Uchburchak - Triangle

Teng yonli uchburchak
Muntazam uchburchak
Turi	Muntazam ko'pburchak
Qirralar va tepaliklar	3
Schläfli belgisi	{3}
Kokseter diagrammasi
Simmetriya guruhi	Ikki tomonlama (D.3), buyurtma 2 × 3
Ichki burchak (daraja )	60°
Ikki tomonlama ko'pburchak	O'zi
Xususiyatlari	Qavariq, tsiklik, teng tomonli, izogonal, izotoksal

Uchburchak
Uchburchak
Qirralar va tepaliklar	3
Schläfli belgisi	{3} (teng tomon uchun)
Maydon	turli xil usullar; pastga qarang
Ichki burchak (daraja )	60 ° (teng qirrali uchun)

Uchburchak = Uch (uch) + burchak

A uchburchak a ko'pburchak uchtasi bilan qirralar va uchta tepaliklar. Bu asosiy narsalardan biridir shakllar yilda geometriya. Uchlari uchburchak A, Bva C bilan belgilanadi ${displaystyle riangle ABC}$.^[1]

Yilda Evklid geometriyasi, har qanday uchta nuqtakollinear, noyob uchburchakni va bir vaqtning o'zida noyobligini aniqlang samolyot (ya'ni ikki o'lchovli Evklid fazosi ). Boshqacha qilib aytganda, bu uchburchakni o'z ichiga olgan bitta tekislik mavjud va har bir uchburchak qandaydir tekislikda joylashgan. Agar butun geometriya faqat Evklid samolyoti, faqat bitta tekislik mavjud va barcha uchburchaklar unda joylashgan; ammo, yuqori o'lchovli evklid bo'shliqlarida bu endi to'g'ri emas. Ushbu maqola Evklid geometriyasidagi uchburchaklar va xususan, Evklid tekisligi haqida, boshqacha ko'rsatmalar bundan mustasno.

Uchburchakning turlari

Eyler diagrammasi teng qirrali uchburchaklar ta'rifidan foydalanib, uchburchak turlari kamida 2 ta teng tomon (ya'ni, teng qirrali uchburchaklar teng yonli).

Yon tomonlarning uzunligi bo'yicha

Uchburchaklar tomonlarining uzunliklari bo'yicha tasniflanishi mumkin:^[2][3]

• An teng qirrali uchburchak bir xil uzunlikdagi uch tomonga ega. Teng yonli uchburchak ham a muntazam ko'pburchak barcha burchaklar bilan 60 °.^[4]
• An yonbosh uchburchak teng uzunlikdagi ikki tomonga ega.^{[1-eslatma][5]} Teng yonli uchburchakda bir xil o'lchamdagi ikkita burchak, ya'ni bir xil uzunlikdagi ikki tomonga qarama-qarshi burchaklar mavjud. Bu haqiqat teng qirrali uchburchak teoremasi tomonidan ma'lum bo'lgan Evklid. Ba'zi matematiklar teng yonli uchburchakni aniq ikkita teng tomonga ega, boshqalari esa teng yonli uchburchakni bitta kamida ikkita teng tomon.^[5] Oxirgi ta'rif barcha teng qirrali uchburchaklarni teng yonli uchburchaklar hosil qiladi. Da paydo bo'lgan 45-45-90 to'rtburchaklar uchburchak tetrakis kvadrat plitkalari, teng yonli
• A skalan uchburchagi har xil uzunlikdagi barcha tomonlariga ega.^[6] Teng ravishda, u har xil o'lchovning barcha burchaklariga ega.

Teng tomonli	Isosceles	Scalene

Lyuk belgilari, Shomil belgilari deb ham ataladi, uchburchaklar va boshqa geometrik figuralarning diagrammalarida teng uzunlikdagi tomonlarni aniqlash uchun foydalaniladi.^[1] Yon tomonga "Shomil" naqshini, qisqa chiziqli segmentlar shaklida belgilanishi mumkin balli belgilar; ikkala tomon ham bir xil naqsh bilan belgilangan bo'lsa, ikki tomon teng uzunlikka ega. Uchburchakda naqsh odatda 3 tadan oshmaydi. Teng yonli uchburchakning har uch tomonida bir xil naqsh, yonbosh uchburchakning atigi 2 tomonida bir xil naqsh mavjud va skalen uchburchagining har ikki tomoni ham har xil, chunki hech bir tomoni teng emas.

Xuddi shunday, teng burchaklarni ko'rsatish uchun burchaklar ichidagi 1, 2 yoki 3 kontsentrik yoylarning naqshlaridan foydalaniladi: teng qirrali uchburchak barcha uch burchakda, yonbosh uchburchakda atigi 2 burchakda va skalen uchburchakda bir xil naqsh mavjud. har qanday burchakka teng bo'lmaganligi sababli barcha burchaklarda har xil naqshlarga ega.

Ichki burchaklar bo'yicha

Uchburchaklar ham ularga qarab tasniflanishi mumkin ichki burchaklar, bu erda o'lchangan daraja.

• A to'g'ri uchburchak (yoki to'g'ri burchakli uchburchak, ilgari a deb nomlangan to'rtburchaklar uchburchak) ichki burchaklaridan biri 90 ° ga teng (a to'g'ri burchak ). To'g'ri burchakka qarama-qarshi tomon bu gipotenuza, uchburchakning eng uzun tomoni. Qolgan ikki tomonga oyoqlari yoki katetiya^[7] (birlik: katetus ) uchburchakning To'g'ri uchburchaklar Pifagor teoremasi: ikki oyoq uzunliklarining kvadratlari yig'indisi gipotenuza uzunligining kvadratiga teng: a² + b² = v², qayerda a va b oyoqlarning uzunligi va v gipotenuzaning uzunligi. Maxsus to'rtburchaklar hisob-kitoblarni osonlashtiradigan qo'shimcha xususiyatlarga ega bo'lgan to'g'ri uchburchaklar. Eng mashhur ikkitadan biri - 3-4 ta to'rtburchak uchburchak, bu erda 3² + 4² = 5². Bunday vaziyatda 3, 4 va 5 a Pifagor uchligi. Ikkinchisi 45 daraja (45-45-90 uchburchak) o'lchamdagi ikkita burchakka ega bo'lgan yonbosh uchburchak.
• 90 ° o'lchamdagi burchakka ega bo'lmagan uchburchaklar deyiladi qiya uchburchaklar.
• Barcha ichki burchaklari 90 ° dan past bo'lgan uchburchak an o'tkir uchburchak yoki o'tkir burchakli uchburchak.^[3] Agar v eng uzun tomonning uzunligi, keyin a² + b² > v², qayerda a va b boshqa tomonlarning uzunliklari.
• Ichki burchagi 90 ° dan yuqori bo'lgan uchburchak an to'mtoq uchburchak yoki yassi burchakli uchburchak.^[3] Agar v eng uzun tomonning uzunligi, keyin a² + b² < v², qayerda a va b boshqa tomonlarning uzunliklari.
• Ichki burchagi 180 ° bo'lgan uchburchak (va kollinear tepaliklar) buzilib ketgan.
• To'g'ri degeneratsiya qilingan uchburchakning kollinear uchlari bor, ularning ikkitasi tasodifiydir.

Bir xil o'lchov bilan ikkita burchakka ega bo'lgan uchburchakning uzunligi ham bir xil bo'lgan ikki tomoni bor va shuning uchun u teng qirrali uchburchakdir. Bundan kelib chiqadiki, barcha burchaklar bir xil o'lchovga ega bo'lgan uchburchakda, uchala tomon ham bir xil uzunlikka ega va shuning uchun teng tomonli bo'ladi.

To'g'ri	O'tkir	O'tkir
	${displaystyle underbrace {qquad qquad qquad qquad qquad qquad qquad} _ {}}$
	Qiyshiq

Asosiy faktlar

Tashqi burchakni ko'rsatadigan uchburchak.

Uchburchaklar ikkio'lchovli samolyot raqamlari, agar kontekst boshqacha ko'rsatmasa (qarang) Yassi bo'lmagan uchburchaklar, quyida). Qattiq muolajalarda uchburchak a deb ataladi 2-oddiy (Shuningdek qarang Polytope ). Uchburchaklar haqidagi elementar faktlar taqdim etildi Evklid, uning 1-4 kitoblarida Elementlar Miloddan avvalgi 300 yillarda yozilgan.

