How to describe a circle around an isosceles trapezoid. Remember and apply the properties of a trapezoid
If a circle is inscribed in a trapezoid, the problem has several paths along which reasoning can be carried out.
1. A circle can be inscribed in a quadrilateral if and only if the sums of the lengths of its opposite sides are equal. It follows that If a circle is inscribed in a trapezoid, then the sum of its bases is equal to the sum of the sides.
AB+CD=AD+BC
2. Tangent segments drawn from one point are equal. It follows that
3. The height of a trapezoid is equal to the length of the diameter of the inscribed circle or two of its radii.
MK is the height of the trapezoid, MK=2r, where r is the radius of the circle inscribed in the trapezoid.
4. The center of the incircle is the intersection point of the bisectors of the angles of the trapezoid.
Let's look at a basic problem.
Find the radius of a circle inscribed in a trapezoid if the point of contact divides the side into segments of length m and n (CF=m, FD=n).
1) ∠ADC+∠BCD=180º (as the sum of internal one-sided angles for parallel lines AD and BC and secant CD);
2) since point O is the point of intersection of the bisectors of the corners of the trapezoid, then ∠ODF+∠OCF=1/2∙(∠ADC+∠BCD)=90º;
3) since the sum of the angles of a triangle is 180º, then in the triangle COD ∠COD=90º;
4) thus, the triangle COD is right-angled, and OF is the altitude drawn to the hypotenuse, CF and FD are the projections of the legs OC and OD to the hypotenuse. Since the height drawn to the hypotenuse is between the projections of the legs onto the hypotenuse,
Hence, the radius of a circle inscribed in a trapezoid is expressed in terms of the lengths of the segments, as the lateral side is divided by the point of contact, as
And since the height of a trapezoid is equal to its diameter, the height of the trapezoid can be expressed in terms of the lengths of these segments.
A trapezoid is a special case of a quadrilateral in which one pair of sides is parallel. The term "trapezoid" comes from the Greek word τράπεζα, meaning "table", "table". In this article we will look at the types of trapezoid and its properties. In addition, we will figure out how to calculate individual elements of this For example, the diagonal of an isosceles trapezoid, the center line, area, etc. The material is presented in the style of elementary popular geometry, i.e. in an easily accessible form.
General information
First, let's figure out 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 for two non-adjacent sides. The main types of quadrilaterals are parallelogram, rectangle, rhombus, square, trapezoid and deltoid.
So let's get back to trapezoids. As we have already said, this figure has two parallel sides. They are called bases. The other two (non-parallel) are the lateral sides. In the materials of exams and various tests, you can often find problems related to trapezoids, 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 more 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. Her two angles are always equal to 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 equal in pairs.
The main principles of the methodology for studying the properties of a trapezoid
The main principle includes 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 (preferably system ones). At the same time, it is very important that the teacher knows what tasks need to be assigned to students at one time or another during the educational process. Moreover, each property of a 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 trapezoid. This implies a return in the learning process to individual features of a given geometric figure. This makes it easier for students to remember them. For example, the property of four points. It can be proven both when studying similarity and subsequently using vectors. And the equivalence of triangles adjacent to the lateral sides of a figure can be proven 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 on an inscribed trapezoid or a right triangle on an inscribed trapezoid, etc.
The use of “extracurricular” features of a geometric figure in the content of a school course is a task-based technology for teaching them. Constantly referring to the properties being studied while going through other topics allows students to gain a deeper knowledge of the trapezoid and ensures the success of solving assigned problems. So, let's start studying this wonderful figure.
Elements and properties of an isosceles trapezoid
As we have already noted, this geometric figure has equal sides. It is also known as the correct trapezoid. Why is it so remarkable and why did it get such a name? The peculiarity of this figure is that not only the sides and angles at the bases are equal, 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 an isosceles one can be described as a circle. This is due to the fact that the sum of the opposite angles of this figure is equal to 180 degrees, and only under this condition can one describe a circle around a quadrilateral. The next property of the geometric figure under consideration is that the distance from the vertex of the base 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. Let us consider a solution to this problem, provided that the dimensions of the sides of the figure are known.
