A trapezoid is a special case of a quadrangle in which one pair of sides is parallel. The term "trapezoid" came from the Greek word τράπεζα, meaning "table", "table". In this article we will consider the types of trapezoid and its properties. In addition, we will figure out how to calculate the individual elements of this geometric figure. For example, the diagonal of an isosceles trapezoid, the middle line, 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 look at what a quadrangle is. This figure is a special case of a polygon containing four sides and four vertices. Two vertices of a quadrangle that are not adjacent are called opposite. The same can be said of two non-adjacent sides. The main types of quadrangles are a parallelogram, a rectangle, a rhombus, a square, a trapezoid and a deltoid.
So, back to the trapezoid. As we have already said, in this figure the two sides are parallel. They are called bases. The other two (non-parallel) are the sides. In the materials of exams and various examinations it is very often possible to meet problems associated with trapezes, the solution of which often requires the student to have knowledge not provided for in 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, in addition to this, the mentioned geometric figure has other features. But about them a little 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. She has two angles always equal to ninety degrees.
2. An isosceles trapezoid is a geometric figure in which the sides are equal. This means that the angles at the bases are also pairwise equal.
The main principles of the study of trapezoid properties
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 system 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 in the learning process. Moreover, each trapezoid property 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 trapezoid. This implies a return to individual features of a given geometric figure in the learning process. This makes it easier for students to remember. For example, the property of four points. It can be proved both in the study of similarity, and subsequently using vectors. And the equidistance of triangles adjacent to the lateral 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 one straight line, but also using the formula S = 1/2 (ab * sinα). In addition, one can work out the sine theorem on an inscribed trapezoid or a right triangle on a described trapezoid, etc.
The use of “extra-curricular” features of the geometric figure in the content of the school course is the task technology for teaching them. The constant appeal to the studied properties while passing through other topics allows students to learn more about 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, in this geometric figure the sides are equal. It is also known as the correct trapezoid. But why is it so remarkable and why did it get such a name? The features of this figure include the fact that it has equal not only the sides and angles at the bases, but also the diagonals. In addition, the sum of the angles of an isosceles trapezoid is 360 degrees. But that's not all! Of all the known trapezoids, only around the isosceles one can describe the circle. 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 we describe the circle around the quadrangle. The next property of the considered geometric figure is that the distance from the top of the base to the projection of the opposite top on the 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.
Decision
Typically, a quadrangle 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 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 N from angle B. As a result, we obtain a right-angled triangle ABN, where AB is the hypotenuse, and BN and AN are legs. We calculate the size of the leg AN: subtract the smaller from the larger base, and divide the result by 2. We write in the form of the formula: (ZY) / 2 = F. Now we use the cos function to calculate the acute angle of the triangle. We get the following record: cos (β) = X / F. Now we calculate the angle: β = arcos (X / F). Further, knowing one angle, we can determine the second, for this we produce an elementary arithmetic action: 180 - β. All angles are defined.
There is a second solution to this problem. At the beginning we lower N. from the angle B to the height. We calculate the value of the leg BN. We know that the square of the hypotenuse of a right triangle is the sum of the squares of the legs. We get: BN = √ (X2-F2). Next, we use the trigonometric function tg. As a result, we have: β = arctan (BN / F). An acute angle is found. Next, we determine the obtuse angle similarly to the first method.
Property of isosceles trapezoid diagonals
First we write down four rules. 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 middle line are equal;
- the area of the trapezoid will be equal to the square of the height (midline, half sum of the bases);
- the square of the diagonal is equal to half the square of the sum of the bases or twice the square of the midline (height).
Now consider the formulas that determine the diagonal of an isosceles trapezoid. This information block can be divided into four parts:
1. The formula for the length of the diagonal through its sides.
We accept that A is the lower base, B is the upper, C are the equal sides, D is the diagonal. In this case, the length can be determined as follows:
D = √ (C2 + A * B).
2. Diagonal length formulas by the cosine theorem.
We assume that A is the lower base, B is the upper, C are the equal sides, D is the diagonal, α (at the lower base) and β (at the upper base) are the angles of the trapezoid. We get the following formulas with which you can calculate the length of the diagonal:
- D = √ (A2 + C2-2A * C * cosα);
- D = √ (A2 + C2-2A * C * cosβ);
- D = √ (B2 + C2-2B * C * cosβ);
- D = √ (B2 + C2-2B * C * cosα).