Uchburchakning ichki burchaklari o'lchovlari har doim 180 gradusgacha qo'shiladi (ularning tengligini ko'rsatish uchun bir xil rang).

The uchburchak ichki burchaklari o'lchovlari yig'indisi yilda Evklid fazosi har doim 180 daraja.^[8][3] Bu fakt Evklidnikiga tengdir parallel postulat. Bu ikki burchak o'lchovini hisobga olgan holda har qanday uchburchakning uchinchi burchagi o'lchovini aniqlashga imkon beradi. An tashqi burchak uchburchak - bu chiziqli juftlik (va shuning uchun) qo'shimcha ) ichki burchakka. Uchburchakning tashqi burchagi o'lchovi unga qo'shni bo'lmagan ikkita ichki burchak o'lchovlari yig'indisiga teng; bu tashqi burchak teoremasi. Har qanday uchburchakning uchta tashqi burchagi (har bir tepalikka bittadan) o'lchovlari yig'indisi 360 daraja.^[2-eslatma]

O'xshashlik va muvofiqlik

Ikki uchburchak deyilgan o'xshash, agar bitta uchburchakning har bir burchagi boshqa uchburchakda mos keladigan burchak bilan bir xil o'lchovga ega bo'lsa. Shunga o'xshash uchburchaklarning mos tomonlari bir xil nisbatda bo'lgan uzunliklarga ega va bu xususiyat o'xshashlikni o'rnatish uchun ham etarli.

Ba'zi asosiy teoremalar shunga o'xshash uchburchaklar haqida:

• Agar shunday bo'lsa ikkita uchburchakning bir juft ichki burchagi bir-birining o'lchoviga teng, yana bir jufti ham bir-biriga teng, uchburchaklar o'xshash.
• Agar va faqat ikkita uchburchakning bir juft juft tomoni mos keladigan tomonlarning boshqa jufti bilan mutanosib bo'lsa va ularning kiritilgan burchaklari bir xil o'lchovga ega bo'lsa, unda uchburchaklar o'xshashdir. (The kiritilgan burchak chunki ko'pburchakning istalgan ikki tomoni bu ikki tomon orasidagi ichki burchakdir.)
• Agar faqat ikkita uchburchakning uchta juft tomoni bir xil nisbatda bo'lsa, u holda uchburchaklar o'xshashdir.^[3-eslatma]

Ikkita uchburchak uyg'un to'liq bir xil o'lcham va shaklga ega:^[4-eslatma] mos keladigan ichki burchaklarning barcha juftlari o'lchov jihatidan teng va barcha mos tomonlarning juftliklari bir xil uzunlikka ega. (Bu jami oltita tenglik, ammo uchtasi ko'pincha muvofiqlikni isbotlash uchun etarli.)

Ba'zilar alohida zarur va etarli shartlar bir juft uchburchak mos kelishi uchun:

• SAS postulati: Uchburchakning ikki tomoni boshqa uchburchakning ikki tomoni bilan bir xil uzunlikka ega va unga kiritilgan burchaklar bir xil o'lchovga ega.
• ASA: Ikki ichki burchak va uchburchakning qo'shilgan tomoni mos ravishda boshqa uchburchakning o'lchamlari va uzunligiga teng. (The kiritilgan tomon chunki ular uchun umumiy bo'lgan tomon juft burchakdir.)
• SSS: Uchburchakning har bir tomoni boshqa uchburchakning mos tomoni bilan bir xil uzunlikka ega.
• AAS: Uchburchakning ikkita burchagi va unga mos keladigan (qo'shilmagan) tomoni mos ravishda boshqa uchburchakning o'lchamlari va uzunligiga teng. (Bunga ba'zan shunday deyiladi AAcorrS va keyin yuqoridagi ASA ni o'z ichiga oladi.)

Ayrim etarli shartlar:

• Gipotenuza-oyoq (HL) teoremasi: to'rtburchak uchburchakdagi gipotenuza va oyoqning uzunligi boshqa to'rtburchaklarnikiga teng. Bunga RHS (to'g'ri burchak, gipotenuza, yon) ham deyiladi.
• Gipotenuza-burchak teoremasi: bitta to'rtburchaklar uchburchakdagi gipotenuza va o'tkir burchak, mos ravishda, boshqa to'rtburchak uchburchakdagi kabi uzunlik va o'lchovga ega. Bu AAS teoremasining ma'lum bir hodisasidir.

Muhim shart:

• Yon tomon-burchak (yoki burchak-yon tomon) sharti: Agar uchburchakning ikki tomoni va unga mos kelmaydigan burchagi mos ravishda boshqa uchburchakdagi kabi uzunlik va o'lchovga ega bo'lsa, u holda bu emas muvofiqlikni isbotlash uchun etarli; ammo berilgan burchak ikki tomonning uzun tomoniga qarama-qarshi bo'lsa, u holda uchburchaklar mos keladi. Gipotenuza-oyoq teoremasi bu mezonning alohida holatidir. Yonma-yon burchak sharti o'z-o'zidan uchburchaklar mos kelishini kafolatlamaydi, chunki bitta uchburchak tekis burchakka, ikkinchisi esa o'tkir burchakka ega bo'lishi mumkin.

To'g'ri uchburchaklar va o'xshashlik tushunchasidan foydalanib, trigonometrik funktsiyalar sinus va kosinusni aniqlash mumkin. Bu funktsiyalar burchak qaysi tergov qilinmoqda trigonometriya.

To'g'ri uchburchaklar

Pifagor teoremasi

Markaziy teorema bu Pifagor teoremasi, bu har qanday narsada ko'rsatilgan to'g'ri uchburchak, uzunligining kvadrati gipotenuza boshqa ikki tomon uzunliklari kvadratlari yig'indisiga teng. Agar gipotenuzaning uzunligi bo'lsa vva oyoqlarning uzunligi bor a va b, keyin teorema buni ta'kidlaydi

${displaystyle a ^ {2} + b ^ {2} = c ^ {2}.}$

Aksincha, to'g'ri: agar uchburchak tomonlarining uzunliklari yuqoridagi tenglamani qondirsa, u holda uchburchakning qarama-qarshi tomoni to'g'ri burchakka ega v.

To'g'ri uchburchaklar haqida ba'zi boshqa ma'lumotlar:

• To‘g‘ri burchakli uchburchakning o‘tkir burchaklari bir-birini to'ldiruvchi.

${displaystyle a + b + 90 ^ {circ} = 180 ^ {circ} O'ng arra a + b = 90 ^ {circ} O'ng arrow a = 90 ^ {circ} -b.}$

• Agar to'rtburchaklar uchburchakning oyoqlari bir xil uzunlikka ega bo'lsa, u holda bu oyoqlarning qarshisidagi burchaklar bir xil o'lchovga ega. Ushbu burchaklar bir-birini to'ldirganligi sababli, ularning har biri 45 darajani tashkil etadi. Pifagor teoremasi bo'yicha, gipotenuzaning uzunligi oyoq vaqtining uzunligidir √2.
• O'tkir burchaklari 30 va 60 daraja bo'lgan to'rtburchaklar uchburchakda gipotenuza qisqaroq tomonning uzunligidan ikki baravar, uzun tomoni qisqaroq tomonlarning uzunligiga teng √3:

${displaystyle c = 2a,}$

${displaystyle b = imes {sqrt {3}}.}$

Barcha uchburchaklar uchun burchaklar va tomonlar kosinuslar qonuni va sinuslar qonuni (deb ham nomlanadi kosinus qoidasi va sinus qoidalar).

Uchburchakning mavjudligi

Yon tomonlarning holati

The uchburchak tengsizligi uchburchakning istalgan ikki tomoni uzunliklari yig'indisi uchinchi tomon uzunligidan katta yoki teng bo'lishi kerakligini bildiradi. Ushbu yig'indisi faqat uchburchak uchburchagi uchli chiziqli, uchlari to'qnashgan bo'lsa, uchinchi tomonning uzunligini tenglashtirishi mumkin. Ushbu summaning uchinchi tomon uzunligidan kam bo'lishi mumkin emas. Berilgan uchta musbat yon uzunlikdagi uchburchak, agar u yon uzunliklar uchburchak tengsizligini qondiradigan bo'lsa, mavjud bo'ladi.

