- (Greek trapezion). 1) in the geometry of a quadrilateral, in which two sides are parallel, but two are not. 2) a figure adapted for gymnastic exercises. Dictionary of foreign words included in the Russian language. Chudinov A.N., 1910. TRAPEZIA ... ... Dictionary of foreign words of the Russian language
Trapeze- Trapeze. TRAPEZIA (from the Greek trapezion, literally a table), a convex quadrilateral in which two sides are parallel (the bases of a trapezoid). The area of a trapezoid is equal to the product of half the sum of the bases (midline) and the height. … Illustrated Encyclopedic Dictionary
Quadrilateral, projectile, crossbar Dictionary of Russian synonyms. trapezium n., number of synonyms: 3 crossbar (21) ... Synonym dictionary
- (from the Greek trapezion, literally a table), a convex quadrilateral in which two sides are parallel (the bases of a trapezoid). The area of a trapezoid is equal to the product of half the sum of the bases (midline) and the height ... Modern Encyclopedia
- (from the Greek trapezion letters. table), a quadrangle in which two opposite sides, called the bases of the trapezoid, are parallel (AD and BC in the figure), and the other two are not parallel. The distance between the bases is called the height of the trapezoid (at ... ... Big Encyclopedic Dictionary
TRAPEZIA A quadrangular flat figure in which two opposite sides are parallel. The area of a trapezoid is half the sum of the parallel sides multiplied by the length of the perpendicular between them... Scientific and technical encyclopedic dictionary
TRAPEZIA, trapezoid, female. (from the Greek trapeza table). 1. Quadrilateral with two parallel and two non-parallel sides (mat.). 2. A gymnastic apparatus consisting of a crossbar suspended on two ropes (sport.). Acrobatic… … Explanatory Dictionary of Ushakov
TRAPEZIA, and, wives. 1. A quadrilateral with two parallel and two non-parallel sides. Bases of a trapezoid (its parallel sides). 2. A circus or gymnastic projectile, a crossbar suspended on two cables. Explanatory dictionary of Ozhegov. FROM … Explanatory dictionary of Ozhegov
Female, geom. a quadrilateral with unequal sides, of which two are postenic (parallel). A trapezoid is a similar quadrilateral in which all sides are apart. Trapezohedron, a body cut by trapezoids. Dahl's Explanatory Dictionary. IN AND. Dal. 1863 1866 ... Dahl's Explanatory Dictionary
- (Trapeze), USA, 1956, 105 min. Melodrama. Aspiring acrobat Tino Orsini enters the circus troupe, where Mike Ribble, a famous trapeze artist in the past, works. Once Mike performed with Tino's father. Young Orsini wants Mike... ... Cinema Encyclopedia
A quadrilateral with two sides parallel and two other sides not parallel. Distance between parallel sides. height T. If the parallel sides and height contain a, b and h meters, then the area T. contains square meters ... Encyclopedia of Brockhaus and Efron
\[(\Large(\text(Arbitrary trapezoid)))\]
Definitions
A trapezoid is a convex quadrilateral in which two sides are parallel and the other two sides are not parallel.
The parallel sides of a trapezoid are called its bases, and the other two sides are called its sides.
The height of a trapezoid is the perpendicular dropped from any point of one base to another base.
Theorems: properties of a trapezoid
1) The sum of the angles at the side is \(180^\circ\) .
2) The diagonals divide the trapezoid into four triangles, two of which are similar and the other two are equal.
Proof
1) Because \(AD\parallel BC\) , then the angles \(\angle BAD\) and \(\angle ABC\) are one-sided at these lines and the secant \(AB\) , therefore, \(\angle BAD +\angle ABC=180^\circ\).
2) Because \(AD\parallel BC\) and \(BD\) is a secant, then \(\angle DBC=\angle BDA\) as lying across.
Also \(\angle BOC=\angle AOD\) as vertical.
Therefore, in two corners \(\triangle BOC \sim \triangle AOD\).