Solution
Typically, 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 equal to X, and the sizes of the bases are equal to Y and Z (smaller and larger, respectively). To carry out the calculation, it is necessary to draw the height H from angle B. The result is a right 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 = F. Now, to calculate the acute angle of the triangle, we use the cos function. We get the following entry: 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 perform an elementary arithmetic operation: 180 - β. All angles are defined.
There is a second solution to this problem. First, we lower it from the corner to height H. We calculate the value of the leg BN. We know that the square of the hypotenuse of a right triangle is equal to 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 has been found. Next, we define it similarly to the first method.
Property of diagonals of an isosceles trapezoid
First, let's 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 midline are equal;
The center of the circle is the point at which ;
If the lateral side is divided by the point of tangency into segments H and M, then it is equal to the square root of the product of these segments;
The quadrilateral that is 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 trapezoids
This topic is very convenient for studying the properties of this For example, the diagonals divide a trapezoid into four triangles, and those adjacent to the bases are similar, and those adjacent to the sides are equal in size. This statement can be called a property of the triangles into which the trapezoid is divided by its diagonals. The first part of this statement is proven through the sign of similarity at 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 are the bases of the trapezoid) is divided by diagonals VD and AC. 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. Triangles SOD and BOS have a common height if the segments BO and OD are their bases. We find 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, triangles BOS and AOB have a common height. We take the segments CO and OA as their bases. We get PBOS/PAOB = CO/OA = K and PAOB = PBOS/K. It follows from this that PSOD = PAOB.
To consolidate the material, students are recommended to find the 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 triangles BOS and AOD have equal areas; it is necessary to find the area of the trapezoid. Since PSOD = PAOB, it means PABSD = PBOS+PAOD+2*PSOD. From the similarity of 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.
Properties of similarity
Continuing to develop this topic, we can prove other interesting features of trapezoids. Thus, using similarity, one can prove the property of a segment that passes through the point formed by the intersection of the diagonals of this geometric figure, parallel to the bases. To do this, let's solve the following problem: we need to find the length of the segment RK that passes through 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/AC=RO/BS=AD/(BS+AD). From here we get that RO=BS*BP/(BS+BP). Similarly, from the similarity of triangles DOC and DBS, it follows that OK = BS*AD/(BS+AD). From here we get that RO=OK and RK=2*BS*AD/(BS+AD). A segment passing through the point of intersection of the diagonals, parallel to the bases and connecting two lateral sides, is divided in half by the point of intersection. Its length is the harmonic mean of the figure's bases.
Consider the following property of a trapezoid, which is called the property of four points. The intersection points of the diagonals (O), the intersection of the continuation of the sides (E), as well as the midpoints of the bases (T and F) always lie on the same line. This can be easily proven by the similarity method. The resulting triangles BES and AED are similar, and in each of them the medians ET and EJ divide the vertex angle E into equal parts. Therefore, points E, T and F lie on the same straight line. In the same way, points T, O, and Zh are located on the same straight line. All this follows from the similarity of triangles BOS and AOD. From here we conclude that all four points - E, T, O and F - will lie on the same straight line.
Using similar trapezoids, you can ask students to find the length of the segment (LS) that divides the figure into two similar ones. This segment must be parallel to the bases. Since the resulting trapezoids ALFD and LBSF are similar, then BS/LF = LF/AD. It follows that LF=√(BS*AD). We find that the segment dividing 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 figures. We assume that the trapezoid ABSD is divided by the segment EH into two similar ones. From vertex B a height is omitted, which is divided by segment EN into two parts - B1 and B2. We get: PABSD/2 = (BS+EN)*B1/2 = (AD+EN)*B2/2 and PABSD = (BS+AD)*(B1+B2)/2. Next, we compose a system whose first equation is (BS+EN)*B1 = (AD+EN)*B2 and the second (BS+EN)*B1 = (BS+AD)*(B1+B2)/2. It follows that B2/B1 = (BS+EN)/(AD+EN) and BS+EN = ((BS+AD)/2)*(1+B2/B1). We find that the length of the segment dividing the trapezoid into two equal ones is equal to the root mean square of the lengths of the bases: √((BS2+AD2)/2).