Burchaklar bo'yicha shartlar

Berilgan uchta burchak, degeneratsiz uchburchakni hosil qiladi (va haqiqatan ham ularning cheksizligi), agar faqat shu ikkala shart bajarilsa: (a) har bir burchak ijobiy bo'lsa va (b) burchaklar 180 ° ga teng bo'lsa. Agar degeneratlangan uchburchaklarga ruxsat berilsa, 0 ° burchakka ruxsat beriladi.

Trigonometrik sharoitlar

Uchta ijobiy burchak a, βva γ, ularning har biri 180 ° dan kam, uchburchakning burchaklari agar va faqat agar quyidagi shartlardan biri:

${displaystyle a {frac {alpha} {2}} an {frac {eta} {2}} + an {frac {eta} {2}} an {frac {gamma} {2}} + an {frac {gamma} {2}} an {frac {alfa} {2}} = 1,}$^[9]

${displaystyle sin ^ {2} {frac {alpha} {2}} + sin ^ {2} {frac {eta} {2}} + sin ^ {2} {frac {gamma} {2}} + 2sin {frac {alfa} {2}} sin {frac {eta} {2}} sin {frac {gamma} {2}} = 1,}$^[9]

${displaystyle sin (2alpha) + sin (2 eta) + sin (2gamma) = 4sin (alfa) sin (eta) sin (gamma),}$

${displaystyle cos ^ {2} alfa + cos ^ {2} eta + cos ^ {2} gamma + 2cos (alfa) cos (eta) cos (gamma) = 1,}$^[10]

${displaystyle an (alfa) + an (eta) + an (gamma) = an (alfa) an (eta) an (gamma),}$

oxirgi tenglik faqat burchaklarning hech biri 90 ° ga teng bo'lmagan taqdirda qo'llaniladi (shuning uchun teginish funktsiyasining qiymati har doim cheklangan bo'ladi).

Uchburchak bilan bog'langan nuqtalar, chiziqlar va doiralar

Uchburchak bilan bog'langan (va ko'pincha ichida) o'ziga xos nuqtani topadigan va noyob xususiyatlarini qondiradigan minglab turli xil qurilishlar mavjud: maqolaga qarang Uchburchak markazlari entsiklopediyasi ularning katalogi uchun. Ko'pincha ular uchta chiziq (yoki tepaliklar) bilan nosimmetrik tarzda bog'langan uchta chiziqni topib, so'ngra uchta chiziqning bitta nuqtada uchrashishini isbotlash orqali quriladi: bularning mavjudligini isbotlash uchun muhim vosita Ceva teoremasi, bu uchta satr qachon bo'lishini aniqlash uchun mezon beradi bir vaqtda. Xuddi shunday, uchburchak bilan bog'langan chiziqlar ko'pincha uchta nosimmetrik tarzda qurilgan nuqta ekanligini isbotlash yo'li bilan quriladi kollinear: Bu yerga Menelaus teoremasi foydali umumiy mezonni beradi. Ushbu bo'limda eng ko'p uchraydigan qurilishlarning bir nechtasi tushuntirilgan.

The aylana uchburchakning uchta tepasidan o'tgan aylananing markazi.

A perpendikulyar bissektrisa uchburchakning tomoni - bu orqali o'tgan to'g'ri chiziq o'rta nuqta yon tomoni va unga perpendikulyar, ya'ni u bilan to'g'ri burchak hosil qiladi. Uchta perpendikulyar bissektrisalar bitta uchburchak uchburchagida to'qnash keladi aylana, odatda tomonidan belgilanadi O; bu nuqta aylana, doira barcha uchta tepaliklardan o'tib ketadi. Ushbu doiraning diametri diametri, yuqorida ko'rsatilgan sinuslar qonunidan topish mumkin. Aylana radiusi deyiladi sirkradius.

Fales teoremasi shuni anglatadiki, agar aylana aylanasi uchburchakning yon tomonida joylashgan bo'lsa, u holda qarama-qarshi burchak o'ng tomonga to'g'ri keladi. Agar aylanma uchburchak uchburchak ichida joylashgan bo'lsa, u holda uchburchak o'tkir; agar aylana aylanasi uchburchakdan tashqarida joylashgan bo'lsa, u holda uchburchak tekis bo'ladi.

Balandliklarning kesishishi quyidagicha ortsentr.

An balandlik uchburchak - bu vertikal va qarama-qarshi tomonga perpendikulyar (ya'ni to'g'ri burchak hosil qiladigan) to'g'ri chiziq. Ushbu qarama-qarshi tomonga tayanch balandlikning balandligi va balandlikning bazani (yoki uning kengaytmasini) kesib o'tadigan nuqtasi deyiladi oyoq balandlik. Balandlikning uzunligi - bu taglik va tepalik orasidagi masofa. Uch balandlik bitta nuqtada kesib o'tiladi, deyiladi ortsentr odatda tomonidan ko'rsatilgan uchburchakning H. Ortsentratsiya uchburchak ichida joylashgan bo'lsa, faqat uchburchak o'tkir bo'lsa.

Burchak bissektrisalarining kesishishi - ning markazi aylana.

An burchak bissektrisasi uchburchak - bu vertikal orqali to'g'ri chiziq bo'lib, mos keladigan burchakni yarmiga qisqartiradi. Uch burchak bissektrisasi bitta nuqtada kesishadi, rag'batlantirish, odatda tomonidan belgilanadi Men, uchburchakning markazi aylana. Aylana - bu uchburchak ichida yotgan va uch tomonga tegadigan aylana. Uning radiusi deyiladi nurlanish. Uchta muhim doiralar mavjud chekkalari; ular uchburchakning tashqarisida yotib, bir tomonga va boshqa ikkitasining kengaytmalariga tegishadi. In-va aylanalarning markazlari an hosil qiladi ortsentrik tizim.

Medianlarning kesishishi quyidagicha centroid.

A o'rtacha uchburchakning a orqali to'g'ri chiziq tepalik va o'rta nuqta qarama-qarshi tomonni va uchburchakni ikkita teng maydonga ajratadi. Uchta median uchburchakning bitta nuqtasida kesishadi centroid yoki geometrik baritsentr, odatda tomonidan belgilanadi G. Qattiq uchburchak buyumning tsentroidi (bir xil zichlikdagi yupqa varaqdan kesilgan) ham unga tegishli massa markazi: ob'ekt bir xil tortishish maydonida uning markazida muvozanatlashishi mumkin. Centroid har bir medianani 2: 1 nisbatda kesib tashlaydi, ya'ni tepalik va sentroid orasidagi masofa centroid va qarama-qarshi tomonning o'rta nuqtasi orasidagi masofadan ikki baravar ko'pdir.

To'qqiz nuqta doirasi oltita nuqta uchburchakning chetida joylashgan simmetriyani namoyish etadi.

Uch tomonning o'rta nuqtalari va uchta balandlikning oyoqlari hammasi bitta uchburchakning aylanasida joylashgan to'qqiz nuqta doirasi. U nomlangan qolgan uchta nuqta - bu tepaliklar va balandliklar orasidagi balandlik qismining o'rta nuqtalari ortsentr. To'qqiz nuqta doiraning radiusi aylananing yarmiga teng. U aylanaga tegadi (da Feyerbaxning fikri ) va uchta chekkalari.

Eyler chizig'i ortsentr (ko'k), to'qqizta doira markazi (qizil), tsentroid (to'q sariq) va aylana (yashil) orqali to'g'ri chiziq.

Orthocenter (ko'k nuqta), to'qqizta doira doirasi markazi (qizil), centroid (to'q sariq) va sirkumentr (yashil) barchasi bitta chiziqda yotadi, ular Eyler chizig'i (qizil chiziq). To'qqiz nuqta doiraning markazi ortsentr va aylana aylanasi o'rtasida o'rta nuqtada joylashgan bo'lib, markaz va aylana aylanasi orasidagi masofa markaz va ortsentr markazlari orasidagi yarmiga teng.

Aylananing markazi umuman Eyler chizig'ida joylashgan emas.

