The use of the area in the Solution problems Текст научной статьи по специальности «Математика»

PEDAGOGICAL SCIENCES

THE USE OF THE AREA IN THE SOLUTION PROBLEMS 1 2 Saipnazarov Sh.A. , Karimov J.K. (Republic of Uzbekistan)

Email: Saipnazarov442@scientifictext.ru

1Saipnazarov Shailovbek Aktamovich - Doctor of physico-mathematical Sciences, Professor; 2Karimov Javlon Kuziyevich - Senior Lecturer, DEPARTMENT HIGHER MATHEMATICS, TASHKENT STATE ECONOMIC UNIVERSITY, TASHKENT, REPUBLIC OF UZBEKISTAN

Abstract: this article discusses a method for calculating the areas of convex polygons, in particular triangles and quadrangles, and also considers several problems in which different formulas are used to calculate the area of a triangle. We want to offer you a material, which is considered way to calculate the areas of convex Polygon (triangles, quadrangles). There is formula calculation of the area of arbitrary convex Quadrangle, which can be called analogue of formula Geron, given their some resemblance. Keywords: area, convex, polygon, circle, triangle, quadrangle.

ИСПОЛЬЗОВАНИЕ ПЛОЩАДИ В РЕШЕНИИ ПРОБЛЕМ 12 Саипназаров Ш.А. , Каримов Ж.К. (Республика Узбекистан)

1Саипназаров Шайловбек Актамович - доктор наук по физике и математике, профессор; 2Каримов Жавлон Кузиевич - старший преподаватель, кафедра высшей математики, Ташкентский государственный экономический университет, г. Ташкент, Республика Узбекистан

Аннотация: в настоящей статье рассматривается способ вычисления площадей выпуклых многоугольников, в частности, треугольников и четырехугольников, а также рассмотрены несколько задач, в которых применяются различные формулы для вычисления площади треугольника. Мы хотим предложить вам материал, который считается способом вычисления площадей выпуклого многоугольника (треугольники, четырехугольники). Существует формула расчета площади произвольного выпуклого четырехугольника, которую можно назвать аналогом формулы Герона, учитывая их некоторое сходство.

Ключевые слова: площадь, выпуклый, многоугольник, окружность, треугольник, четырехугольник.

We want to offer you a material, which is considered way to calculate the areas of convex Polygon (triangles, quadrangles). There is formula calculation of the area of arbitrary convex Quadrangle, which can be called analogue of formula Geron, given their some resemblance. We prove the following theorem: the area of arbitrary convex Quadrangle can be determined by the formula:

S = p - a)(p - b)(p - c)(p - d)- abcd cos2

where a,b,c,d - the length of the parties, formula p - semiperimeter, a and ft -

opposing angles Quadrangle. [1]

The proof of. Let in a quadrilateral ABCD, AB = a, BC = b, CD = c, DA = d, ZABC = ft ZADC = a (Figure 1.)

Fig. ¡.Opposite angles Quadrangle

Of the triangle ABC by virtue of theorem cosines AC = a + b - 2ab cos P. Of the Triangle ADC:

AC2 = c2 + d2 - 2 cd cos a

Equating the right side of these expressions, we get

a2 + b2 - 2 ab cos P = c2 + d2 - 2 cd cos a or

a2 + b2 - c2 - d2 = 2 ab cos P — 2 cd cos a (1) We find the area Quadrangle ABCD as the sum of squares triangles ABC and

ADC:

5 =1 ab sin P +1 CD sin P

2 2 (2) In equalites (1) and (2) we will square both sides, and then it up by term:

(a2 + b2 -c2 -d2) +1652 = (2abcosP-2cdcosa)2 +(2absinP + 2cdsina)2

Carry out the equivalent conversion, we obtain [2]

1652 = 4a2b2 + 4c2d2 - (a2 + b2 - c2 - d2 )2 - 8abcd cos (a + P),

1652 = (2ab + 2cd)2 - (a2 + b2 - c2 - d2 )2 - 8abcd - 8abcd cos (a + P), 1652 = (2ab + 2cd)2 -(a2 + b2 -c2 - d2 )2 - 16abcdcos2 a+P, 1652 = (2ab + 2cd - a2 + b2 - c2 - d2) • (2ab + 2cd + a2 + b2 - c2 - d2) -16abcd ca + P (c + d )2-(a - b)2 • (a + b )2-(c - d )2

1652 =

- 16abcd cos2

! cos

a + ß

2

1652 = (c + d + b - a )•( c + d + a - b)-( a + b + d - c )-(a + b + c - d )-16abcd cos2 a + P,

1652 = 2 (p - a )• 2 (p - b)- 2 (p - d )-16abcd cos2^-^^ from here

5 = p - a)(p - b)( p - c)(p - d) - abcd cos2

a + ß 2 (3)

the theorem is proved.

