Samarqand davlat universiteti raqamli texnologiyalar fakulteti dasturiy injiniring yo



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Eshmo\'minov Sanjar 1- hisobot
Kuzgi semistr 2 kurslar imtihon jadvali
> with(geometry):

triangle(ABC, [point(A,10,3), point(B,-4,7), point(C,4,8)]):

median(mA, A, ABC); mA

> form(mA); line2d

> detail(mA);

name of the object: mA

form of the object: line2d

equation of the line: 75-9/2*x-10*y = 0

> median(mA, A, ABC, E);

> form(mA); segmebt2d

> coordinates(E);

> detail(mA);

name of the object: mA

form of the object: segment2d

the two ends of the segment: [[10, 3], [0, 15/2]]

ABC uchburchk og‘irlik markazi:

> with(geometry):

s:=[point(A,10,3), point(B,-4,7), point(C,4,8)];

> centroid(G,s); G

> form(G); point2d

> coordinates(G);

> detail(G);name of the object: G

form of the object: point2d

coordinates of the point: [10/3, 6]

Uchlari uchburchak tomonlarining ortalrida bo‘lgan uchburchak.

> with(geometry):

triangle(T, [point(A,10,3), point(B,-4,7), point(C,4,8)]):

medial(mT,T); mT

> detail(mT); name of the object: mT

form of the object: triangle2d

method to define the triangle: points

the three vertices: [[3, 5], [7, 11/2], [0, 15/2]]

> draw({T(color=blue),mT(color=red)},style=line,axes=NONE, rinttext=true);


Endi AN bissektirsa tenglamasin ikki xil usulda topish mumkin:

a)

tenglamga asosan aniqlaymiz.

Bu bissektirsa AB va AC tomonlar orasida bo‘lagligi uchun bu tomonlar tenglamalari

AB: 2x + 7y – 41 = 0 va AC: 5x + 6y – 68 = 0.

ga asosan:



bundan:



(A) , (B)

tenglamlardan qaysi biri A burchakning ichki burchagining bissektirsasi AD ekanligini aniqlaymiz. (A) tenglamaga B va C nuqtalarning koordinatalarini qo‘yganda chap va o‘ng kasirlar ishoralari har xil bo‘ladi. Bundan (A) tenglama ABC ning A ichki burchagining bissektrisasi bo‘ladi. (B) tenglamaga B va C nuqtalarning koordinatalari qo‘yilganda bir xil ishorali bo‘lgani uchun (B) tenglama qo‘shni burchak bissektrisasi bo‘ladi.



Ichki burchak bissektrisasi:

> with(geometry):

triangle(ABC, [point(A,10,3), point(B,-4,7), point(C,4,8)]):

define the ``line'' bisector bA



> bisector(bA, A, ABC); bA

> Equation(bA,[x,y]);

> detail(bA); name of the object: bA

form of the object: line2d

equation of the line: 1630.535418-104.0420976*x-196.7048141*y = 0

Qo‘shni burchakning (qo‘shma) bissektrisasi;

> with(geometry):

triangle(ABC,[point(A,10.,3),point(B,-4,7),point(C,4,8)]):

define the external bisector bA



> ExternalBisector(bA, A, ABC); bA

> Equation(bA,[x,y]);

> detail(bA); name of the object: bA

form of the object: line2d

equation of the line: 1654.921848-196.7048141*x+104.0420976*y = 0

> bisector(ibA,A,ABC):

ArePerendicular(bA,ibA); true

b) A burchak bissektrisasini BC tomon bilan kesishish nuqtasi N ning koordinatalarini topamiz. Geometriya kursidan mahlumki, uchburchak burchagining bissektrisasi burchak qarshisidagi tomonni burchakka yopishpgan tomonlarga proportsional bo‘laklarga bo‘ladi.

Demak: dan  dani son qiymatini topib(= ),

,

formulalarga asosan N(x, y) nuqtani topamiz va ikki nuqtadan o‘tuvchi to‘g‘ri chiziq formulasiga asosan: AN bissektrisa tenglamisini tuzamiz:



.

bA bissektrisani BC tomon bilan kesishish nuqtasi N koordinatalari:

> restart;

> with(geometry):

triangle(ABC, [point(A,10.,3.), point(B,-4,7), point(C,4,8)]):

define the ``segment'' bisector bA



> bisector(bA, A, ABC,N); bA

> form(bA); segmebt2d

> OnSegment(N, B, C, sqrt(212/61.));

> coordinates(N);

> detail(bA); name of the object: bA

form of the object: segment2d

the two ends of the segment: [[10., 3.], [1.206942951, 7.650867870]]

5)  ABC ning yuzini quyidagicha hisoblaymiz.

a) , (bunda a- asos, h-asosga tushirilgan balandlik) formulasiga asosan hisoblaymiz. Asos uchun BC tomon uzunligi:

=8.062 ni

va balandlik uchun 3) punktdagi AD uzunligi ekanidan,

 ABC ning yuzasi: ,

b)

formulaga uchburchak ABC ning uchlarining koordinatalarini qo‘yib uni yuzasini topamiz.



> with(geometry):

triangle(ABC, [point(A,10,3), point(B,-4,7), point(C,4,8)]):

area(ABC); 23

A nuqtadan BC tamonga parallel:

> with(geometry):

triangle(ABC, [point(A,10,3), point(B,-4,7), point(C,4,8)]): line(BC, [B,C]); BC

> arallelLine(lA,A,BC); lA

> Equation(lA,[x,y]);

> detail(lA); name of the object: lA

form of the object: line2d

equation of the line: -14-x+8*y = 0

ABC uchburchakni qurish:




ABC uchburchakni tamonlari tenglamalariga asosan qurish;

> with(plots):

> plot([(41-2*x)/7,(68-5*x)/6,(60+x)/8,(75-9/2*x)/10, (14+x)/8,(197*x-1655.)/104.1, (-104.1*x+1631.)/187, (83-8*x)], x=-6..12, color=[red,red,red,blue,blue,green,green, brown], style=[line], thickness=2, view=[-7..12,-18..16]);



ABC uchburchakni uchining koordinatalariga asosan kesmalari bo‘yicha qurish:

> triangle(ABC,[point(A,10,3),point(B,-4,7),point(C,4,8)]): altitude(hA1,A,ABC), median(mA, A, ABC), bisector(bA, A, ABC),ExternalBisector(bqA, A, ABC);



> draw([ABC(color=red),hA1(color=brown),mA(color=blue), bA(color=green), bqA(color=green)],title=`Uchburchakni qurish`,style=patch,thickness=2, rinttext=true, view=[-4..12,0..8]);

Tekislikda to’g’ri chiziq tenglamalariga doir masalalar




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