Agar bir xil tepalikdan o'tgan burchak bissektrisasida medianani aks ettirsa, a ga ega bo'ladi simmedian. Uch simmedian bitta nuqtada kesishadi simmedian nuqtasi uchburchakning

Yon va burchaklarni hisoblash

Yon uzunligini yoki burchak o'lchovini hisoblashning turli xil standart usullari mavjud. To'g'ri burchakli uchburchakda qiymatlarni hisoblash uchun ma'lum usullar mos keladi; boshqa vaziyatlarda yanada murakkab usullar talab qilinishi mumkin.

To'g'ri uchburchaklardagi trigonometrik nisbatlar

A to'g'ri uchburchak har doim 90 ° (π / 2 radian) burchakni o'z ichiga oladi, bu erda S yorlig'i bilan A va B burchaklar o'zgarishi mumkin. Trigonometrik funktsiyalar to'rtburchaklar uchburchakning yon uzunliklari va ichki burchaklari orasidagi munosabatlarni aniqlaydi.

Yilda to'g'ri uchburchaklar, sinus, kosinus va tangensning trigonometrik nisbatlaridan noma'lum burchaklar va noma'lum tomonlarning uzunliklarini topish uchun foydalanish mumkin. Uchburchakning tomonlari quyidagicha ma'lum:

• The gipotenuza bu to'g'ri burchakka qarama-qarshi tomon yoki bu holda to'rtburchak uchburchakning eng uzun tomoni sifatida aniqlanadi h.
• The qarama-qarshi tomon bizni qiziqtirgan burchakka qarama-qarshi tomon, bu holda a.
• The qo'shni tomon bizni qiziqtirgan burchak va to'g'ri burchak bilan aloqa qiladigan tomon, shuning uchun uning nomi. Bu holda qo'shni tomon b.

Sinus, kosinus va tangens

The sinus burchakning qarama-qarshi tomoni uzunligining gipotenuza uzunligiga nisbati. Bizning holatlarimizda

${displaystyle sin A = {frac {ext {qarshi tomon}} {ext {hypotenuse}}} = {frac {a} {h}} ,.}$

Ushbu nisbat, burchakni o'z ichiga olgan holda, tanlangan ma'lum to'rtburchakka bog'liq emas A, chunki bu uchburchaklar hammasi o'xshash.

The kosinus burchak - bu qo'shni tomon uzunligining gipotenuza uzunligiga nisbati. Bizning holatlarimizda

${displaystyle cos A = {frac {ext {ulashgan tomon}} {ext {hypotenuse}}} = {frac {b} {h}} ,.}$

The teginish burchakning qarama-qarshi tomoni uzunligining qo'shni tomonning uzunligiga nisbati. Bizning holatlarimizda

${displaystyle a A = {frac {ext {qarama-qarshi tomon}} {ext {qo'shni tomon}}} = {frac {a} {b}} = {frac {sin A} {cos A}},}$

Qisqartma "SOH-CAH-TOA "foydali mnemonik ushbu nisbatlar uchun.

Teskari funktsiyalar

The teskari trigonometrik funktsiyalar har qanday ikki tomonning uzunligi bo'lgan to'g'ri burchakli uchburchak uchun ichki burchaklarni hisoblash uchun ishlatilishi mumkin.

Arcsin yordamida qarama-qarshi tomon uzunligidan va gipotenuza uzunligidan burchakni hisoblash mumkin.

${displaystyle heta = arcsin chap ({frac {ext {qarshi tomon}} {ext {hypotenuse}}} ight)}$

Arkos yordamida qo'shni tomonning uzunligi va gipotenuzaning uzunligidan burchakni hisoblash mumkin.

${displaystyle heta = arccos chapda ({frac {ext {ulashgan tomon}} {ext {hypotenuse}}} ight)}$

Arktandan qarama-qarshi tomonning uzunligi va qo'shni tomonning uzunligidan burchakni hisoblash uchun foydalanish mumkin.

${displaystyle heta = arktan chap ({frac {ext {qarshi tomon}} {ext {ulashgan tomon}}} ight)}$

Kirish geometriyasi va trigonometriya kurslarida gunoh yozuvlari⁻¹, cos⁻¹va boshqalar, ko'pincha arcsin, arccos va boshqalar o'rnida ishlatiladi, ammo trigonometrik funktsiyalar odatda kuchga ko'tarilgan yuqori matematikada arcsin, arccos va boshqalar yozuvlari standart hisoblanadi, chunki bu chalkashliklarni oldini oladi. multiplikativ teskari va kompozitsion teskari.

Sinus, kosinus va tangens qoidalari

Uzunliklari a, b va c hamda a, b va γ burchaklari mos ravishda uchburchak.

The sinuslar qonuni yoki sinus qoidalar,^[11] tomon uzunligining unga mos keladigan qarama-qarshi burchagi sinusiga nisbati doimiy ekanligini bildiradi, ya'ni

${displaystyle {frac {a} {sin alfa}} = {frac {b} {sin eta}} = {frac {c} {sin gamma}}.}$

Ushbu nisbat berilgan uchburchakning aylanasi doirasining diametriga teng. Ushbu teoremaning yana bir talqini shundaki, burchaklari a, b va g bo'lgan har bir uchburchak yon uzunliklari sin a, sin β va sin p ga teng bo'lgan uchburchakka o'xshaydi. Ushbu uchburchakni avval 1-diametrli aylana qurish va uning ichiga uchburchakning ikkala burchagini yozish orqali qurish mumkin. Ushbu uchburchak tomonlarining uzunligi sin a, sin β va sin p bo'ladi. Uzunligi sin a bo'lgan tomon, o'lchovi a bo'lgan burchakka qarama-qarshi va boshqalar.

The kosinuslar qonuni, yoki kosinus qoidasi, uchburchakning noma'lum tomoni uzunligini boshqa tomonlari uzunligiga va noma'lum tomoniga qarama-qarshi burchakka bog'laydi.^[11] Qonunga muvofiq:

Tomonlari uzunligi uchburchak uchun a, b, v va uchburchakning ma'lum bo'lgan ikkita uzunligini hisobga olgan holda a, b, g burchaklari a va b, va ikkala ma'lum tomonlar orasidagi burchak γ (yoki noma'lum tomonga qarama-qarshi burchak) v), uchinchi tomonni hisoblash uchun v, quyidagi formuladan foydalanish mumkin:

${displaystyle c ^ {2} = a ^ {2} + b ^ {2} -2abcos (gamma)}$

${displaystyle b ^ {2} = a ^ {2} + c ^ {2} -2accos (eta)}$

${displaystyle a ^ {2} = b ^ {2} + c ^ {2} -2bccos (alfa)}$

Agar har qanday uchburchakning uch tomonining uzunligi ma'lum bo'lsa, uchta burchakni hisoblash mumkin:

${displaystyle alfa = arccos chap ({frac {b ^ {2} + c ^ {2} -a ^ {2}} {2bc}} ight)}$

${displaystyle eta = arccos qoldi ({frac {a ^ {2} + c ^ {2} -b ^ {2}} {2ac}} ight)}$

${displaystyle gamma = arccos qoldi ({frac {a ^ {2} + b ^ {2} -c ^ {2}} {2ab}} ight)}$

The tangents qonuni, yoki teginish qoidasi yordamida ikki tomon va burchak yoki ikkita burchak va yon ma'lum bo'lganida yon yoki burchak topish mumkin. Unda:^[12]

${displaystyle {frac {ab} {a + b}} = {frac {an [{frac {1} {2}} (alfa - eta)]} {an [{frac {1} {2}} (alfa +) eta)]}}.}$

Uchburchaklar echimi

"Uchburchaklar echimi" asosiy hisoblanadi trigonometrik Muammo: uchburchakning etishmayotgan xususiyatlarini (uchta burchak, uch tomonning uzunligi va boshqalar) ushbu xususiyatlardan kamida uchtasi berilganida topish. Uchburchak a da joylashgan bo'lishi mumkin samolyot yoki a soha. Ushbu muammo ko'pincha turli trigonometrik dasturlarda, masalan geodeziya, astronomiya, qurilish, navigatsiya va boshqalar.

Uchburchakning maydonini hisoblash

Uchburchakning maydoni, masalan, yordamida ko'rsatilishi mumkin uchburchaklar uyg'unligi, a maydonining yarmiga teng parallelogram uzunligi va balandligi bir xil.

Formulaning grafik hosilasi ${displaystyle T = {frac {h} {2}} b}$ bu uchburchakning maydonini ikki baravar oshirish va keyin uni yarmiga qisqartirishning odatiy tartibidan qochadi.

Maydonni hisoblash T uchburchak - har xil vaziyatlarda tez-tez uchraydigan elementar muammo. Eng yaxshi ma'lum va sodda formula:

${displaystyle T = {frac {1} {2}} bh,}$

qayerda b - uchburchak asosining uzunligi va h uchburchakning balandligi yoki balandligi. "Baza" atamasi istalgan tomonni, "balandlik" esa poydevorga qarama-qarshi bo'lgan vertikaldan asosni o'z ichiga olgan chiziqqa perpendikulyar uzunligini bildiradi. Milodiy 499 yilda Aryabhata, ushbu tasvirlangan usuldan Aryabhatiya (2.6-bo'lim).^[13]

Oddiy bo'lsa ham, bu formulani faqat balandlikni osongina topish mumkin bo'lganda foydalidir, bu har doim ham shunday emas. Masalan, uchburchak maydonni o'lchashda har bir tomonning uzunligini o'lchash nisbatan osonroq, ammo «balandlik» ni qurish nisbatan qiyin bo'lishi mumkin. Amaliyotda uchburchak haqida ma'lum bo'lgan narsalarga qarab turli xil usullardan foydalanish mumkin. Quyida uchburchak maydoni uchun tez-tez ishlatiladigan formulalar tanlangan.^[14]

Trigonometriyadan foydalanish

Balandlikni topish uchun trigonometriyani qo'llash h.

Uchburchkning balandligini dastur yordamida topish mumkin trigonometriya.

SASni bilish: O'ngdagi rasmdagi yorliqlardan foydalanib, balandlik h = a gunoh ${displaystyle gamma}$. Buni formulaga almashtirish ${displaystyle T = {frac {1} {2}} bh}$ Yuqorida keltirilgan uchburchakning maydoni quyidagicha ifodalanishi mumkin:

${displaystyle T = {frac {1} {2}} absin gamma = {frac {1} {2}} bcsin alfa = {frac {1} {2}} casin eta}$

(bu erda a - ichki burchak A, β - ichki burchak B, ${displaystyle gamma}$ ning ichki burchagi C va v bu chiziq AB).

Bundan tashqari, chunki sin a = sin (π - a) = gunoh (b +) ${displaystyle gamma}$) va shunga o'xshash boshqa ikkita burchak uchun:

${displaystyle T = {frac {1} {2}} absin (alfa + eta) = {frac {1} {2}} bcsin (eta + gamma) = {frac {1} {2}} kasin (gamma + alfa) ).}$

AASni bilish:

${displaystyle T = {frac {b ^ {2} (sin alfa) (sin (alfa + eta))} {2sin eta}},}$

va shunga o'xshash, agar ma'lum bo'lgan tomon bo'lsa a yoki v.

ASA ni bilish:^[2]

${displaystyle T = {frac {a ^ {2}} {2 (cot eta + cot gamma)}} = {frac {a ^ {2} (sin eta) (sin gamma)} {2sin (eta + gamma)}) },}$

va shunga o'xshash, agar ma'lum bo'lgan tomon bo'lsa b yoki v.

Heron formulasidan foydalanish

Uchburchakning shakli tomonlarning uzunliklari bilan belgilanadi. Shuning uchun maydonni tomonlarning uzunliklaridan ham olish mumkin. By Heron formulasi:

${displaystyle T = {sqrt {s (s-a) (s-b) (s-c)}}}$

qayerda ${displaystyle s = {frac {a + b + c} {2}}}$ bo'ladi semiperimetr, yoki uchburchak perimetrining yarmi.

Heron formulasini yozishning yana uchta ekvivalent usuli

${displaystyle T = {frac {1} {4}} {sqrt {(a ^ {2} + b ^ {2} + c ^ {2}) ^ {2} -2 (a ^ {4} + b ^ {4} + c ^ {4})}}}$

${displaystyle T = {frac {1} {4}} {sqrt {2 (a ^ {2} b ^ {2} + a ^ {2} c ^ {2} + b ^ {2} c ^ {2} ) - (a ^ {4} + b ^ {4} + c ^ {4})}}}$

${displaystyle T = {frac {1} {4}} {sqrt {(a + b-c) (a-b + c) (- a + b + c) (a + b + c)}}.}$

Vektorlardan foydalanish

A maydoni parallelogram uch o'lchovli ichiga o'rnatilgan Evklid fazosi yordamida hisoblash mumkin vektorlar. Vektorlarga ruxsat bering AB va AC mos ravishda dan A ga B va dan A ga C. Parallelogramma maydoni ABDC keyin

${displaystyle | mathbf {AB} imes mathbf {AC} |,}$

ning kattaligi o'zaro faoliyat mahsulot vektorlar AB va AC. ABC uchburchagi maydoni uning yarmiga teng,

${displaystyle {frac {1} {2}} | mathbf {AB} imes mathbf {AC} |.}$

Uchburchakning maydoni ABC bilan ham ifodalanishi mumkin nuqta mahsulotlari quyidagicha:

${displaystyle {frac {1} {2}} {sqrt {(mathbf {AB} cdot mathbf {AB}) (mathbf {AC} cdot mathbf {AC}) - (mathbf {AB} cdot mathbf {AC}) ^ { 2}}} = {frac {1} {2}} {sqrt {| mathbf {AB} | ^ {2} | mathbf {AC} | ^ {2} - (mathbf {AB} cdot mathbf {AC}) ^ {2}}}.,}$

Ikki o'lchovli Evklid fazosida, vektorni ifodalaydi AB kabi dekart fazosidagi erkin vektor ga teng (x₁,y₁) va AC kabi (x₂,y₂), buni quyidagicha yozish mumkin:

${displaystyle {frac {1} {2}}, | x_ {1} y_ {2} -x_ {2} y_ {1} |.,}$

Koordinatalardan foydalanish

Agar vertex bo'lsa A a ning kelib chiqishi (0, 0) da joylashgan Dekart koordinatalar tizimi va qolgan ikkita tepaning koordinatalari quyidagicha berilgan B = (x_B, y_B) va C = (x_C, y_C), keyin maydonni quyidagicha hisoblash mumkin¹⁄₂ marta mutlaq qiymat ning aniqlovchi

${displaystyle T = {frac {1} {2}} chap | det {egin {pmatrix} x_ {B} & x_ {C} y_ {B} & y_ {C} end {pmatrix}} ight | = {frac {1 } {2}} | x_ {B} y_ {C} -x_ {C} y_ {B} |.}$

Uchta umumiy tepalik uchun tenglama:

${displaystyle T = {frac {1} {2}} chap | det {egin {pmatrix} x_ {A} & x_ {B} & x_ {C} y_ {A} & y_ {B} & y_ {C} 1 & 1 & 1end {pmatrix }} ight | = {frac {1} {2}} {ig |} x_ {A} y_ {B} -x_ {A} y_ {C} + x_ {B} y_ {C} -x_ {B} y_ {A} + x_ {C} y_ {A} -x_ {C} y_ {B} {ig |},}$

sifatida yozilishi mumkin

${displaystyle T = {frac {1} {2}} {ig |} (x_ {A} -x_ {C}) (y_ {B} -y_ {A}) - (x_ {A} -x_ {B} ) (y_ {C} -y_ {A}) {ig |}.}$

Agar nuqtalar soat sohasi farqli o'laroq ketma-ket belgilanadigan bo'lsa, yuqoridagi aniqlovchi iboralar ijobiy bo'ladi va mutlaq qiymat belgilari qoldirilishi mumkin.^[15] Yuqoridagi formulalar oyoq kiyimining formulasi yoki o'lchovchi formulasi.

Agar biz tepaliklarni murakkab tekislikda joylashtirsak va ularni soat millariga teskari ketma-ketlikda quyidagicha belgilasak a = x_A + y_Amen, b = x_B + y_Bmenva v = x_C + y_Cmenva ularning murakkab konjugatlarini quyidagicha belgilang ${displaystyle {ar {a}}}$, ${displaystyle {ar {b}}}$va ${displaystyle {ar {c}}}$, keyin formula

${displaystyle T = {frac {i} {4}} {egin {vmatrix} a & {ar {a}} & 1 b & {ar {b}} & 1 c & {ar {c}} & 1end {vmatrix}}}$

poyabzal formulasiga tengdir.

Uch o'lchovda umumiy uchburchakning maydoni A = (x_A, y_A, z_A), B = (x_B, y_B, z_B) va C = (x_C, y_C, z_C) bo'ladi Pifagor summasi uchta asosiy tekislikdagi tegishli proektsiyalar maydonlarining (ya'ni. x = 0, y = 0 va z = 0):

${displaystyle T = {frac {1} {2}} {sqrt {{egin {vmatrix} x_ {A} & x_ {B} & x_ {C} y_ {A} & y_ {B} & y_ {C} 1 & 1 & 1end {vmatrix }} ^ {2} + {egin {vmatrix} y_ {A} & y_ {B} & y_ {C} z_ {A} & z_ {B} & z_ {C} 1 & 1 & 1end {vmatrix}} ^ {2} + {egin {vmatrix} z_ {A} & z_ {B} & z_ {C} x_ {A} & x_ {B} & x_ {C} 1 & 1 & 1end {vmatrix}} ^ {2}}}.}$

Lineer integrallardan foydalanish

Uchburchak kabi har qanday yopiq egri chiziq ichidagi maydon chiziqli integral egri chiziqning egri chiziqli to'g'ri chiziqdan egri chiziqning algebraik yoki imzolangan masofasi atrofida L. O'ng tomonga ishora qiladi L yo'naltirilganligi sababli salbiy masofada bo'lishi kerak L, integralning og'irligi parallel ravishda yoy uzunligining tarkibiy qismi sifatida qabul qilinadi L yoy uzunligining o'zi emas.

Ushbu usul o'zboshimchalik bilan maydonni hisoblash uchun juda mos keladi ko'pburchak. Qabul qilish L bo'lish x-aksis, ketma-ket vertikallar orasidagi chiziqli integral (x_men,y_men) va (x_men+1,y_men+1) tayanch vaqtlari o'rtacha balandlik bilan beriladi, ya'ni (x_men+1 − x_men)(y_men + y_men+1)/2. Maydonning belgisi - harakatlanish yo'nalishining umumiy ko'rsatkichi, salbiy maydon esa teskari harakatlanishni bildiradi. Keyin uchburchakning maydoni uch qirrali ko'pburchakka o'xshab tushadi.

Chiziqli integral usul boshqa koordinatalarga asoslangan usullar bilan o'zboshimchalik bilan koordinata tizimini tanlashga o'xshash bo'lsa, boshqalarnikidan farqli o'laroq uchburchakning uchini kelib chiqishi yoki yon tomoni asos sifatida o'zboshimchalik bilan tanlamaydi. Bundan tashqari, tomonidan belgilangan koordinata tizimini tanlash L odatdagi uchdan ko'ra atigi ikki daraja erkinlikni amalga oshiradi, chunki vazn mahalliy masofa (masalan,) x_men+1 − x_men yuqorida) qayerdan usul normal o'qni tanlashni talab qilmaydi L.

Ishlayotganda qutb koordinatalari ga aylantirish shart emas Dekart koordinatalari chiziqli integratsiyadan foydalanish, chunki ketma-ket vertikalar orasidagi chiziqli integral (r_men, θ_men) va (r_men+1, θ_men+1) ko'pburchak to'g'ridan-to'g'ri tomonidan berilgan r_menr_men+1gunoh (θ_men+1 - θ_men)/2. Bu $ Delta $ ning barcha qiymatlari uchun amal qiladi, | | | | bo'lganda sonning aniqligi pasayadi π dan kattaroq kattalikdagi ko'p tartiblardir. Ushbu formulada salbiy maydon soat yo'nalishi bo'yicha harakatlanishni bildiradi, bu qutb va dekart koordinatalarini aralashtirishda esda tutilishi kerak. Xuddi tanlov kabi y-aksis (x = 0) dekart koordinatalarida chiziqli integratsiya uchun ahamiyatsiz, shuning uchun nol sarlavhani tanlash (b = 0) bu erda ahamiyatsiz.

Heron formulasiga o'xshash formulalar

Uchta formulalar Heron formulasi bilan bir xil tuzilishga ega, ammo har xil o'zgaruvchilar bilan ifodalanadi. Birinchidan, medianlarni yon tomondan belgilash a, bva v navbati bilan m_a, m_bva m_v va ularning yarim yig'indisi (m_a + m_b + m_v)/2 σ sifatida bizda bor^[16]

${displaystyle T = {frac {4} {3}} {sqrt {sigma (sigma -m_ {a}) (sigma -m_ {b}) (sigma -m_ {c})}}.}$

Keyinchalik, balandliklarni yon tomondan belgilang a, bva v navbati bilan h_a, h_bva h_v, va balandliklarning o'zaro ta'sirining yarim yig'indisini quyidagicha belgilang ${displaystyle H = (h_ {a} ^ {- 1} + h_ {b} ^ {- 1} + h_ {c} ^ {- 1}) / 2}$ bizda ... bor^[17]

${displaystyle T ^ {- 1} = 4 {sqrt {H (H-h_ {a} ^ {- 1}) (H-h_ {b} ^ {- 1}) (H-h_ {c} ^ {- 1})}}.}$

Va burchaklar sinuslarining yarim yig'indisini quyidagicha belgilang S = [(sin a) + (sin β) + (sin ph)] / 2, bizda ... bor^[18]

${displaystyle T = D ^ {2} {sqrt {S (S-sin alfa) (S-sin eta) (S-sin gamma)}}}$

qayerda D. aylananing diametri: ${displaystyle D = {frac {a} {sin alfa}} = {frac {b} {sin eta}} = {frac {c} {sin gamma}}.}$

Pik teoremasidan foydalanish

Qarang Pik teoremasi har qanday ixtiyoriy maydonni topish texnikasi uchun panjara ko'pburchagi (vertikal va gorizontal ravishda tutashgan panjara nuqtalari teng masofada, uchlari esa panjara bilan chizilgan).

Teorema:

${displaystyle T = I + {frac {1} {2}} B-1}$

qayerda ${displaystyle I}$ ichki panjara nuqtalarining soni va B ko'pburchak chegarasida yotgan panjaralar soni.

Boshqa maydon formulalari

Kabi ko'plab boshqa maydon formulalari mavjud

${displaystyle T = rcdot s,}$

qayerda r bo'ladi nurlanish va s bo'ladi semiperimetr (aslida, bu formula uchun amal qiladi barchasi tangensial ko'pburchaklar ) va^{[19]:Lemma 2}

${displaystyle T = r_ {a} (s-a) = r_ {b} (s-b) = r_ {c} (s-c)}$

qayerda ${displaystyle r_ {a} ,, r_ {b} ,, r_ {c}}$ ning radiuslari chekkalari yon tomonlarga teginish a, b, c navbati bilan.

Bizda ham bor

${displaystyle T = {frac {1} {2}} D ^ {2} (sin alfa) (sin eta) (sin gamma)}$

va^[20]

${displaystyle T = {frac {abc} {2D}} = {frac {abc} {4R}}}$

sirkumametr uchun D.; va^[21]

${displaystyle T = {frac {alfa} {4}} (b ^ {2} + c ^ {2} -a ^ {2})}$

a ≠ 90 ° burchak uchun.

Maydonni quyidagicha ifodalash mumkin^[22]

${displaystyle T = {sqrt {rr_ {a} r_ {b} r_ {c}}}.}$

1885 yilda Beyker^[23] uchburchak uchun yuzdan ortiq aniq maydon formulalari to'plamini berdi. Bunga quyidagilar kiradi:

${displaystyle T = {frac {1} {2}} [abch_ {a} h_ {b} h_ {c}] ^ {1/3},}$

${displaystyle T = {frac {1} {2}} {sqrt {abh_ {a} h_ {b}}},}$

${displaystyle T = {frac {a + b} {2 (h_ {a} ^ {- 1} + h_ {b} ^ {- 1})}},}$

${displaystyle T = {frac {Rh_ {b} h_ {c}} {a}}}$

sirkradius uchun (aylana radiusi) Rva

${displaystyle T = {frac {h_ {a} h_ {b}} {2sin gamma}}.}$

Maydonning yuqori chegarasi

Hudud T of any triangle with perimeter p qondiradi

${displaystyle Tleq { frac {p^{2}}{12{sqrt {3}}}},}$

with equality holding if and only if the triangle is equilateral.^[24][25]:657

Other upper bounds on the area T tomonidan berilgan^[26]:p.290

${displaystyle 4{sqrt {3}}Tleq a^{2}+b^{2}+c^{2}}$

va

${displaystyle 4{sqrt {3}}Tleq {frac {9abc}{a+b+c}},}$

both again holding if and only if the triangle is equilateral.

Bisecting the area

Cheksiz ko'p lines that bisect the area of a triangle.^[27] Three of them are the medians, which are the only area bisectors that go through the centroid. Three other area bisectors are parallel to the triangle's sides.

Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter. There can be one, two, or three of these for any given triangle.

Further formulas for general Euclidean triangles

The formulas in this section are true for all Euclidean triangles.

Medians, angle bisectors, perpendicular side bisectors, and altitudes

The medians and the sides are related by^[28]:70-bet

${displaystyle {frac {3}{4}}(a^{2}+b^{2}+c^{2})=m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}$

va

${displaystyle m_{a}={frac {1}{2}}{sqrt {2b^{2}+2c^{2}-a^{2}}}={sqrt {{frac {1}{2}}(a^{2}+b^{2}+c^{2})-{frac {3}{4}}a^{2}}}}$,

and equivalently for m_b va m_v.

For angle A opposite side a, the length of the internal angle bisector is given by^[29]

${displaystyle w_{A}={frac {2{sqrt {bcs(s-a)}}}{b+c}}={sqrt {bcleft[1-{frac {a^{2}}{(b+c)^{2}}}ight]}}={frac {2bc}{b+c}}cos {frac {A}{2}},}$

yarim semimetr uchun s, where the bisector length is measured from the vertex to where it meets the opposite side.

The interior perpendicular bisectors are given by

${displaystyle p_{a}={frac {2aT}{a^{2}+b^{2}-c^{2}}},}$

${displaystyle p_{b}={frac {2bT}{a^{2}+b^{2}-c^{2}}},}$

${displaystyle p_{c}={frac {2cT}{a^{2}-b^{2}+c^{2}}},}$

where the sides are ${displaystyle ageq bgeq c}$ and the area is ${displaystyle T.}$^{[30]:Thm 2}

The altitude from, for example, the side of length a bu

${displaystyle h_{a}={frac {2T}{a}}.}$

Circumradius and inradius

The following formulas involve the circumradius R va nurlanish r:

${displaystyle R={sqrt {frac {a^{2}b^{2}c^{2}}{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}};}$

${displaystyle r={sqrt {frac {(-a+b+c)(a-b+c)(a+b-c)}{4(a+b+c)}}};}$

${displaystyle {frac {1}{r}}={frac {1}{h_{a}}}+{frac {1}{h_{b}}}+{frac {1}{h_{c}}}}$

qayerda h_a etc. are the altitudes to the subscripted sides;^[28]:79-bet

${displaystyle {frac {r}{R}}={frac {4T^{2}}{sabc}}=cos alpha +cos eta +cos gamma -1;}$^[10]

va

${displaystyle 2Rr={frac {abc}{a+b+c}}}$.

The product of two sides of a triangle equals the altitude to the third side times the diameter D. of the circumcircle:^[28]:64-bet

${displaystyle ab=h_{c}D,quad quad bc=h_{a}D,quad ca=h_{b}D.}$

Adjacent triangles

Suppose two adjacent but non-overlapping triangles share the same side of length f and share the same circumcircle, so that the side of length f is a chord of the circumcircle and the triangles have side lengths (a, b, f) va (v, d, f), with the two triangles together forming a tsiklik to'rtburchak with side lengths in sequence (a, b, v, d). Keyin^[31]:84

${displaystyle f^{2}={frac {(ac+bd)(ad+bc)}{(ab+cd)}}.,}$

Centroid

Ruxsat bering G be the centroid of a triangle with vertices A, Bva Cva ruxsat bering P be any interior point. Then the distances between the points are related by^[31]:174

${displaystyle (PA)^{2}+(PB)^{2}+(PC)^{2}=(GA)^{2}+(GB)^{2}+(GC)^{2}+3(PG)^{2}.,}$

The sum of the squares of the triangle's sides equals three times the sum of the squared distances of the centroid from the vertices:

${displaystyle AB^{2}+BC^{2}+CA^{2}=3(GA^{2}+GB^{2}+GC^{2}).}$^[32]

Ruxsat bering q_a, q_bva q_v be the distances from the centroid to the sides of lengths a, bva v. Keyin^[31]:173

${displaystyle {frac {q_{a}}{q_{b}}}={frac {b}{a}},quad quad {frac {q_{b}}{q_{c}}}={frac {c}{b}},quad quad {frac {q_{a}}{q_{c}}}={frac {c}{a}},}$

va

${displaystyle q_{a}cdot a=q_{b}cdot b=q_{c}cdot c={frac {2}{3}}T,}$

for area T.

Circumcenter, incenter, and orthocenter

Karnot teoremasi states that the sum of the distances from the circumcenter to the three sides equals the sum of the circumradius and the inradius.^[28]:p.83 Here a segment's length is considered to be negative if and only if the segment lies entirely outside the triangle. This method is especially useful for deducing the properties of more abstract forms of triangles, such as the ones induced by Yolg'on algebralar, that otherwise have the same properties as usual triangles.

Eyler teoremasi states that the distance d between the circumcenter and the incenter is given by^[28]:85-bet

${displaystyle displaystyle d^{2}=R(R-2r)}$

yoki unga teng ravishda

${displaystyle {frac {1}{R-d}}+{frac {1}{R+d}}={frac {1}{r}},}$

qayerda R is the circumradius and r is the inradius. Thus for all triangles R ≥ 2r, with equality holding for equilateral triangles.

If we denote that the orthocenter divides one altitude into segments of lengths siz va v, another altitude into segment lengths w va x, and the third altitude into segment lengths y va z, keyin uv = wx = yz.^[28]:94-bet

The distance from a side to the circumcenter equals half the distance from the opposite vertex to the orthocenter.^[28]:p.99

The sum of the squares of the distances from the vertices to the orthocenter H plus the sum of the squares of the sides equals twelve times the square of the circumradius:^[28]:102-bet

${displaystyle AH^{2}+BH^{2}+CH^{2}+a^{2}+b^{2}+c^{2}=12R^{2}.}$

Burchaklar

Ga qo'shimcha ravishda sinuslar qonuni, kosinuslar qonuni, law of tangents, va trigonometric existence conditions given earlier, for any triangle

${displaystyle a=bcos C+ccos B,quad b=ccos A+acos C,quad c=acos B+bcos A.}$

Morlining trisektor teoremasi

The Morley triangle, resulting from the trisection of each interior angle. Bu a cheklangan bo'linish qoidasi.

Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, called the Morley triangle.

Figures inscribed in a triangle

Koniklar

As discussed above, every triangle has a unique inscribed circle (incircle) that is interior to the triangle and tangent to all three sides.

Every triangle has a unique Steiner inellipse which is interior to the triangle and tangent at the midpoints of the sides. Marden's theorem shows how to find the foci of this ellipse.^[33] This ellipse has the greatest area of any ellipse tangent to all three sides of the triangle.

The Mandart inellipse of a triangle is the ellipse inscribed within the triangle tangent to its sides at the contact points of its excircles.

For any ellipse inscribed in a triangle ABC, let the foci be P va Q. Keyin^[34]

${displaystyle {frac {{overline {PA}}cdot {overline {QA}}}{{overline {CA}}cdot {overline {AB}}}}+{frac {{overline {PB}}cdot {overline {QB}}}{{overline {AB}}cdot {overline {BC}}}}+{frac {{overline {PC}}cdot {overline {QC}}}{{overline {BC}}cdot {overline {CA}}}}=1.}$

Qavariq ko'pburchak

Every convex polygon with area T can be inscribed in a triangle of area at most equal to 2T. Equality holds (exclusively) for a parallelogram.^[35]

Olti burchakli

The Lemoin olti burchakli a cyclic hexagon with vertices given by the six intersections of the sides of a triangle with the three lines that are parallel to the sides and that pass through its simmedian nuqtasi. In either its simple form or its self-intersecting form, the Lemoine hexagon is interior to the triangle with two vertices on each side of the triangle.

Kvadratchalar

Every acute triangle has three inscribed squares (squares in its interior such that all four of a square's vertices lie on a side of the triangle, so two of them lie on the same side and hence one side of the square coincides with part of a side of the triangle). In a right triangle two of the squares coincide and have a vertex at the triangle's right angle, so a right triangle has only two aniq inscribed squares. An obtuse triangle has only one inscribed square, with a side coinciding with part of the triangle's longest side. Within a given triangle, a longer common side is associated with a smaller inscribed square. If an inscribed square has side of length q_a and the triangle has a side of length a, part of which side coincides with a side of the square, then q_a, a, the altitude h_a from the side a, and the triangle's area T ga bog'liqdir^[36][37]

${displaystyle q_{a}={frac {2Ta}{a^{2}+2T}}={frac {ah_{a}}{a+h_{a}}}.}$

The largest possible ratio of the area of the inscribed square to the area of the triangle is 1/2, which occurs when a² = 2T, q = a/2, and the altitude of the triangle from the base of length a ga teng a. The smallest possible ratio of the side of one inscribed square to the side of another in the same non-obtuse triangle is ${displaystyle 2{sqrt {2}}/3=0.94....}$^[37] Both of these extreme cases occur for the isosceles right triangle.

Uchburchaklar

From an interior point in a reference triangle, the nearest points on the three sides serve as the vertices of the pedal triangle shu nuqtadan. If the interior point is the circumcenter of the reference triangle, the vertices of the pedal triangle are the midpoints of the reference triangle's sides, and so the pedal triangle is called the midpoint triangle or medial triangle. The midpoint triangle subdivides the reference triangle into four congruent triangles which are similar to the reference triangle.

The Gergonne triangle or intouch triangle of a reference triangle has its vertices at the three points of tangency of the reference triangle's sides with its incircle. The extouch triangle of a reference triangle has its vertices at the points of tangency of the reference triangle's excircles with its sides (not extended).

Figures circumscribed about a triangle

The tangential triangle of a reference triangle (other than a right triangle) is the triangle whose sides are on the chiziqli chiziqlar to the reference triangle's circumcircle at its vertices.

As mentioned above, every triangle has a unique circumcircle, a circle passing through all three vertices, whose center is the intersection of the perpendicular bisectors of the triangle's sides.

Further, every triangle has a unique Shtayner atrofi, which passes through the triangle's vertices and has its center at the triangle's centroid. Of all ellipses going through the triangle's vertices, it has the smallest area.

The Kiepert hyperbola is the unique conic which passes through the triangle's three vertices, its centroid, and its circumcenter.

Of all triangles contained in a given convex polygon, there exists a triangle with maximal area whose vertices are all vertices of the given polygon.^[38]

Specifying the location of a point in a triangle

One way to identify locations of points in (or outside) a triangle is to place the triangle in an arbitrary location and orientation in the Dekart tekisligi, and to use Cartesian coordinates. While convenient for many purposes, this approach has the disadvantage of all points' coordinate values being dependent on the arbitrary placement in the plane.

Two systems avoid that feature, so that the coordinates of a point are not affected by moving the triangle, rotating it, or reflecting it as in a mirror, any of which give a congruent triangle, or even by rescaling it to give a similar triangle:

• Uch chiziqli koordinatalar specify the relative distances of a point from the sides, so that coordinates ${displaystyle x:y:z}$ indicate that the ratio of the distance of the point from the first side to its distance from the second side is ${displaystyle x:y}$ , va boshqalar.
• Barycentric coordinates shaklning ${displaystyle alpha : eta :gamma }$ specify the point's location by the relative weights that would have to be put on the three vertices in order to balance the otherwise weightless triangle on the given point.

Non-planar triangles

A non-planar triangle is a triangle which is not contained in a (flat) plane. Some examples of non-planar triangles in non-Euclidean geometries are spherical triangles yilda sferik geometriya va giperbolik uchburchaklar yilda giperbolik geometriya.

While the measures of the internal angles in planar triangles always sum to 180°, a hyperbolic triangle has measures of angles that sum to less than 180°, and a spherical triangle has measures of angles that sum to more than 180°. A hyperbolic triangle can be obtained by drawing on a negatively curved surface, such as a saddle surface, and a spherical triangle can be obtained by drawing on a positively curved surface such as a soha. Thus, if one draws a giant triangle on the surface of the Earth, one will find that the sum of the measures of its angles is greater than 180°; in fact it will be between 180° and 540°.^[39] In particular it is possible to draw a triangle on a sphere such that the measure of each of its internal angles is equal to 90°, adding up to a total of 270°.

Specifically, on a sphere the sum of the angles of a triangle is

180° × (1 + 4f),

qayerda f is the fraction of the sphere's area which is enclosed by the triangle. For example, suppose that we draw a triangle on the Earth's surface with vertices at the North Pole, at a point on the equator at 0° longitude, and a point on the equator at 90° West longitude. The katta doira line between the latter two points is the equator, and the great circle line between either of those points and the North Pole is a line of longitude; so there are right angles at the two points on the equator. Moreover, the angle at the North Pole is also 90° because the other two vertices differ by 90° of longitude. So the sum of the angles in this triangle is 90° + 90° + 90° = 270°. The triangle encloses 1/4 of the northern hemisphere (90°/360° as viewed from the North Pole) and therefore 1/8 of the Earth's surface, so in the formula f = 1/8; thus the formula correctly gives the sum of the triangle's angles as 270°.

From the above angle sum formula we can also see that the Earth's surface is locally flat: If we draw an arbitrarily small triangle in the neighborhood of one point on the Earth's surface, the fraction f of the Earth's surface which is enclosed by the triangle will be arbitrarily close to zero. In this case the angle sum formula simplifies to 180°, which we know is what Euclidean geometry tells us for triangles on a flat surface.

Triangles in construction

The Flatiron binosi in New York is shaped like a uchburchak prizma

To'rtburchaklar have been the most popular and common geometric form for buildings since the shape is easy to stack and organize; as a standard, it is easy to design furniture and fixtures to fit inside rectangularly shaped buildings. But triangles, while more difficult to use conceptually, provide a great deal of strength. As computer technology helps me'morlar design creative new buildings, triangular shapes are becoming increasingly prevalent as parts of buildings and as the primary shape for some types of skyscrapers as well as building materials. In Tokyo in 1989, architects had wondered whether it was possible to build a 500-story tower to provide affordable office space for this densely packed city, but with the danger to buildings from zilzilalar, architects considered that a triangular shape would be necessary if such a building were to be built.^[40]

Yilda Nyu-York shahri, kabi Broadway crisscrosses major avenues, the resulting blocks are cut like triangles, and buildings have been built on these shapes; one such building is the triangularly shaped Flatiron binosi which real estate people admit has a "warren of awkward spaces that do not easily accommodate modern office furniture" but that has not prevented the structure from becoming a landmark icon.^[41] Designers have made houses in Norvegiya using triangular themes.^[42] Triangle shapes have appeared in churches^[43] as well as public buildings including colleges^[44] as well as supports for innovative home designs.^[45]

Triangles are sturdy; while a rectangle can collapse into a parallelogram from pressure to one of its points, triangles have a natural strength which supports structures against lateral pressures. A triangle will not change shape unless its sides are bent or extended or broken or if its joints break; in essence, each of the three sides supports the other two. A rectangle, in contrast, is more dependent on the strength of its joints in a structural sense. Some innovative designers have proposed making g'isht not out of rectangles, but with triangular shapes which can be combined in three dimensions.^[46] It is likely that triangles will be used increasingly in new ways as architecture increases in complexity. It is important to remember that triangles are strong in terms of rigidity, but while packed in a tessellating arrangement triangles are not as strong as olti burchakli under compression (hence the prevalence of hexagonal forms in tabiat ). Tessellated triangles still maintain superior strength for konsol however, and this is the basis for one of the strongest man made structures, the tetrahedral truss.

Shuningdek qarang

Izohlar

Adabiyotlar

Tashqi havolalar

• Ivanov, A.B. (2001) [1994], "Uchburchak", Matematika entsiklopediyasi, EMS Press
• Klark Kimberling: Uchburchak markazlari entsiklopediyasi. Har qanday uchburchak bilan bog'liq bo'lgan 5200 ta qiziqarli fikrlarni sanab o'tadi.