Evidence of an inaccessible and rectangular trapezium. Trapeze

In this article we will try as far as possible to fully reflect the properties of the trapezium. In particular, we will talk about the general signs and properties of the trapezium, as well as the properties of the inscribed trapezium and the circle inscribed in the trapezium. We affect the properties of an inaccessible and rectangular trapezium.

An example of solving the problem using the considered properties will help you decompose on places in my head and it is better to remember the material.

Trapeze and all-all-all

To begin with, it is briefly remember what a trapezium is and what other concepts are connected with it.

So, the trapezium - the Figure-quadrilateral, two of the sides of which are parallel to each other (this is the foundation). And two are not parallel - these are the sides.

In the trapezium, the height can be lowered - perpendicular to the grounds. The middle line and diagonal were performed. And also from any angle of trapezium is possible to carry out bisector.

About various properties associated with all these elements and their combinations, we will talk now.

Properties of diagonals trapezium

In order to be clearer while reading, sketch yourself on a sheet of acklace and spend the diagonal in it.

1. If you find the middle of each of the diagonals (we denote these points x and t) and connect them, it turns out a segment. One of the properties of the diagonals of the trapezium lies in the fact that the segment of HT lies on the midline. And its length can be obtained by separating the difference in the bases for two: HT \u003d (A - B) / 2.
2. Before us, all the same trapezium acme. The diagonals intersect at the point of O. Let's look at the triangles of Oo and the IOC formed by the segments of diagonals along with the bases of the trapezium. These triangles are similar. The likeness ratio K triangles is expressed through the ratio of the bases of the trapezoid: k \u003d ae / km.
   The ratio of the areas of triangles, and the IOC is described by the K 2 coefficient.
3. All the same trapezium, the same diagonals intersecting at the point O. Only this time we will consider the triangles that cut the diagonals formed together with the sides of the trapezium. The area of \u200b\u200bthe Triangles of AKO and EMO are equal - their squares are the same.
4. Another property of the trapezium includes the construction of diagonals. So, if you continue the sides of the AK and Me in the direction of a smaller base, then sooner or later they will cross to some point. Further, through the middle of the bases of the trapezium will spend direct. It crosses the foundations at points X and T.
   If we now extend direct HT, it will connect together the point of intersection of the diagonals of the trapezoid, the point in which the continuation of the side and the middle of the bases X and T.
5. Through the intersection point of the diagonals, we carry out a segment that connects the base of the trapezion (t lies on a smaller base of the CM, x - on the larger AE). The intersection point of diagonals divides this segment in the following ratio: Then / oh \u003d km / ae.
6. And now through the intersection point of diagonals, we will carry out the parallel bases of the trapezoid (A and B) of the segment. The intersection point will divide it into two equal parts. You can find the length of the segment by the formula 2ab / (A + B).

Properties of the middle line

Middle line in the trapezium parallel to its grounds.

1. The length of the midline of the trapezium can be calculated if the base lengths are folded and to divide them in half: m \u003d (A + B) / 2.
2. If you spend through both bases a trapezium of any segment (height, for example), the middle line will divide it into two equal parts.

Property bisector trapezium

Choose any angle of the trapeze and swivectris. Take, for example, the angle of our trapezing acme. After making the construction yourself, you can easily make sure the bisector is cut off from the base (or its continuation on the straight out of the figure itself) the segment of the same length as the side side.

Properties of the corners of the trapezium

1. What of the two pairs of the adjacent to the side of the corners you did not choose, the sum of the angles in the pair is always 180 0: α + β \u003d 180 0 and γ + δ \u003d 180 0.
2. Connect the middle of the bases of the trapezoid segment TX. Now look at the corners at the bases of the trapezium. If the sum of the angles at any of them is 90 0, the length of the segment TX is easy to calculate on the basis of the difference in the base lengths, in half: TX \u003d (AE - km) / 2.
3. If through the side of the angle of the trapezium to carry out parallel straight lines, they separate the side of the angle on proportional segments.