3. Formulas for the length of the diagonals of an isosceles trapezoid.
We assume that A is the lower base, B is the upper base, D is the diagonal, M is the midline, H is the height, P is the area of the trapezoid, α and β are the angles between the diagonals. We determine the length by the following formulas:
- D = √ (M2 + H2);
- D = √ (H2 + (A + B) 2/4);
- D = √ (H (A + B) / sinα) = √ (2P / sinα) = √ (2M * H / sinα).
For this case, the equality holds: sinα = sinβ.
4. Diagonal length formulas through sides and height.
We accept that A is the lower base, B is the upper, C is the sides, D is the diagonal, H is the height, α is the angle at the lower base.
We determine the length by the following formulas:
- D = √ (H2 + (AP * ctgα) 2);
- D = √ (H2 + (B + P * ctgα) 2);
- D = √ (A2 + C2-2A * √ (C2-H2)).
Elements and properties of a rectangular trapezoid
Let's look at how interesting this geometric figure is. As we already said, a rectangular trapezoid has two right angles.
In addition to the classical definition, there are others. For example, a rectangular trapezoid is a trapezoid with one side perpendicular to the bases. Or a figure having right angles on the side. In this type of trapezoid, the height is equal to the lateral side, which is perpendicular to the bases. The midline is the line connecting the midpoints of the two sides. The property of the mentioned element is that it is parallel to the bases and equal to half their sum.
Now let's look at the basic formulas that define this geometric shape. For this, we accept that A and B are grounds; C (perpendicular to the bases) and D are the sides of a rectangular trapezoid, M is the midline, α is an acute angle, P is the area.
1. The lateral side perpendicular to the bases is equal to the height of the figure (C = H), and is equal to the product of the length of the second lateral side D and the sine of the angle α with a larger base (C = D * sinα). In addition, it is equal to the product of the tangent of the acute angle α and the difference of the bases: C = (AB) * tgα.
2. The lateral side D (not perpendicular to the bases) is equal to the partial difference between A and B and the cosine (α) of the acute angle or the partial height of the figure H and the sine of the acute angle: D = (A-B) / cos α = C / sinα.
3. The lateral side, which is perpendicular to the bases, is equal to the square root of the difference of the square D - the second side - and the square of the difference of the bases:
C = √ (D2- (A-B) 2).
4. Side D of a rectangular trapezoid is equal to the square root of the sum of the square of side C and the square of the difference of the bases of the geometric figure: D = √ (C2 + (A-B) 2).
5. The lateral side C is equal to the quotient of dividing the double area by the sum of its bases: C = P / M = 2P / (A + B).
6. The area is determined by the product of M (the middle line of a rectangular trapezoid) by a height or a side perpendicular to the bases: P = M * H = M * C.
7. Side C is equal to the quotient of dividing the doubled area of the figure by the product of the sine of the acute angle and the sum of its bases: C = P / M * sinα = 2P / ((A + B) * sinα).
8. Formulas of the lateral side of a rectangular trapezoid through its diagonals and the angle between them:
- sinα = sinβ;
- C = (D1 * D2 / (A + B)) * sinα = (D1 * D2 / (A + B)) * sinβ,
where D1 and D2 are the diagonals of the trapezoid; α and β are the angles between them.
9. Formulas of the side through the angle at the lower base and other sides: D = (AB) / cosα = C / sinα = H / sinα.
Since a trapezoid with a right angle is a special case of a trapezoid, the rest of the formulas defining these figures will correspond to a rectangular one.
Inscribed circle properties
If the condition says that a circle is inscribed in a rectangular trapezoid, then the following properties can be used:
- the sum of the bases is equal to the sum of the sides;
- the distances from the top of the rectangular figure to the points of tangency of the inscribed circle are always equal;
- the height of the trapezoid is equal to the side, perpendicular to the bases, and equal to the diameter of the circle ;
- the center of the circle is the point at which the bisectors of the angles intersect;
- if the side is divided by the point of tangency into the segments H and M, then the radius of the circle is equal to the square root of the product of these segments;
- the quadrangle formed by the points of tangency, the top of the trapezoid and the center of the inscribed circle is a square whose side is equal to the radius;
- the area of the figure is equal to the product of the bases and the product of half the sum of the bases to its height.
Similar trapezoid
This topic is very convenient for studying the properties of this geometric figure. For example, the diagonals divide the trapezoid into four triangles, and adjacent to the bases are similar, and to the sides - is equal. This statement can be called a property of triangles into which the trapezoid is divided by its diagonals. The first part of this statement is proved through the sign of similarity in two angles. To prove the second part, it is better to use the method below.