Let's prove that \(S_(\triangle AOB)=S_(\triangle COD)\). Let \(h\) be the height of the trapezoid. Then \(S_(\triangle ABD)=\frac12\cdot h\cdot AD=S_(\triangle ACD)\). Then: \
Definition
The midline of a trapezoid is a segment that connects the midpoints of the sides.
Theorem
The median line of the trapezoid is parallel to the bases and equal to half their sum.
Proof*
1) Let's prove the parallelism.
Draw a line \(MN"\parallel AD\) (\(N"\in CD\) ) through the point \(M\) ). Then, by the Thales theorem (because \(MN"\parallel AD\parallel BC, AM=MB\)) the point \(N"\) is the midpoint of the segment \(CD\)... Hence, the points \(N\) and \(N"\) will coincide.
2) Let's prove the formula.
Let's draw \(BB"\perp AD, CC"\perp AD\) . Let \(BB"\cap MN=M", CC"\cap MN=N"\).
Then, by the Thales theorem, \(M"\) and \(N"\) are the midpoints of the segments \(BB"\) and \(CC"\), respectively. So \(MM"\) is the middle line \(\triangle ABB"\) , \(NN"\) is the middle line \(\triangle DCC"\) . That's why: \
Because \(MN\parallel AD\parallel BC\) and \(BB", CC"\perp AD\) , then \(B"M"N"C"\) and \(BM"N"C\) are rectangles. By the Thales theorem, \(MN\parallel AD\) and \(AM=MB\) imply that \(B"M"=M"B\) . Hence, \(B"M"N"C"\) and \(BM"N"C\) are equal rectangles, hence \(M"N"=B"C"=BC\) .
In this way:
\ \[=\dfrac12 \left(AB"+B"C"+BC+C"D\right)=\dfrac12\left(AD+BC\right)\]
Theorem: property of an arbitrary trapezoid
The midpoints of the bases, the point of intersection of the diagonals of the trapezoid and the point of intersection of the extensions of the lateral sides lie on the same straight line.
Proof*
It is recommended that you familiarize yourself with the proof after studying the topic “Similar Triangles”.
1) Let us prove that the points \(P\) , \(N\) and \(M\) lie on the same straight line.
Draw a line \(PN\) (\(P\) is the point of intersection of the extensions of the sides, \(N\) is the midpoint of \(BC\) ). Let it intersect the side \(AD\) at the point \(M\) . Let us prove that \(M\) is the midpoint of \(AD\) .
Consider \(\triangle BPN\) and \(\triangle APM\) . They are similar in two angles (\(\angle APM\) - common, \(\angle PAM=\angle PBN\) as corresponding at \(AD\parallel BC\) and \(AB\) secant). Means: \[\dfrac(BN)(AM)=\dfrac(PN)(PM)\]
Consider \(\triangle CPN\) and \(\triangle DPM\) . They are similar in two angles (\(\angle DPM\) - common, \(\angle PDM=\angle PCN\) as corresponding at \(AD\parallel BC\) and \(CD\) secant). Means: \[\dfrac(CN)(DM)=\dfrac(PN)(PM)\]
From here \(\dfrac(BN)(AM)=\dfrac(CN)(DM)\). But \(BN=NC\) , hence \(AM=DM\) .
2) Let us prove that the points \(N, O, M\) lie on one straight line.
Let \(N\) be the midpoint of \(BC\) , \(O\) be the intersection point of the diagonals. Draw a line \(NO\) , it will intersect the side \(AD\) at the point \(M\) . Let us prove that \(M\) is the midpoint of \(AD\) .
