knowledge of the trapezoid and ensure success in solving problems on the application of its properties.
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With such a form as a trapezoid, we meet in life quite often. For example, any bridge that is made of concrete blocks is a prime example. A more visual option can be considered the steering of each vehicle And so on. The properties of the figure were already known in Ancient Greece
, which was described in more detail by Aristotle in his scientific work"Start". And the knowledge that was developed thousands of years ago is still relevant today. Therefore, we will get acquainted with them in more detail.
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Basic concepts
Figure 1. The classic shape of a trapezoid.
A trapezoid is essentially a quadrilateral, consisting of two segments that are parallel and two others that are not parallel. Speaking about this figure, it is always necessary to remember such concepts as: bases, height and middle line. Two segments of a quadrilateral which are called bases to each other (segments AD and BC). The height is called the segment perpendicular to each of the bases (EH), i.e. intersect at an angle of 90° (as shown in Fig. 1).
If we add up all the degree measures of the internal, then the sum of the angles of the trapezoid will be equal to 2π (360 °), like any quadrilateral. A segment whose ends are the midpoints of the sidewalls (IF) called the middle line. The length of this segment is the sum of the bases BC and AD divided by 2.
There are three types of geometric shapes: straight, regular and isosceles. If at least one angle at the vertices of the base is right (for example, if ABD = 90 °), then such a quadrilateral is called a right trapezoid. If the side segments are equal (AB and CD), then it is called isosceles (respectively, the angles at the bases are equal).
How to find the area
For that, to find the area of a quadrilateral ABCD use the following formula:
Figure 2. Solving the problem of finding the area
For more good example Let's solve an easy problem. For example, let the upper and lower bases be equal to 16 and 44 cm, respectively, and the sides are 17 and 25 cm. Let's build a perpendicular segment from the vertex D so that DE II BC (as shown in Figure 2). Hence we get that
Let DF - will be. From ΔADE (which will be equilateral), we get the following:
That is, to express plain language, we first found the height ΔADE, which is also the height of the trapezoid. From here we calculate the area of the quadrilateral ABCD, with the already known value of the height DF, using the already known formula.
Hence, the desired area ABCD is 450 cm³. That is, it can be said with certainty that To calculate the area of a trapezoid, you need only the sum of the bases and the length of the height.
Important! When solving the problem, it is not necessary to find the value of the lengths separately; it is quite possible if other parameters of the figure are applied, which, with appropriate proof, will be equal to the sum of the bases.
Types of trapezium
Depending on which sides the figure has, what angles are formed at the bases, there are three types of quadrilateral: rectangular, sided and equilateral.
Versatile
There are two forms: acute and obtuse. ABCD is acute only if the base angles (AD) are acute and the side lengths are different. If the value of one angle is the number Pi / 2 more (the degree measure is more than 90 °), then we get an obtuse angle.
If the sides are equal in length
Figure 3. View of an isosceles trapezoid
If non-parallel sides are equal in length, then ABCD is called isosceles (correct). Moreover, for such a quadrilateral, the degree measure of the angles at the base is the same, their angle will always be less than the right one. It is for this reason that the isosceles is never divided into acute and obtuse. A quadrilateral of this shape has its own specific differences, which include:
- The segments connecting opposite vertices are equal.
- Acute angles with a larger base are 45 ° (an illustrative example in Figure 3).
- If you add the degrees of opposite angles, then in total they will give 180 °.
- Around any regular trapezoid can be built.
- If you add the degree measure of opposite angles, then it is equal to π.
Moreover, due to their geometric arrangement of points, there are basic properties of an isosceles trapezoid:
Angle value at base 90°
The perpendicularity of the lateral side of the base is a capacious characteristic of the concept of "rectangular trapezium". There cannot be two sides with corners at the base, because otherwise it will be already a rectangle. In quadrilaterals of this type, the second side will always form sharp corner with a large base, and with a smaller one - obtuse. In this case, the perpendicular side will also be the height.