Similarity findings
Thus, we have proven that:
1. The segment connecting the midpoints of the lateral sides of a 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 dividing the trapezoid into similar ones has the length of the geometric mean of the bases BS and AD.
4. An element dividing a figure into two equal ones has the length of the root mean square of the numbers AD and BS.
To consolidate the material and understand the connection between the considered segments, the student needs to construct 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 located? This answer will lead the student to the discovery of the desired relationship between average values.
A segment connecting the midpoints of the diagonals of a trapezoid
Consider the following property of this figure. We assume that the segment MH is parallel to the bases and bisects the diagonals. Let's call the intersection points Ш and Ш. This segment will be equal to half the difference of the bases. Let's look at this in more detail. MS is the middle line of the ABS triangle, it is equal to BS/2. MSH is the middle line of triangle ABD, it is equal to AD/2. Then we get that ShShch = MSh-MSh, therefore, ShShch = 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? You need to add the lower base to the upper base - in any direction, for example, to the right. And we extend the lower one by the length of the upper one to the left. Next, we connect them diagonally. The point of intersection of this segment with the midline 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 be inscribed in a circle only 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.
Corollaries of the incircle:
1. The height of the described trapezoid is always equal to two radii.
2. The side of the described trapezoid is observed from the center of the circle at a right angle.
The first corollary is obvious, but to prove the second it is necessary to establish that the angle SOD is right, which, in fact, is also not difficult. But knowledge of this property will allow you to use a right triangle when solving problems.
Now let us specify these consequences for an isosceles trapezoid inscribed in a circle. We find that the height is the geometric mean of the bases of the figure: H=2R=√(BS*AD). While practicing the basic technique for solving problems for trapezoids (the principle of drawing two heights), the student must solve the following task. We assume that BT is the height of the isosceles figure ABSD. It is necessary to find the 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 vertex B to the base AD. Since the circle is inscribed in a trapezoid, then BS+AD = 2AB or AB = (BS+AD)/2. From triangle ABN we find sinα = BN/AB = 2*BN/(BS+AD). PABSD = (BS+BP)*BN/2, BN=2R. We get PABSD = (BS+BP)*R, it follows that R = PABSD/(BS+BP).
All formulas for 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 = (A+B)/2.
2. Through height, base and corners:
M = A-H*(ctgα+ctgβ)/2;
M = B+N*(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/2N = D1*D2*sinβ/2N.
4. Through area and height: M = P/N.
How to find the circumradius of a trapezoid?
Depending on the conditions, this can be done in different ways. There is no ready-made formula for the radius of a circle circumscribed about a trapezoid.
I. The radius of a circumscribed circle around a trapezoid as the radius of a circumscribed circle around a triangle whose vertices are the vertices of the trapezoid
The circumcircle of a trapezoid passes through all its vertices, therefore, it is circumscribed for any of the triangles whose vertices are the vertices of the trapezoid.
In general, it can be found using one of the formulas
where a is the side of the triangle, α is the angle opposite it;
or by formula
where a, b, c are the sides, S is the area of the triangle.
For a trapezoid ABCD, the radius can be found, for example, as the radius of a circle circumscribed about triangle ABD:
where the sine of angle A can be found from right triangle ABF:
III. The radius of a circle circumscribed about a trapezoid as the distance to the point of intersection of the bisector perpendiculars
The radius of the circumcircle is the point of intersection of the perpendicular bisectors with the sides of the trapezoid. (You can reason differently: in the isosceles triangle AOD (AO=OD=R), the height ON is also the median. For the triangle BOC, the same is true.)
If the height of the trapezoid is known KN=h, the bases AD=a, BC=b can be designated ON=x.
If the center of the circle lies inside the trapezoid, OK=h-x, from the right triangles ANO and BKO we can express
and equate the right sides
By solving these equations for x, you can find R.