\(\triangle BNO\sim \triangle DMO\) at two angles (\(\angle OBN=\angle ODM\) as lying at \(BC\parallel AD\) and \(BD\) secant; \(\angle BON=\angle DOM\) as vertical). Means: \[\dfrac(BN)(MD)=\dfrac(ON)(OM)\]
Similarly \(\triangle CON\sim \triangle AOM\). Means: \[\dfrac(CN)(MA)=\dfrac(ON)(OM)\]
From here \(\dfrac(BN)(MD)=\dfrac(CN)(MA)\). But \(BN=CN\) , hence \(AM=MD\) .
\[(\Large(\text(Isosceles trapezoid)))\]
Definitions
A trapezoid is called rectangular if one of its angles is right.
A trapezoid is called isosceles if its sides are equal.
Theorems: properties of an isosceles trapezoid
1) An isosceles trapezoid has equal base angles.
2) The diagonals of an isosceles trapezoid are equal.
3) The two triangles formed by the diagonals and the base are isosceles.
Proof
1) Consider an isosceles trapezoid \(ABCD\) .
From the vertices \(B\) and \(C\) we drop to the side \(AD\) the perpendiculars \(BM\) and \(CN\), respectively. Since \(BM\perp AD\) and \(CN\perp AD\) , then \(BM\parallel CN\) ; \(AD\parallel BC\) , then \(MBCN\) is a parallelogram, hence \(BM = CN\) .
Consider right triangles \(ABM\) and \(CDN\) . Since they have equal hypotenuses and the leg \(BM\) is equal to the leg \(CN\) , these triangles are congruent, therefore, \(\angle DAB = \angle CDA\) .
2) 
Because \(AB=CD, \angle A=\angle D, AD\)- general, then on the first sign. Therefore, \(AC=BD\) .
3) Because \(\triangle ABD=\triangle ACD\), then \(\angle BDA=\angle CAD\) . Therefore, the triangle \(\triangle AOD\) is isosceles. It can be proved similarly that \(\triangle BOC\) is isosceles.
Theorems: signs of an isosceles trapezoid
1) If the angles at the base of a trapezoid are equal, then it is isosceles.
2) If the diagonals of a trapezoid are equal, then it is isosceles.
Proof
Consider a trapezoid \(ABCD\) such that \(\angle A = \angle D\) .
Let's complete the trapezoid to the triangle \(AED\) as shown in the figure. Since \(\angle 1 = \angle 2\) , then the triangle \(AED\) is isosceles and \(AE = ED\) . The angles \(1\) and \(3\) are equal as corresponding with parallel lines \(AD\) and \(BC\) and the secant \(AB\) . Similarly, the angles \(2\) and \(4\) are equal, but \(\angle 1 = \angle 2\) , then \(\angle 3 = \angle 1 = \angle 2 = \angle 4\), therefore, the triangle \(BEC\) is also isosceles and \(BE = EC\) .
Eventually \(AB = AE - BE = DE - CE = CD\), i.e. \(AB = CD\) , which was to be proved.
2) Let \(AC=BD\) . Because \(\triangle AOD\sim \triangle BOC\), then we denote their similarity coefficient by \(k\) . Then if \(BO=x\) , then \(OD=kx\) . Similar to \(CO=y \Rightarrow AO=ky\) .
Because \(AC=BD\) , then \(x+kx=y+ky \Rightarrow x=y\) . So \(\triangle AOD\) is isosceles and \(\angle OAD=\angle ODA\) .
Thus, according to the first sign \(\triangle ABD=\triangle ACD\) (\(AC=BD, \angle OAD=\angle ODA, AD\)- general). So \(AB=CD\) , so.
In this article, we will try to reflect the properties of the trapezoid as fully as possible. In particular, we will talk about the general signs and properties of a trapezoid, as well as about the properties of an inscribed trapezoid and about a circle inscribed in a trapezoid. We will also touch on the properties of an isosceles and rectangular trapezoid.
An example of solving a problem using the considered properties will help you sort things out in your head and better remember the material.
Trapeze and all-all-all
To begin with, let's briefly recall what a trapezoid is and what other concepts are associated with it.
So, a trapezoid is a quadrilateral figure, two of the sides of which are parallel to each other (these are the bases). And two are not parallel - these are the sides.