If our square parallelogram, that is a = c, b = d, a = P than it follows from (3) we obtain

5 = ab sina

and if trapezoid, that is cos _+P = cos 90° = 0

5 = <s/(P -a)(P -b)(P -c)(P -d)

If in our formula d = 0, then our Polygon become a Triangle, that is

5 = V( p - a)(p - b)( p - c)

Hence a number of the consequences of.

A consequence of -1. Area of arbitrary Quadrangle, inscribed in a circle, is calculated by the formula:

5 = ^( p - a )( p - b )( p - c )(p - d )

Proof immediately follows from theorem discussed above, taking info account the fact the amount of opposed angles inscribed in a Circle Quadrangle is equal to 1800, that is

a + P = 180°, cosa+P = cos 90° = 0 2

Therefore

5 = V(p - a)(p -b)(p - c)(p -d)

A consequence of -2. Area of arbitrary Quadrangle described around the Circle is calculated by the formula:

5 = 4 labcd sin2 a + P

V 2

The proof of. Since described Quadrangle the amount of opposite sides are equal, that is a + c = b + d the p - a = c; p - b = d, p - c = a, p - d = b. We have

5 =. labcd - abcd cos2 OlZ =. Lcd fi - cos2 = 4 abcd sin2

= -Jabcd • sin a + P 2

A consequence of -3 . Area Quadrangle, inscribed in a circle and described around the Circle, can be calculated by the formula:

5 =Vabcd

Let us consider several the tasks which are used different formula to calculate the area of

the Triangle. Using the formula 5 2 h'a

(where h - height, and a - of the base of the triangle) easy to prove the following statement:

if segments

AB and CD lie on the same line not passing through the point of M , and 51 and 52 - area triangles MAB and MCD , respectively, the

2

2

A.

S

AB CD

Fig. 2. The ratio of the areas of such triangles

Task 1. In the Triangle ABC on the side of AC taken point of M , and on the side of BC point N . The segment BM and AN intersect at the point O (Figure 2). Get the area of the Triangle CMN, if the area of triangles AOM, AOB and BON

respectively S15 S2, S3. (Figure 3)

Fig. 3. Corresponding triangle area

S _ AO _ S2

Solution. Let samon = q\ > s&cmn = qi of (4) it follows that Q, = ON ~ S2 ' where S ■ So

q2 _cn_q1+q+sl

Q2 =

Q2+Q,S, + QS3+S, ■ S3

Semilarly q + S3 nb s2 + S3 , where S2 - Qj

Substituting the last formula found the value of Qi, we obtain the Q2.

S, ■ S3 (S2 + S,)■(S2 + S3)

ans: S2(S22 - Si ■ S3)

Task 2. In the Triangle ABC on the side of BC selected point of BC selected point of D , so that BD : CD = 1: 2 . What respect to the median CE divides the segment

AD ?

European science № 10 (42) ■ 54

Solution. Decisions based on the comparison of the areas of triangles. Let m — point of intersection of AD and CE (Figure 4). We denote S2, S3, S4 area triangles

AEM, EMB, MBD, CMD respectively. Using the formula (4) we obtain:

S1 = S2 ' S4 = 2S3 '

S = 2S + s =1 S s = 3S + S =1 S

saabd 2S1 ^ S3 ^ °AABC , saebc 3s3 ^ S1 2 s^abc

Fig. 4. Calculate the area of a triangle in space using vectors

From here S^bc = 3 (2S2 + S3) = 2 (3S3 + Si) or 3S3 = 4Si.

Nov is easy to find the desired ratio:

AM: MD = S^abm : «W = 2S1: S3 = 3:2

Список литературы /References

1. Вавилов В.В. и др. Задачи по математике, начала анализа. справ. пособие. М.: Наука, 1990. 608 с. ISBN 5.02-014201-8.

2. Виленкин Н.Я. и др. Математика. Учебное пособие для студентов пед. институтов. М.: Просвещение, 1977. 320 с.