The properties of an equilibrium (equal) trapezium

1. In an equilibrium trapezium, the angles are equal to any of the grounds.
2. Now build a trapeze again to imagine what it is about. Look carefully on the basis of AE - the top of the opposite base M is projected into a species on a straight line that contains ae. The distance from the top A to the point of projection of the vertex M and the average line of an equifiable trapezium is equal.
3. A few words about the property of the diagonals of an equifiable trapezium - their lengths are equal. As well as the same angles of tilt these diagonals to the base of the trapezium.
4. Only about an equilibrium trapezium can be described in the circumference, since the sum of the opposite angles of the quadrangle 180 0 is a prerequisite for this.
5. From the previous paragraph, the property of an equilibrium trapezium follows - if you can describe the circle near the trapezion, it is an equifiable.
6. Of the features of an equilibrium trapezium, the height of the trapezium flows: if its diagonals intersect at right angles, the length of the height is equal to half the amount of the grounds: h \u003d (A + B) / 2.
7. Again, spend the segment of TX through the middle of the bases of the trapezium - in an equilibrium trapezium it is perpendicular to the grounds. And at the same time TX - the axis of the symmetry of an in-beamed trapezium.
8. This time omit to a greater base (we denote it a) the height of the opposite vertex of the trapezium. It turns out two segments. The length of one can be found if the base lengths are folded and divided by half: (A + B) / 2. The second we will get when, from a larger base, the smaller and the resulting difference is divided into two: (A - B) / 2.

The properties of the trapezium included in the circle

Once, it was about the trapezoid inscribed in the circle, we will focus on this issue. In particular, where the center of the circle is located in relation to the trapezium. Here it is also recommended not to be lazy to take a pencil into your hands and draw something about what will be discussed below. So you will understand faster and remember better.

1. The location of the center of the circle is determined by the angle of inclination of the trapezoid diagonal to its side. For example, a diagonal can exit the vertex of a trapezium at a right angle to the side. In this case, the larger base crosses the center of the circle described exactly in the middle (R \u003d ½AE).
2. Diagonal and side can occur under an acute angle - then the center of the circle is inside the trapezium.
3. The center of the described circle may be beyond the trapezoids, for its large base, if between the trapezoid diagonal and the side is a stupid angle.
4. The angle formed by the diagonal and the large base of the acme trapezium (inscribed angle) is half of that central angle, which corresponds to it: May \u003d ½M..
5. Briefly about two ways to find the radius of the described circle. Method First: Look carefully on your drawing - what do you see? You can easily notice that the diagonal breaks the trapeze into two triangles. The radius can be found through the ratio of the side of the triangle to the sinus of the opposite angle, multiplied by two. For example, R \u003d AE / 2 * SINAME. Similarly, the formula can be painted for any of the sides of both triangles.
6. Method of the second: we find the radius of the described circle through the area of \u200b\u200bthe triangle formed by the diagonal, the side and the base of the trapezium: R \u003d AM * ME * AE / 4 * S.

The properties of the trapezium described near the circle

You can enter a circuit in a trapezium if one condition is observed. More about it below. And together this combination of figures has a number of interesting properties.

1. If a circle is inscribed in the trapezion, the length of its midline can be easily found by folding the length of the sides and dividing the amount obtained in half: m \u003d (C + D) / 2.
2. A acme trapezium described near the circumference, the sum of the base lengths is equal to the sum of the lengths of the sides: AK + ME \u003d km + ae.
3. From this property of the base of the trapezium implies an inverse statement: the circle can be entered into that trapezium, the sum of the bases of which is equal to the sum of the sides.
4. The point of touching the circle with a radius of R, inscribed in the trapezium, breaks the side by two segments, let's call them a and b. The radius of the circle can be calculated by the formula: r \u003d √ab.
5. And one more property. In order not to get confused, this example also draw yourself. We have an old-good trapezion of AKME, described near the circle. It was diagonally intersecting at the point of O. The triangles of the aok and the sides formed by the cuts of diagonals and sides are rectangular.
   The heights of these triangles, lowered on hypotenuses (that is, the sides of the trapezium) coincide with the radius of the inscribed circle. And the height of the trapezium - coincides with the diameter of the inscribed circle.

Properties of a rectangular trapezium

Rectangular call a trapezium, one of the corners of which is direct. And its properties arise from this circumstance.

1. A rectangular trapezium has one of the lateral sides perpendicular to the grounds.
2. Height and side of the trapezium, adjacent to the straight corner, are equal. This allows you to calculate the area of \u200b\u200bthe rectangular trapezium (general formula S \u003d (A + B) * H / 2) Not only through the height, but also through the side side, adjacent to direct corner.
3. For rectangular trapezium, the general properties of the trapezium diagonals are relevant above.