In a trapezoid, the height can be omitted - perpendicular to the bases. The middle line and diagonals are drawn. And also from any angle of the trapezoid it is possible to draw a bisector.
About the various properties associated with all these elements and their combinations, we will now talk.
Properties of the diagonals of a trapezoid
To make it clearer, while reading, sketch out the ACME trapezoid on a piece of paper and draw diagonals in it.
- If you find the midpoints of each of the diagonals (let's call these points X and T) and connect them, you get a segment. One of the properties of the diagonals of a trapezoid is that the segment XT lies on the midline. And its length can be obtained by dividing the difference of the bases by two: XT \u003d (a - b) / 2.
- Before us is the same ACME trapezoid. The diagonals intersect at point O. Let's consider the triangles AOE and IOC formed by the segments of the diagonals together with the bases of the trapezoid. These triangles are similar. The similarity coefficient of k triangles is expressed in terms of the ratio of the bases of the trapezoid: k = AE/KM.
The ratio of the areas of triangles AOE and IOC is described by the coefficient k 2 . - All the same trapezoid, the same diagonals intersecting at point O. Only this time we will consider triangles that the diagonal segments formed together with the sides of the trapezoid. The areas of triangles AKO and EMO are equal - their areas are the same.
- Another property of a trapezoid includes the construction of diagonals. So, if we continue the sides of AK and ME in the direction of the smaller base, then sooner or later they will intersect to some point. Next, draw a straight line through the midpoints of the bases of the trapezoid. It intersects the bases at points X and T.
If we now extend the line XT, then it will join together the point of intersection of the diagonals of the trapezoid O, the point at which the extensions of the sides and the midpoints of the bases of X and T intersect. - Through the point of intersection of the diagonals, we draw a segment that will connect the bases of the trapezoid (T lies on the smaller base of KM, X - on the larger AE). The intersection point of the diagonals divides this segment in the following ratio: TO/OH = KM/AE.
- And now through the point of intersection of the diagonals we draw a segment parallel to the bases of the trapezoid (a and b). The intersection point will divide it into two equal parts. You can find the length of a segment using the formula 2ab/(a + b).
Properties of the midline of a trapezoid
Draw the middle line in the trapezium parallel to its bases.
- The length of the midline of a trapezoid can be calculated by adding the lengths of the bases and dividing them in half: m = (a + b)/2.
- If you draw any segment (height, for example) through both bases of the trapezoid, the middle line will divide it into two equal parts.
Property of the bisector of a trapezoid
Pick any angle of the trapezoid and draw a bisector. Take, for example, the angle KAE of our trapezoid ACME. Having completed the construction on your own, you can easily see that the bisector cuts off from the base (or its continuation on a straight line outside the figure itself) a segment of the same length as the side.
Trapezoid angle properties
- Whichever of the two pairs of angles adjacent to the side you choose, the sum of the angles in a pair is always 180 0: α + β = 180 0 and γ + δ = 180 0 .
- Connect the midpoints of the bases of the trapezoid with a segment TX. Now let's look at the angles at the bases of the trapezoid. If the sum of the angles for any of them is 90 0, the length of the TX segment is easy to calculate based on the difference in the lengths of the bases, divided in half: TX \u003d (AE - KM) / 2.
- If parallel lines are drawn through the sides of the angle of a trapezoid, they will divide the sides of the angle into proportional segments.
Properties of an isosceles (isosceles) trapezoid
- In an isosceles trapezoid, the angles at any of the bases are equal.
- Now build a trapezoid again to make it easier to imagine what it is about. Look carefully at the base of AE - the vertex of the opposite base of M is projected to a certain point on the line that contains AE. The distance from vertex A to the projection point of vertex M and the midline of an isosceles trapezoid are equal.
- A few words about the property of the diagonals of an isosceles trapezoid - their lengths are equal. And also the angles of inclination of these diagonals to the base of the trapezoid are the same.