Segment between the middle of the sidewalls
If we connect the midpoints of the sides, and the resulting segment will be parallel to the bases, and equal in length to half their sum, then the formed straight line will be the middle line. The value of this distance is calculated by the formula:
For a more illustrative example, consider a problem using the middle line.
A task. The median line of the trapezoid is 7 cm, it is known that one of the sides is 4 cm larger than the other (Fig. 4). Find the lengths of the bases.
Figure 4. Solving the problem of finding base lengths
Solution. Let the smaller base of DC be equal to x cm, then the larger base will be equal to (x + 4) cm, respectively. From here, using the formula for the middle line of the trapezoid, we get:
It turns out that the smaller base of DC is 5 cm, and the larger one is 9 cm.
Important! The concept of the median line is the key to solving many problems in geometry. Based on its definition, many proofs for other figures are built. Using the concept in practice, perhaps more rational solution and search for the required value.
Determination of height, and how to find it
As noted earlier, the height is a segment that intersects the bases at an angle of 2Pi / 4 and is the shortest distance between them. Before finding the height of the trapezoid, it is necessary to determine what input values are given. For a better understanding, consider the problem. Find the height of the trapezoid, provided that the bases are 8 and 28 cm, the sides are 12 and 16 cm, respectively.
Figure 5. Solving the problem of finding the height of a trapezoid
Let's draw segments DF and CH at right angles to the base AD. According to the definition, each of them will be the height of a given trapezoid (Fig. 5). In this case, knowing the length of each sidewall, using the Pythagorean theorem, we find what the height in triangles AFD and BHC is.
The sum of the segments AF and HB is equal to the difference of the bases, i.e.:
Let the length of AF be equal to x cm, then the length of the segment HB = (20 - x) cm. As it was established, DF=CH , hence .
Then we get the following equation:
It turns out that the segment AF in the triangle AFD is 7.2 cm, from here we calculate the height of the trapezoid DF using the same Pythagorean theorem:
Those. the height of the ADCB trapezoid will be 9.6 cm. As you can see, the height calculation is a more mechanical process, and is based on the calculations of the sides and angles of triangles. But, in a number of problems in geometry, only degrees of angles can be known, in which case the calculations will be made through the ratio of the sides of the inner triangles.
Important! In essence, a trapezoid is often thought of as two triangles, or as a combination of a rectangle and a triangle. To solve 90% of all problems found in school textbooks, the properties and characteristics of these figures. Most of the formulas for this GMT are derived relying on the "mechanisms" for these two types of figures.
How to quickly calculate the length of the base
Before you find the base of the trapezoid, you need to determine what parameters are already given, and how to use them rationally. A practical approach is to extract the length of the unknown base from the midline formula. For a clearer perception of the picture, we will show how this can be done using an example of a task. Let it be known that the middle line of the trapezoid is 7 cm, and one of the bases is 10 cm. Find the length of the second base.
Solution: Knowing that the middle line is equal to half the sum of the bases, it can be argued that their sum is 14 cm.
(14cm=7cm×2). From the condition of the problem, we know that one of is equal to 10 cm, hence the smaller side of the trapezoid will be equal to 4 cm (4 cm = 14 - 10).
Moreover, for a more comfortable solution of problems of this kind, we recommend that you learn well such formulas from the trapezoid area as:
- middle line;
- area;
- height;
- diagonals.
Knowing the essence (precisely the essence) of these calculations, you can easily find out the desired value.
Video: trapezoid and its properties
Video: trapezoid features
Output
From the considered examples of problems, we can draw a simple conclusion that the trapezoid, in terms of calculating problems, is one of the simplest figures in geometry. For successful solution tasks, first of all, it is not worth deciding what information is known about the object being described, in what formulas they can be applied, and deciding what needs to be found. By executing this simple algorithm, no task using this geometric figure will be effortless.
