# 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 Yüklə 182,49 Kb.Dostları ilə paylaş:

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