- Only near an isosceles trapezoid can a circle be described, since the sum of the opposite angles of a quadrilateral 180 0 is a prerequisite for this.
- The property of an isosceles trapezoid follows from the previous paragraph - if a circle can be described near a trapezoid, it is isosceles.
- From the features of an isosceles trapezoid, the property of the height of a trapezoid follows: if its diagonals intersect at right angles, then the length of the height is equal to half the sum of the bases: h = (a + b)/2.
- Draw the line TX again through the midpoints of the bases of the trapezoid - in an isosceles trapezoid it is perpendicular to the bases. And at the same time, TX is the axis of symmetry of an isosceles trapezoid.
- This time lower to the larger base (let's call it a) the height from the opposite vertex of the trapezoid. You will get two cuts. The length of one can be found if the lengths of the bases are added and divided in half: (a+b)/2. We get the second one when we subtract the smaller one from the larger base and divide the resulting difference by two: (a – b)/2.
Properties of a trapezoid inscribed in a circle
Since we are already talking about a trapezoid inscribed in a circle, let's dwell on this issue in more detail. In particular, where is the center of the circle in relation to the trapezoid. Here, too, it is recommended not to be too lazy to pick up a pencil and draw what will be discussed below. So you will understand faster, and remember better.
- The location of the center of the circle is determined by the angle of inclination of the diagonal of the trapezoid to its side. For example, a diagonal may emerge from the top of a trapezoid at right angles to the side. In this case, the larger base intersects the center of the circumscribed circle exactly in the middle (R = ½AE).
- The diagonal and the side can also meet at an acute angle - then the center of the circle is inside the trapezoid.
- The center of the circumscribed circle may be outside the trapezoid, beyond its large base, if there is an obtuse angle between the diagonal of the trapezoid and the lateral side.
- The angle formed by the diagonal and the large base of the trapezoid ACME (inscribed angle) is half of the central angle that corresponds to it: MAE = ½MY.
- Briefly about two ways to find the radius of the circumscribed circle. Method one: look carefully at your drawing - what do you see? You will easily notice that the diagonal splits the trapezoid into two triangles. The radius can be found through the ratio of the side of the triangle to the sine of the opposite angle, multiplied by two. For example, R \u003d AE / 2 * sinAME. Similarly, the formula can be written for any of the sides of both triangles.
- Method two: we find the radius of the circumscribed circle through the area of the triangle formed by the diagonal, side and base of the trapezoid: R \u003d AM * ME * AE / 4 * S AME.
Properties of a trapezoid circumscribed about a circle
You can inscribe a circle in a trapezoid if one condition is met. More about it below. And together this combination of figures has a number of interesting properties.
- If a circle is inscribed in a trapezoid, the length of its midline can be easily found by adding the lengths of the sides and dividing the resulting sum in half: m = (c + d)/2.
- For a trapezoid ACME, circumscribed about a circle, the sum of the lengths of the bases is equal to the sum of the lengths of the sides: AK + ME = KM + AE.
- From this property of the bases of a trapezoid, the converse statement follows: a circle can be inscribed in that trapezoid, the sum of the bases of which is equal to the sum of the sides.
- The tangent point of a circle with radius r inscribed in a trapezoid divides the lateral side into two segments, let's call them a and b. The radius of a circle can be calculated using the formula: r = √ab.
- And one more property. In order not to get confused, draw this example yourself. We have the good old ACME trapezoid, circumscribed around a circle. Diagonals are drawn in it, intersecting at the point O. The triangles AOK and EOM formed by the segments of the diagonals and the sides are rectangular.
The heights of these triangles, lowered to the hypotenuses (i.e., the sides of the trapezoid), coincide with the radii of the inscribed circle. And the height of the trapezoid is the same as the diameter of the inscribed circle.
Properties of a rectangular trapezoid
A trapezoid is called rectangular, one of the corners of which is right. And its properties stem from this circumstance.