Proof of the theorem
We accept that the ABSD figure (HELL and BS - the basis of the trapezoid) is divided by the diagonals of the VD and AS. The point of their intersection is O. We get four triangles: AOS - at the lower base, BOS - at the upper base, ABO and SOD at the sides. The triangles SOD and BOS have a common height if the segments BO and OD are their bases. We get that the difference in their areas (P) is equal to the difference of these segments: CBE / PSOD = BO / OD = K. Therefore, PSOD = CBE / K. Similarly, the BFB and AOB triangles have a common height. We take for their bases segments of CO and OA. We get CBE / PAOB = CO / OA = K and PAOB = CBE / K. It follows 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, having solved the following problem. It is known that the area of the triangles BOS and AOD is equal, it is necessary to find the area of the trapezoid. Since PSOD = PAOB, it means that PABSD = CBE + PAOD + 2 * PSOD. From the similarity of the triangles BOS and AOD, it follows that BO / OD = √ (TBC / PAOD). Consequently, CBE / PSOD = BO / OD = √ (CBE / PAOD). We get PSOD = √ (CBOS * PAOD). Then PABSD = TBC + PAOD + 2 * √ (TBC * PAOD) = (√ TBC + √ PAOD) 2.
Similarity Properties
Continuing to develop this topic, one can prove other interesting features of the trapezoid. So, using similarity, we 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 will solve the following problem: it is necessary to find the length of the segment of the RK that passes through the point O. From the similarity of the triangles AOD and BOS it follows that AO / OS = AD / BS. From the similarity of the triangles AOR and ASB, it follows that AO / AC = PO / BS = AD / (BS + AD). From this we obtain that PO = BS * AD / (BS + AD). Similarly, from the similarity of triangles DOK and DBS, it follows that OK = BS * HELL / (BS + HELL). From this we obtain that PO = OK and PK = 2 * BS * HELL / (BS + HELL). A line passing through the diagonal intersection point parallel to the bases and connecting the two sides is divided in half by the intersection point. Its length is the harmonic mean of the base of the figure.
Consider the following quality of the trapezoid, which is called the property of four points. The intersection points of the diagonals (O), the intersection of the extension of the sides (E), as well as the middle of the bases (T and G) always lie on the same line. This is easily proved by the similarity method. The resulting triangles BEC and AED are similar, and in each of them the medians ET and EJ divide the angle at the vertex E into equal parts. Consequently, the points E, T and M lie on the same line. In the same way, the points T, O, and G. are located on one 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), which divides the figure into two similar ones. This segment should be parallel to the bases. Since the resulting trapezoid ALPD and LBSF are similar, BS / LF = LF / AD. It follows that LF = √ (BS * HELL). We get that the segment dividing the trapezoid into two similar ones has a length equal to the geometric mean of the lengths of the base of the figure.
Consider the following similarity property. It is based on a segment that divides the trapezoid into two equal figures. We accept that the trapezoid ABSD is divided by the EN segment 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 = (BS + ) * 1 / 2 = (HELL + ) * 2 / 2 and PABSD = (BS + AD) * (1 + 2) / 2. Next, we compose a system whose first equation (BS + EH) * B1 = (HELL + EH) * B2 and the second (BS + EH) * B1 = (BS + HELL) * (B1 + B2) / 2. It follows that B2 / B1 = (BS + EH) / (AD + EH) and BS + EH = ((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 root mean square of the base lengths: √ ((BS2 + AD2) / 2).
Similarity conclusions
Thus, we have proved that:
1. The segment connecting the middle of the sides at the trapezoid is parallel to HELL and BS and equal to the arithmetic mean of BS and HELL (trapezoid base length).
2. The line passing through the point O of the intersection of the diagonals parallel to HELL and BS will be equal to the harmonic mean of the numbers HELL and BS (2 * BS * HELL / (BS + HELL)).
3. The segment dividing the trapezoid into similar ones has the length of the geometric mean of the BS and HELL bases.
4. The element dividing the figure into two equal, 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 middle line and the segment that passes through point O - the intersection of the diagonals of the figure - parallel to the bases. But where will the third and fourth be? This answer will lead the student to discover the desired relationship between the mean values.
The line connecting the midpoints of the diagonals of the trapezoid
Consider the following property of this figure. We accept that the segment MN is parallel to the bases and divides the diagonals in half. We call the intersection points W and U. This segment will be equal to the half-difference of the bases. We will analyze this in more detail. MS - the middle line of the triangle ABS, it is equal to BS / 2. MSC - the middle line of the triangle ABD, it is equal to AD / 2. Then we get that = -, therefore, = / 2-BS / 2 = (AD + BC) / 2.
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