IV. If the diagonal of a trapezoid is perpendicular to the side, the center of the circumscribed circle lies at the middle of the larger base and the radius is half of the larger base.
Project work “Interesting properties of a trapezoid” Completed by: 10th grade students Kudzaeva Ellina Bazzaeva Diana MCOU Secondary School s. N.Batako Head: Gagieva A.O. November 20, 2015
Purpose of the work: To consider the properties of the trapezoid, which are not studied in the school geometry course, but when solving geometric problems of the Unified State Exam from the expanded part C 4, it may be necessary to know and be able to apply precisely these properties.
Properties of a trapezoid: If a trapezoid is divided by a line parallel to its bases equal to a and b, into two equal trapezoids. Then the segment to this line, enclosed between the lateral sides, is equal to a B to
Property of a segment passing through the point of intersection of the diagonals of a trapezoid. The segment parallel to the bases passing through the point of intersection of the diagonals is equal to: a in c
Properties of a trapezoid: A straight line segment parallel to the bases of a trapezoid, enclosed inside the trapezoid, is divided into three parts by its diagonals. Then the segments adjacent to the sides are equal to each other. MP=OK R M O K
Properties of an isosceles trapezoid: If a circle can be inscribed in a trapezoid, then the radius of the circle is the average proportional to the segments into which the tangent point divides the side. O S V A D. E O
Properties of an isosceles trapezoid: If the center of the circumscribed circle lies at the base of the trapezoid, then its diagonal is perpendicular to the side O A B C D
Properties of an isosceles trapezoid: A circle can be inscribed in an isosceles trapezoid if the side side is equal to its midline. S V A D h
1) If the problem statement says that a circle is inscribed in a rectangular trapezoid, you can use the following properties: 1. The sum of the bases of the trapezoid is equal to the sum of the sides. 2. The distances from the vertex of the trapezoid to the tangent points of the inscribed circle are equal. 3. The height of a rectangular trapezoid is equal to its smaller side and equal to the diameter of the inscribed circle. 4. The center of the inscribed circle is the point of intersection of the bisectors of the angles of the trapezoid. 5. If the tangent point divides the side into segments m and n, then the radius of the inscribed circle is equal to
Properties of a rectangular trapezoid into which a circle is inscribed: 1) A quadrilateral formed by the center of the inscribed circle, the points of contact and the vertex of the trapezoid - a square whose side is equal to the radius. (AMOE and BKOM are squares with side r). 2) If a circle is inscribed in a rectangular trapezoid, then the area of the trapezoid is equal to the product of its bases: S=AD*BC
Proof: The area of a trapezoid is equal to the product of half the sum of its bases and its height: Let us denote CF=m, FD=n. Since the distances from the vertices to the tangent points are equal, the height of the trapezoid is equal to two radii of the inscribed circle, and
I. The bisectors of the angles at the lateral side of the trapezoid intersect at an angle of 90º. 1)∠ABC+∠BAD=180º (as internal one-sided with AD∥BC and secant AB). 2) ∠ABK+∠KAB=(∠ABC+∠BAD):2=90º (since the bisectors bisect the angles). 3) Since the sum of the angles of a triangle is 180º, in triangle ABK we have: ∠ABK+∠KAB+∠AKB=180º, hence ∠AKB=180-90=90º. Conclusion: The angle bisectors on the lateral side of a trapezoid intersect at right angles. This statement is used when solving problems on a trapezoid into which a circle is inscribed.
I I. The point of intersection of the bisectors of the trapezoid adjacent to the lateral side lies on the midline of the trapezoid. Let the bisector of angle ABC intersect side AD at point S. Then triangle ABS is isosceles with base BS. This means that its bisector AK is also a median, that is, point K is the midpoint of BS. If M and N are the midpoints of the lateral sides of the trapezoid, then MN is the midline of the trapezoid and MN∥AD. Since M and K are the midpoints of AB and BS, then MK is the midline of the triangle ABS and MK∥AS. Since only one line parallel to this one can be drawn through point M, point K lies on the midline of the trapezoid.
III. The point of intersection of the bisectors of acute angles at the base of a trapezoid belongs to another base. In this case, triangles ABK and DCK are isosceles with bases AK and DK, respectively. Thus, BC=BK+KC=AB+CD. Conclusion: If the bisectors of the acute angles of a trapezoid intersect at a point belonging to the smaller base, then the smaller base is equal to the sum of the lateral sides of the trapezoid. An isosceles trapezoid in this case has a smaller base twice the size of its side.
I V. The point of intersection of the bisectors of obtuse angles at the base of the trapezoid belongs to another base. In this case, triangles ABF and DCF are isosceles with bases BF and CF, respectively. Hence AD=AF+FD=AB+CD. Conclusion: If the bisectors of the obtuse angles of a trapezoid intersect at a point belonging to the larger base, then the larger base is equal to the sum of the lateral sides of the trapezoid. In this case, an isosceles trapezoid has a larger base that is twice as large as its side.
If an isosceles trapezoid with sides a, b, c, d can be inscribed and circles can be drawn around it, then the area of the trapezoid is
A trapezoid is a geometric figure with four angles. When constructing a trapezoid, it is important to take into account that two opposite sides are parallel, and the other two, on the contrary, are not parallel relative to each other. This word came to modern times from Ancient Greece and sounded like “trapedzion”, which meant “table”, “dining table”.
This article talks about the properties of a trapezoid circumscribed about a circle. We will also look at the types and elements of this figure.
Elements, types and characteristics of the geometric figure trapezoid
The parallel sides in this figure are called the bases, and those that are not parallel are called the sides. Provided that the sides are the same length, the trapezoid is considered isosceles. A trapezoid whose sides lie perpendicular to the base at an angle of 90° is called rectangular.
This seemingly simple figure has a considerable number of properties inherent in it, emphasizing its characteristics:
- If you draw a middle line along the sides, it will be parallel to the bases. This segment will be equal to 1/2 the difference of the bases.
- When constructing a bisector from any corner of a trapezoid, an equilateral triangle is formed.
- From the properties of a trapezoid described around a circle, it is known that the sum of the parallel sides must be equal to the sum of the bases.
- When constructing diagonal segments, where one of the sides is the base of a trapezoid, the resulting triangles will be similar.
- When constructing diagonal segments, where one of the sides is lateral, the resulting triangles will have equal area.
- If we continue the side lines and construct a segment from the center of the base, then the formed angle will be equal to 90°. The segment connecting the bases will be equal to 1/2 of their difference.
Properties of a trapezoid circumscribed about a circle
It is possible to enclose a circle in a trapezoid only under one condition. This condition is that the sum of the sides must be equal to the sum of the bases. For example, when constructing a trapezoid AFDM, AF + DM = FD + AM is applicable. Only in this case can a circle be enclosed in a trapezoid.
So, more about the properties of a trapezoid described around a circle:
- If a circle is enclosed in a trapezoid, then in order to find the length of its line that intersects the figure in half, it is necessary to find 1/2 of the sum of the lengths of the sides.
- When constructing a trapezoid circumscribed about a circle, the formed hypotenuse is identical to the radius of the circle, and the height of the trapezoid is also the diameter of the circle.
- Another property of an isosceles trapezoid circumscribed about a circle is that its side is immediately visible from the center of the circle at an angle of 90°.
A little more about the properties of a trapezoid enclosed in a circle
Only an isosceles trapezoid can be inscribed in a circle. This means that it is necessary to meet the conditions under which the constructed AFDM trapezoid will meet the following requirements: AF + DM = FD + MA.
Ptolemy's theorem states that in a trapezoid enclosed in a circle, the product of the diagonals is identical and equal to the sum of the opposite sides multiplied. This means that when constructing a circle circumscribed about the trapezoid AFDM, the following applies: AD × FM = AF × DM + FD × AM.
Quite often in school exams there are problems that require solving problems with a trapezoid. A large number of theorems need to be memorized, but if you can’t learn them right away, it doesn’t matter. It is best to periodically resort to hints in textbooks, so that this knowledge will fit into your head by itself, without much difficulty.
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