- A rectangular trapezoid has one of the sides perpendicular to the bases.
- The height and side of the trapezoid adjacent to the right angle are equal. This allows you to calculate the area of a rectangular trapezoid (general formula S = (a + b) * h/2) not only through the height, but also through the side adjacent to the right angle.
- For a rectangular trapezoid, the general properties of the trapezoid diagonals already described above are relevant.
Proofs of some properties of a trapezoid
Equality of angles at the base of an isosceles trapezoid:
- You probably already guessed that here we again need the ACME trapezoid - draw an isosceles trapezoid. Draw a line MT from vertex M parallel to the side of AK (MT || AK).
The resulting quadrilateral AKMT is a parallelogram (AK || MT, KM || AT). Since ME = KA = MT, ∆ MTE is isosceles and MET = MTE.
AK || MT, therefore MTE = KAE, MET = MTE = KAE.
Where AKM = 180 0 - MET = 180 0 - KAE = KME.
Q.E.D.
Now, based on the property of an isosceles trapezoid (equality of diagonals), we prove that trapezium ACME is isosceles:
- To begin with, let's draw a straight line МХ – МХ || KE. We get a parallelogram KMHE (base - MX || KE and KM || EX).
∆AMH is isosceles, since AM = KE = MX, and MAX = MEA.
MX || KE, KEA = MXE, therefore MAE = MXE.
It turned out that the triangles AKE and EMA are equal to each other, because AM \u003d KE and AE is the common side of the two triangles. And also MAE \u003d MXE. We can conclude that AK = ME, and hence it follows that the trapezoid AKME is isosceles.
Task to repeat
The bases of the trapezoid ACME are 9 cm and 21 cm, the side of the KA, equal to 8 cm, forms an angle of 150 0 with a smaller base. You need to find the area of the trapezoid.
Solution: From vertex K we lower the height to the larger base of the trapezoid. And let's start looking at the angles of the trapezoid.
Angles AEM and KAN are one-sided. Which means they add up to 1800. Therefore, KAN = 30 0 (based on the property of the angles of the trapezoid).
Consider now the rectangular ∆ANK (I think this point is obvious to readers without further proof). From it we find the height of the trapezoid KH - in a triangle it is a leg, which lies opposite the angle of 30 0. Therefore, KN \u003d ½AB \u003d 4 cm.
The area of the trapezoid is found by the formula: S AKME \u003d (KM + AE) * KN / 2 \u003d (9 + 21) * 4/2 \u003d 60 cm 2.
Afterword
If you carefully and thoughtfully studied this article, were not too lazy to draw trapezoids for all the above properties with a pencil in your hands and analyze them in practice, you should have mastered the material well.
Of course, there is a lot of information here, varied and sometimes even confusing: it is not so difficult to confuse the properties of the described trapezoid with the properties of the inscribed one. But you yourself saw that the difference is huge.
Now you have a detailed summary of all the general properties of a trapezoid. As well as specific properties and features of isosceles and rectangular trapezoids. It is very convenient to use to prepare for tests and exams. Try it yourself and share the link with your friends!
site, with full or partial copying of the material, a link to the source is required.
Circumscribed circle and trapezoid. Hello! For you, another publication in which we will consider problems with trapezoids. Assignments are part of the math exam. Here they are combined into a group, not just one trapezoid is given, but a combination of bodies - a trapezoid and a circle. Most of these problems are solved orally. But there are some that need special attention, for example, problem 27926.
What theory should be kept in mind? It:
Tasks with trapezoids that are available on the blog can be viewed here.
27924. A circle is circumscribed near a trapezoid. The perimeter of the trapezoid is 22, the midline is 5. Find the side of the trapezoid.
Note that a circle can only be circumscribed about an isosceles trapezoid. We are given the middle line, so we can determine the sum of the bases, that is:
So the sum of the sides will be equal to 22–10=12 (perimeter minus the base). Since the sides of an isosceles trapezoid are equal, one side will be equal to six.