Evidence of some trapezium properties

Equality of angles at the basis of an inaccessible trapezium:

• You have probably already guessed themselves that there will again need a trapezoid akme - draw an equally chagrin trapecy. Spend from the vertex of the MT straight MT parallel to the side of the AK (MT || AK).

The resulting quadrilateral AKMT - parallelogram (AK || MT, km || at). Since me \u003d ka \u003d MT, Δ MTa is a chaired and meth \u003d MTT.

AK || MT, therefore, MTa \u003d kae, met \u003d mTa \u003d ka.

From where AKM \u003d 180 0 - MET \u003d 180 0 - KATE \u003d KME.

Q.E.D.

Now, on the basis of the property of an equilibrium trapezium (equality of diagonals), we prove that akme trapezium is an equifiable:

• To begin with, we will spend direct MX - MX || Ke. We get a parallelogram of KMCH (base - MX || KE and KM || EX).

ΔAMh is a waste, since AM \u003d KE \u003d MX, and MAX \u003d Mea.

MX || Ke, Kea \u003d moss, therefore, may \u003d moss.

We have it turned out that the triangles of Ake and EMA are equal to each other, because AM \u003d KE and AE - the common side of two triangles. As well as May \u003d moss. We can conclude that AK \u003d IU, and hence it follows and that the akme trapezium is a waste.

Task for repetition

The base of trapezing acme is 9 cm and 21 cm, the side side of 8 cm, forms an angle of 150 0 with a smaller base. It is required to find an area of \u200b\u200btrapezium.

Solution: from the top to lower the height to a greater base of the trapezium. And let's start considering the corners of the trapezium.

The angles of the AEM and Kahn are one-sided. And this means, in the amount they give 180 0. Therefore, Kan \u003d 30 0 (based on the properties of the angles of the trapezoid).

We now consider the rectangular δank (I suppose this moment is obvious to readers without additional evidence). From it we will find the height of the KN trapezion - in a triangle it is a cathet, which lies opposite the angle of 30 0. Therefore, kn \u003d ½AV \u003d 4 cm.

The area of \u200b\u200bthe trapezium is found by the formula: S acme \u003d (km + ae) * KN / 2 \u003d (9 + 21) * 4/2 \u003d 60 cm 2.

Afterword

If you carefully and thoughtfully studied this article, I did not have a pencil with a pencil in the hands of drawing a trapezium for all given properties and disassemble them in practice, the material was to be understood well.

Of course, there is a lot of information here, diverse and in places even confusing: it is not so difficult to confuse the properties of the described trapezoid with the properties inscribed. But you yourself made sure that the difference is huge.

Now you have a detailed summary of all the common trapezel properties. As well as specific properties and signs of the trapezoids of an isolated and rectangular. They are very convenient to use to prepare for control and exams. Try it yourself and share a link with friends!

the site, with full or partial copying of the material reference to the original source is required.

The section contains geometry tasks (section planning) about trapezes. If you have not found solutions to the task - write about it on the forum. The course will certainly be supplemented.

Trapezium. Definition, Formulas and Properties

The trapezium (from Dr.-Greek. Τραπέέιι - - "table"; τράπεζα - "table, food") - a quadricle, in which exactly one pair of opposite sides is parallel.

The trapezium is a quadricon, which has a parallel parallel.

Note. In this case, the parallelogram is a private trapezoid case.

The parallel opposite parties are called the bases of the trapezium, and the other two are sideways.

The trapezium are:

- versatile ;

- equalobocy;

- rectangular

.
Red and brown flowers indicate the side sides, green and blue - the base of the trapezium.

A - Equality (an equifiable, equique) trapezium
B - rectangular trapezium
C - a versatile trapezium

In a versatile trapezium, all sides of different lengths, and the base is parallel.

The sides are equal, and the base is parallel.

At the base is parallel, one side is perpendicular to the bases, and the second lateral side is inclined to the bases.

Properties of trapezium

• Medium line trapezium parallel to the grounds and equal to their half
• Cut connecting the middle diagonalsIt is equal to half the difference in the base and lies on the midline. His length
• Parallel straight lines crossing the parties of any angle of trapezoids, cut down from the side of the angle proportional segments (see Falez theorem)
• Point of intersection of diagonals of trapeziumThe point of intersection of the continuation of its side sides and the middle of the bases lie on one straight line (see also the properties of the quadrangle)
• Triangles lying on the grounds The trapezoids whose vertices are the intersection point of its diagonals are similar. The ratio of areas of such triangles is equal to the square of the ratio of the bases of the trapezium
• Triangles lying on the sides The trapezoids whose vertices are the intersection point of its diagonals are equal (equal in area)
• In a trapezium you can enter the circleIf the sum of the base lengths of the trapezium is equal to the sum of the lengths of its side sides. The middle line in this case is equal to the sum of the side of the sides, divided by 2 (since the average line of the trapezium is equal to the base of the base)
• Cut parallel base and passing through the intersection point of diagonals, it is divided by the last in half and is equal to the doubled product of the bases divided by the amount of 2ab / (A + B) (Burakova formula)

Corners of the trapezium

Corners of the trapezium there are sharp, straight and stupid.
There are only two angle straight.

Rectangular trapezium has two corners direct, and two others are sharp and stupid. Other types of trapeziums are: two sharp corners and two stupid.

The stupid corners of the trapezium belong to the smaller under the length of the base, and sharp - greater Base.

Any trapezium can be considered like a trungle trunclewhich has a cross-sectional line parallel to the base of the triangle.
Important. Please note that in this way (additional trapezing to the triangle) may solve some tasks about the trapezoid and some theorems are proved.

How to find parties and diagonal trapez

Finding the sides and diagonals of the trapezium are made using the formulas below:

The specified formulas apply notation as in the figure.

a - Less from the bases of the trapezium
b - more from the bases of the trapez
C, D - Side
H 1 H 2 - Diagonal

The sum of the squares of the diagonals of the trapezium is equal to the double product of the base of the trapezium plus the sum of the squares of the side of the sides (Formula 2)

Polygon - part of a plane bounded by a closed broken line. The corners of the polygon are denoted by the points of the peaks of the broken. The vertices of the corners of the polygon and the tops of the polygon are the coincident points.

Definition. The parallelogram is a quadrangle, which has opposite parties parallel.

Properties of parallelogram

1. The opposite parties are equal.
In fig. eleven AB = CD; BC. = AD.

2. The opposite angles are equal (two sharp and two stupid angle).
In fig. 11 ∠ A. = ∠C.; ∠B. = ∠D..

3 diagonals (cuts, connecting two opposite vertices) intersect and the intersection point is divided in half.

In fig. 11 segments AO. = OC.; BO. = OD..

Definition. The trapezium is a quadrangle, which has two opposite parties parallel, and two others - no.

Parallel sides called it basins, and two other parties - sideways.

Types of trapezium

1. Trapezewhere the sides are not equal,
called versatile (Fig. 12).

2. The trapezium, which has the sides are equal, called equally (Fig. 13).

3. The trapezium, which one side is a straight corner with the bases, is called rectangular (Fig. 14).

The segment connecting the middle sides of the trapezoid (Fig. 15) is called the middle line of the trapezoid ( MN.). The middle line of the trapezium is parallel to the grounds and is equal to half a half.

The trapezium can be called a truncated triangle (Fig. 17), therefore, the names of the trapezium are similar to the names of triangles (triangles are versatile, equal, rectangular).

Square parallelogram and trapezium

Rule. Square Pollogram It is equal to the product of its side to the height carried out to this side.

Trapeze - This is a quadrilateral having two parallel sides, which are grounds and two non-parallel sides, which are side parties.

Also there are such names like equaloboca or equality.

- This is a trapezium, which has the corners with the side of the straight.

Elements of trapezium

a, B - founding of a trapezium (A parallel b),

m, N - side sides trapezium

d 1, D 2 - diagonal trapezium

h - height trapezium (segment connecting bases and at the same time perpendicular to them)

Mn - middle line (Cut connecting mid-sides).

Square trapezium

1. By half the bases A, B and height H: S \u003d \\ FRAC (A + B) (2) \\ CDOT H
2. Through the middle line Mn and height H: S \u003d Mn \\ Cdot H
3. Through the diagonals D 1, D 2 and the angle (\\ sin \\ varphi) between them: S \u003d \\ FRAC (d_ (1) d_ (2) \\ sin \\ varphi) (2)

Properties of trapezium

Medium line trapezium

middle line Parallel to the grounds, equal to half a half and separates each segment with the ends, located on direct, which contain bases, (for example, the height of the figure) in half:

Mn || a, mn || b, Mn \u003d \\ FRAC (A + B) (2)

The sum of the corners of the trapezium

The sum of the corners of the trapeziumadjacent to each side, equal to 180 ^ (\\ CIRC):

\\ Alpha + \\ Beta \u003d 180 ^ (\\ CIRC)

\\ Gamma + \\ Delta \u003d 180 ^ (\\ CIRC)

Isometric triangles trapezium

IsometricThat is, having equal areas, are cuts of diagonals and triangles AOB and Doc formed by side sides.

Like formed trapezoid triangles

Similar triangles are AOD and COB, which are formed by their bases and segments of diagonals.

\\ Triangle AOD \\ SIM \\ Triangle COB

Similarity coefficient K is located by the formula:

k \u003d \\ FRAC (AD) (BC)

Moreover, the ratio of the areas of these triangles is k ^ (2).

The ratio of lengths of segments and bases

Each segment connecting the bases and passing through the intersection point of the diagonals of the trapezium is divided into this point in relation to:

\\ FRAC (OX) (OY) \u003d \\ FRAC (BC) (AD)

It will be fair and for height with the diagonals themselves.

1. The segment connecting the middle of the diagonals of the trapezium is equal to half the difference
2. Triangles formed by the bases of the trapezoid and segments of diagonals to the point of their intersection - like
3. Triangles formed by the segments of the trapezium diagonals, the sides of which lie on the side sides of the trapezoid - areometric (have the same area)
4. If you extend the sides of the trapezion towards a smaller base, they will cross at one point with a straight line connecting the middle of the bases
5. Cut connecting the base of the trapezium, and passing through the intersection point of the diagonals of the trapezoid, divides this point in proportion equal to the ratio of the bases of the trapezium
6. Cut, parallel bases of trapezium, and spent over the intersection of diagonals, divides this point in half, and its length is 2ab / (a \u200b\u200b+ b), where a and b - the base of the trapezium

Segate properties connecting the middle of the diagonals of the trapezium

Connect the middle of the diagonals of the ABCD trapezoid, as a result of which the LM segment appears.
Cut connecting the middle of the diagonals of the trapezium, lies on the middle line of the trapez.

This cut paralleled to the grounds of the trapezium.

The length of the segment connecting the middle of the diagonal diagonals is equal to the highness of its bases.

LM \u003d (AD - BC) / 2
or
LM \u003d (A-B) / 2

Properties of triangles formed by trapezium diagonals

Triangles, which are formed by the grounds of trapezoid and the intersection point of the diagonals of the trapezium - are similar.
Boc and AOD triangles are similar. Because Boc and AOD angles are vertical - they are equal.
OCB and OAD angles are internal climbs lying with parallel direct AD and BC (the base of the trapezion are parallel between themselves) and the secant direct AC, therefore, they are equal.
The angles of OBC and ODA are equal for the same reason (internal closures lying).

Since all three angle of one triangle is equal to the appropriate corners of another triangle, then these triangles are similar.

What follows from this?

To solve the geometry problems, the similarity of triangles is used as follows. If we know the lengths of the lengths of the two relevant elements of such triangles, then we find the likeness ratio (divide one thing). From where the length of all other elements correspond to each other in exactly the same meaning.

Properties of triangles lying on the side and diagonals of the trapezium

Consider two triangles lying on the side sides of the AB and CD trapezoid. These are AOB and COD triangles. Despite the fact that the size of individual parties in these triangles can be completely different, but square of triangles formed by side sides and point of intersection of diagonals of trapezium are equalThat is, triangles are equal.

If you extend the sides of the trapeze towards a smaller base, the intersection point of the parties will be coincide with a straight line that passes through the middle of the bases.

Thus, any trapezium can be completed to the triangle. Wherein:

• Triangles formed by the grounds of trapezoids with the total vertex at the point of intersection of extended side sides are similar
• Direct, connecting the middle of the bases of the trapezium, is, at the same time, the median triangle built

Cut properties connecting the base of the trapezium

If you spend a segment, the ends of which lie on the grounds of the trapezoid, which lies at the intersection point of the diagonals of the trapezoid (KN), then the ratio of the components of its segments from the base side to the intersection point of the diagonals (KO / ON) will be equal to the ratio of the bases of the trapezium (BC / AD).

KO / ON \u003d BC / AD

This property follows from the similarity of the respective triangles (see above).

Cut properties, parallel to the bases of the trapezium

If you spend a segment, parallel the bases of the trapezoid and passing through the intersection point of the trapezium diagonals, then it will have the following properties:

• Specified Cut (KM) divides the intersection point of the trapezium diagonals in half
• Length Cutpassing through the intersection point of the diagonals of the trapezoid and parallel bases is equal to Km \u003d 2ab / (A + B)

Formulas for finding diagonals

a, B. - The founding of the trapezium

c, D. - Side sides of the trapezium

d1 D2. - Diagonal trapezium

α β - angles with a larger foundation

Formulas finding diagonals of trapezium through bases, side sides and angles at the base

The first group of formulas (1-3) reflects one of the main properties of the diagonals of the trapezoid:

1. The sum of the squares of the diagonals of the trapezium is equal to the sum of the squares of the side of the side plus a twice product of its bases. This property of the diagonals of the trapezium can be proved as a separate theorem

2 . This formula is obtained by converting the previous formula. The square of the second diagonal is overdone through the sign of equality, after which the square root is extracted from the left and right of the expression.

3 . This formula for finding the length of the diagonal of the trapezoid is similar to the previous one, with the difference that another diagonal is left in the left part of the expression

The following group of formulas (4-5) is similar in meaning and expresses similar ratio.

A group of formulas (6-7) allows you to find a diagonal of the trapezoid, if a greater base of the trapezium is known, one side and angle at the base.

Formulas finding diagonals of trapezium through height

Note. In this lesson, the solution of the geometry of trapezes is given. If you did not find the solution to the geometry task you are interested in - ask a question on the forum.

A task.
The diagonally of the ABCD trapezoid (AD | | Sun) intersect at the point of O. Find the length of the base of the trapezium, if the base is AD \u003d 24 cm, the length of JSC \u003d 9cm, the length of the OS \u003d 6 cm.

Decision.
The solution of this task on ideology is absolutely identical to previous tasks.

AOD and Boc triangles are similar in three corners - AOD and Boc are vertical, and the remaining corners are in pairs, because they are formed by the intersection of one straight and two parallel straight lines.

Since the triangles are similar, all their geometric dimensions are among themselves, as the geometrically sizes known to us under the condition of the task of segments AO and OC. I.e

AO / OC \u003d AD / BC
9/6 \u003d 24 / BC
BC \u003d 24 * 6/9 \u003d 16

Answer: 16 cm

A task .
In the ABCD trapezion, it is known that ad \u003d 24, Sun \u003d 8, AC \u003d 13, BD \u003d 5√17. Find the square of the trapezoid.

Decision .
To find the height of the trapezium from the peaks of a smaller base B and C, to larger the base two heights. Since the trapezium is not equal - then we denote the length am \u003d a, the length kd \u003d b ( not to be confused with designations in the formula Looking for an area of \u200b\u200btrapezium). Since the base of the trapezium is parallel, and we lowered two heights perpendicular to greater base, then MBCK is a rectangle.

So
Ad \u003d am + bc + kd
a + 8 + b \u003d 24
a \u003d 16 - b

DBM and ACK triangles are rectangular, so their direct angles are formed by the heights of the trapezoid. Denote the height of the trapezium through H. Then, by the Pythagora theorem

H 2 + (24 - a) 2 \u003d (5√17) 2
and
H 2 + (24 - b) 2 \u003d 13 2

Take into account 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

We substitute the value of the square of the height into the second equation obtained on the Pythagorean theorem. We get:
425 - (8 + b) 2 + (24 - b) 2 \u003d 169
- (64 + 16b + b) 2 + (24 - b) 2 \u003d -256
-64 - 16B - b 2 + 576 - 48b + b 2 \u003d -256
-64b \u003d -768
B \u003d 12.

Thus, kd \u003d 12
From
H 2 \u003d 425 - (8 + b) 2 \u003d 425 - (8 + 12) 2 \u003d 25
H \u003d 5.

Find the square of the trapezium through its height and half the grounds
where a b is the base of the trapezium, h - the height of the trapezium
S \u003d (24 + 8) * 5/2 \u003d 80 cm 2

Answer: The area of \u200b\u200bthe trapezium is 80 cm 2.