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Definition
Trapeze is a quadrilateral $A B C D$, two sides of which are parallel and the other two are not parallel (Fig. 1).
The parallel sides of the trapezoid ($B C$ and $A D$) are called bases of a trapezoid, not parallel ($A B$ and $C D$) - sides. The perpendicular ($B H$) drawn from any point of one base to another base or its continuation is called the height of the trapezoid.
property of a trapezoid
The sum of angles adjacent to the lateral side is $180^(\circ)$:
$\angle A+\angle B=180^(\circ), \angle C+\angle D=180^(\circ)$ (Figure 1)
The segment connecting the midpoints of the sides of the trapezoid is called the midline of the trapezoid. The median line of the trapezoid is parallel to the bases and equal to their half sum:
$$M N=\frac(A D+B C)(2)$$
Among all trapezoids, two special classes of trapezoids can be chosen: rectangular and isosceles trapezoids.
Definition
Rectangular A trapezoid is called if one of its angles is a right angle.
equilateral is called a trapezoid, in which the sides are equal.
Properties of an isosceles trapezoid
- In an isosceles trapezoid, the angles at the base are pairwise equal to $\angle A=\angle D, \angle B=\angle C$.
- The diagonals of an isosceles trapezoid are equal to $A C=B D$.
Signs of an isosceles trapezium
- If the angles at the base of a trapezoid are equal, then the trapezoid is isosceles.
- If the diagonals of a trapezoid are equal, then it is isosceles.
Trapezium area:
$$S=\frac(a+b)(2) \cdot h$$
where $a$ and $b$ are the bases of the trapezoid and $h$ is its height.
Examples of problem solving
Example
Exercise. The height of an isosceles trapezoid drawn from an obtuse angle divides the base into segments 5 cm and 11 cm long. Find the perimeter of the trapezoid if its height is 12 cm.
Decision. Let's make a drawing (Fig. 3)
$ABCD$ - isosceles trapezium, $BH$ - height, $BH = 12$ cm, $AH = 5$ cm, $HD = 11$ cm.
Consider $\Delta A B H$, it is rectangular ($\angle H=90^(\circ)$). According to the Pythagorean theorem
$$A B=\sqrt(B H^(2)+A H^(2))$$
substituting the initial data, we get
$A B=\sqrt(12^(2)+5^(2))$
$A B=\sqrt(144+25)=\sqrt(169) \Rightarrow A B=13$ (cm)
Since the trapezoid $A B C D$ is isosceles, then its sides are equal: $A B=C D=13$ cm. The larger base of the trapezoid is equal to: $A D=A H+H D$, $A D=5+11=16$ (cm). The smaller base of the trapezoid will be: $B C=A D-2 A H, B C=16-2 \cdot 5=6$ (cm). The perimeter of a trapezoid is:
$P_(A B C D)=A B+B C+C D+A D$
$P_(A B C D)=13+6+13+16$
$P_(A B C D)=48$ (cm)
Answer.$P_(A B C D)=48$ cm
Example
Exercise. In a rectangular trapezoid, two smaller sides are equal to 2 dm, and one of the angles is $45^(\circ)$. Find the area of the trapezoid.
Decision. Let's make a drawing (Fig. 4)
$K L M N$ - rectangular trapezoid, $K L=L M=2$ dm, $L K \perp K N$, $\angle M L K=45^(\circ)$. From the vertex $M$ we lower the height $MP$ to the base $KN$. Consider $\Delta M N P$, it is rectangular ($\angle M P N=90^(\circ)$). Since $\angle M L K=45^(\circ)$, then
$\angle N M P=180^(\circ)-\angle M P N-\angle M L K$
$\angle N M P=180^(\circ)-90^(\circ)-45^(\circ)=45^(\circ)$
Thus, $\angle M L K=\angle N M P$ and $\Delta M N P$ is also isosceles. Hence $M P=P N$. Since $L K=M P=2$ dm, therefore $P N=2$ dm. The larger base is $K N=K P+P N$, since $L M=K P$, we get $K N=2+2=4$ (dm).
The area of the trapezoid is calculated by the formula:
$$S=\frac(a+b)(2) \cdot h$$
In our case, it will take the form:
$$S_(K L M N)=\frac(L M+K N)(2) \cdot M P$$
Substituting the known values, we get
$S_(K L M N)=\frac(2+4)(2) \cdot 2=6$ (dm 2)
Answer.$S_(K L M N)=6$ dm 2
- The segment connecting the midpoints of the diagonals of a trapezoid is equal to half the difference of the bases
- The triangles formed by the bases of the trapezoid and the segments of the diagonals up to the point of their intersection are similar
- Triangles formed by segments of the diagonals of a trapezoid, the sides of which lie on the sides of the trapezoid - equal area (have the same area)
- If we extend the sides of the trapezoid towards the smaller base, then they will intersect at one point with the straight line connecting the midpoints of the bases
- The segment connecting the bases of the trapezoid, and passing through the point of intersection of the diagonals of the trapezoid, is divided by this point in a proportion equal to the ratio of the lengths of the bases of the trapezoid
- A segment parallel to the bases of the trapezoid and drawn through the intersection point of the diagonals is bisected by this point, and its length is 2ab / (a + b), where a and b are the bases of the trapezoid
Properties of a segment connecting the midpoints of the diagonals of a trapezoid
Connect the midpoints of the diagonals of the trapezoid ABCD, as a result of which we will have a segment LM.
A line segment that joins the midpoints of the diagonals of a trapezoid lies on the midline of the trapezium.
This segment parallel to the bases of the trapezium.
The length of the segment connecting the midpoints of the diagonals of a trapezoid is equal to the half-difference of its bases.
LM = (AD - BC)/2
or
LM = (a-b)/2
Properties of triangles formed by the diagonals of a trapezoid
The triangles that are formed by the bases of the trapezoid and the point of intersection of the diagonals of the trapezoid - are similar.
Triangles BOC and AOD are similar. Because the angles BOC and AOD are vertical, they are equal.
Angles OCB and OAD are internal crosswise lying at parallel lines AD and BC (the bases of the trapezium are parallel to each other) and the secant line AC, therefore, they are equal.
Angles OBC and ODA are equal for the same reason (internal cross-lying).
Since all three angles of one triangle are equal to the corresponding angles of another triangle, these triangles are similar.
What follows from this?
To solve problems in geometry, the similarity of triangles is used as follows. If we know the lengths of the two corresponding elements of similar triangles, then we find the similarity coefficient (we divide one by the other). From where the lengths of all other elements are related to each other by exactly the same value.
Properties of triangles lying on the lateral side and diagonals of a trapezoid
Consider two triangles lying on the sides of the trapezoid AB and CD. These are triangles AOB and COD. Despite the fact that the sizes of individual sides of these triangles can be completely different, but the areas of the triangles formed by the sides and the point of intersection of the diagonals of the trapezoid are, that is, the triangles are equal.
If the sides of the trapezoid are extended towards the smaller base, then the point of intersection of the sides will be coincide with a straight line that passes through the midpoints of the bases.
Thus, any trapezoid can be extended to a triangle. Wherein:
- The triangles formed by the bases of a trapezoid with a common vertex at the point of intersection of the extended sides are similar
- The straight line connecting the midpoints of the bases of the trapezoid is, at the same time, the median of the constructed triangle
Properties of a segment connecting the bases of a trapezoid
If you draw a segment whose ends lie on the bases of the trapezoid, which lies at the intersection point of the diagonals of the trapezoid (KN), then the ratio of its constituent segments from the side of the base to the intersection point of the diagonals (KO / ON) will be equal to the ratio of the bases of the trapezoid(BC/AD).
KO/ON=BC/AD
This property follows from the similarity of the corresponding triangles (see above).
Properties of a segment parallel to the bases of a trapezoid
If you draw a segment parallel to the bases of the trapezoid and passing through the intersection point of the diagonals of the trapezoid, then it will have the following properties:
- Preset distance (KM) bisects the point of intersection of the diagonals of the trapezoid
- Cut length, passing through the point of intersection of the diagonals of the trapezoid and parallel to the bases, is equal to KM = 2ab/(a + b)
Formulas for finding the diagonals of a trapezoid
a, b- bases of a trapezoid
c, d- sides of the trapezoid
d1 d2- diagonals of a trapezoid
α β - angles with a larger base of the trapezoid
Formulas for finding the diagonals of a trapezoid through the bases, sides and angles at the base
The first group of formulas (1-3) reflects one of the main properties of the trapezoid diagonals:
1. The sum of the squares of the diagonals of a trapezoid is equal to the sum of the squares of the sides plus twice the product of its bases. This property of the diagonals of a trapezoid can be proved as a separate theorem
2 . This formula is obtained by transforming the previous formula. The square of the second diagonal is thrown over the equal sign, after which the square root is extracted from the left and right sides of the expression.
3 . This formula for finding the length of the diagonal of a trapezoid is similar to the previous one, with the difference that another diagonal is left on the left side of the expression
The next group of formulas (4-5) is similar in meaning and expresses a similar relationship.
The group of formulas (6-7) allows you to find the diagonal of a trapezoid if you know the larger base of the trapezoid, one side and the angle at the base.
Formulas for finding the diagonals of a trapezoid in terms of height
Task.
The diagonals of the trapezoid ABCD (AD | | BC) intersect at point O. Find the length of the base BC of the trapezoid if the base AD = 24 cm, length AO = 9 cm, length OS = 6 cm.
Decision.
The solution of this task is absolutely identical to the previous tasks in terms of ideology.
Triangles AOD and BOC are similar in three angles - AOD and BOC are vertical, and the remaining angles are pairwise equal, since they are formed by the intersection of one line and two parallel lines.
Since the triangles are similar, then all their geometric dimensions are related to each other, as the geometric dimensions of the segments AO and OC known to us by the condition of the problem. I.e
AO/OC=AD/BC
9 / 6 = 24 / B.C.
BC = 24 * 6 / 9 = 16
Answer: 16 cm
Task .
In the trapezoid ABCD it is known that AD=24, BC=8, AC=13, BD=5√17. Find the area of the trapezoid. 
Decision .
To find the height of a trapezoid from the vertices of the smaller base B and C, we lower two heights onto the larger base. Since the trapezoid is unequal, we denote the length AM = a, the length KD = b ( not to be confused with the symbols in the formula finding the area of a trapezoid). Since the bases of the trapezoid are parallel and we have omitted two heights perpendicular to the larger base, then MBCK is a rectangle.
Means
AD=AM+BC+KD
a + 8 + b = 24
a = 16 - b
Triangles DBM and ACK are right-angled, so their right angles are formed by the heights of the trapezoid. Let's denote the height of the trapezoid as h. Then by the Pythagorean theorem
H 2 + (24 - a) 2 \u003d (5√17) 2
and
h 2 + (24 - b) 2 \u003d 13 2
Consider that a \u003d 16 - b, then in the first equation
h 2 + (24 - 16 + b) 2 \u003d 425
h 2 \u003d 425 - (8 + b) 2
Substitute the value of the square of the height into the second equation, obtained by the Pythagorean Theorem. We get:
425 - (8 + b) 2 + (24 - b) 2 = 169
-(64 + 16b + b) 2 + (24 - b) 2 = -256
-64 - 16b - b 2 + 576 - 48b + b 2 = -256
-64b = -768
b = 12
Thus, KD = 12
Where
h 2 \u003d 425 - (8 + b) 2 \u003d 425 - (8 + 12) 2 \u003d 25
h = 5
Find the area of a trapezoid using its height and half the sum of the bases
, where a b - the bases of the trapezoid, h - the height of the trapezoid
S \u003d (24 + 8) * 5 / 2 \u003d 80 cm 2
Answer: the area of a trapezoid is 80 cm2.
Consider several directions for solving problems in which a trapezoid is inscribed in a circle.
When can a trapezoid be inscribed in a circle? A quadrilateral can be inscribed in a circle if and only if the sum of its opposite angles is 180º. Hence it follows that only an isosceles trapezoid can be inscribed in a circle.
The radius of a circle circumscribed about a trapezoid can be found as the radius of a circle circumscribed about one of the two triangles into which the trapezoid divides its diagonal.
Where is the center of the circle circumscribed about the trapezoid? It depends on the angle between the diagonal of the trapezoid and its side.
If the diagonal of a trapezoid is perpendicular to its lateral side, then the center of the circle circumscribed about the trapezoid lies in the middle of its larger base. The radius of the circle described near the trapezoid in this case is equal to half of its larger base:
If the diagonal of a trapezoid forms an acute angle with the lateral side, then the center of the circle circumscribed about the trapezoid lies inside the trapezoid.
If the diagonal of a trapezoid forms an obtuse angle with the lateral side, then the center of the circle circumscribed about the trapezoid lies outside the trapezoid, behind the large base.
The radius of a circle circumscribed about a trapezoid can be found from the corollary of the sine theorem. From triangle ACD
From triangle ABC
Another option to find the radius of the circumscribed circle is −
The sines of angle D and angle CAD can be found, for example, from right triangles CFD and ACF:
When solving problems for a trapezoid inscribed in a circle, you can also use the fact that the inscribed angle is equal to half of the corresponding central angle. For example,
By the way, you can use COD and CAD angles to find the area of a trapezoid. According to the formula for finding the area of a quadrilateral through its diagonals
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 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 the materials of exams and various tests, one can often 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 of the 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 trapezoid. This implies a return in the learning process to the 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 of a school course is a task technology for teaching them. 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.
Decision
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 use the 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 of a right triangle is equal to the sum of the squares of the legs. We get: BN \u003d √ (X2-F2). Next, we use the 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 the square root of the product 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 relationship between the areas of the triangles obtained, 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 of trapeziums. 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 bisected by the point of intersection. 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/BP. 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. Applying 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.