27925. The lateral side of an isosceles trapezoid is equal to its smaller base, the angle at the base is 60 0, the larger base is 12. Find the radius of the circumscribed circle of this trapezoid.
If you solved problems with a circle and a hexagon inscribed in it, then immediately voice the answer - the radius is 6. Why?
Look: an isosceles trapezoid with a base angle of 60 0 and equal sides AD, DC and CB is half a regular hexagon:
In such a hexagon, the segment connecting opposite vertices passes through the center of the circle. *The center of the hexagon and the center of the circle are the same, more
That is, the larger base of this trapezoid coincides with the diameter of the circumscribed circle. So the radius is six.
*Of course, you can consider the equality of triangles ADO, DOC and OCB. Prove that they are equilateral. Further, conclude that the angle AOB is equal to 180 0 and the point O is equidistant from the vertices A, D, C and B, which means AO=OB=12/2=6.
27926. The bases of an isosceles trapezoid are 8 and 6. The radius of the circumscribed circle is 5. Find the height of the trapezoid.
Note that the center of the circumscribed circle lies on the axis of symmetry, and if you build the height of the trapezoid passing through this center, then when it intersects with the bases, it will divide them in half. Let's show this on the sketch, also connect the center to the vertices:
The segment EF is the height of the trapezoid, we need to find it.
In a right triangle OFC we know the hypotenuse (this is the radius of the circle), FC=3 (because DF=FC). Using the Pythagorean theorem, we can calculate OF:
In a right triangle OEB, we know the hypotenuse (this is the radius of the circle), EB=4 (because AE=EB). Using the Pythagorean theorem, we can calculate OE:
Thus EF=FO+OE=4+3=7.
Now an important nuance!
In this problem, the figure clearly shows that the bases lie on opposite sides of the center of the circle, so the problem is solved in this way.
And if the sketch had not been given in the condition?
Then the problem would have two answers. Why? Look carefully - in any circle you can inscribe two trapezoids with given bases:
*That is, given the bases of the trapezium and the radius of the circle, there are two trapezoids.
And the solution will be "second option" will be next.
Using the Pythagorean theorem, we calculate OF:
Let's also calculate OE:
Thus EF=FO–OE=4–3=1.
Of course, in a problem with a short answer to the USE, there cannot be two answers, and a similar problem without a sketch will not be given. Therefore, pay special attention to the sketch! Namely: how the bases of the trapezoid are located. But in tasks with a detailed answer, this was present in past years (with a slightly more complicated condition). Those who considered only one option for the location of the trapezoid lost a point on this task.
27937. A trapezoid is circumscribed about a circle, the perimeter of which is 40. Find its midline.
Here we should immediately recall the property of a quadrilateral circumscribed about a circle:
The sums of the opposite sides of any quadrilateral circumscribed about a circle are equal.
Good evening! Oh, these circumscribed, or inscribed circles, geometric figures. It's so hard to get confused. what and when.
Let's try to figure it out first with the wording. We are given a circle circumscribed about . In other words, this trapezoid is inscribed in a circle.
Let's remember that we can only describe a circle around. And an isosceles trapezoid, in turn, is a trapezoid whose sides are equal.
Let's try to solve the problem. We know that the bases of the isosceles trapezoid ADCB are 6 (DC) and 4 (AB). And the radius of the circumscribed circle is 4. You need to find the height of the trapezoid FK.
FK is the height of the trapezoid. we need to find it, but before that, remember that the point O is the center of the circle. And OS, OD, OA, OB are known radii.
In OFC we know the hypotenuse, which is the radius of the circle, and the leg FC = half the base DC = 3 cm (because DF = FC).
Now let's find OF:
And in a right triangle OKB, we also know the hypotenuse, since this is the radius of the circle. And KB is half of AB; KB = 2 cm. And, using the Pythagorean theorem, we calculate the segment OK:
