HOSILANI GEOMETRIK MASALALARNI YECHISHDA TADBIQI

INTERNATIONAL JOURNAL OF SCIENTIFIC RESEARCHERS ISSN: 3030-332X Impact factor: 8,293 Volume 14, issue 2, November 2025 https://wordlyknowledge.uz/index.php/IJSR worldly knowledge Index: google scholar, research gate, research bib, zenodo, open aire. https://scholar.google.com/scholar?hl=ru&as_sdt=0%2C5&q=wosjournals.com&btnG https://www.researchgate.net/profile/Worldly-Knowledge https://journalseeker.researchbib.com/view/issn/3030-332X 46 HOSILANI GEOMETRIK MASALALARNI YECHISHDA TADBIQI Umirov Baxriddin Xayrillayevich, Xayrullayev Dilshod Baxritdinovich Shahrisabz “Temurbeklar maktabi” harbiy akademik litseyi matematika fani o‘qituvchilari Annotatsiya: Hosila mavzusi o‘rta maktab va akademik litseylarda Algebra fanidan o‘tiladi. Lekin geometriya fanida gapirilmaydi. Hozirgi kunda matematikadan milliy sertifikat, hamda attestatsiya savollarida geometriya fanidan qo‘yilgan savollarni yechish masalasi hosila mavzusida borib taqaladi. Shu maqsadda ushbu maqolada geometriyadan masalalar yechishda hosilani tadbiqiga doir masalalar yechib ko‘rsatilgan. Kalit so‘zlar: Hosila, grafik, masala, yechim, eng katta, hajim. Hosilani hayotda tadbiqi juda ko‘p uchraydi. Geometrik jismlarni o‘lchamlarini yasamasdan oldin hosila yordamida aniqlanadi. Shuning uchun geometriya fanidan masalalarni yechishda hosiladan foydalanamiz. Masala-1. Tomonlari 21 va 16 ga teng bo‘lgan to‘g‘ri to‘rtburchak shaklidagi tunkaning to‘rtta uchidan tomoni h ga teng bo‘lgan kvadrat qirqib olindi quti shakliga keltirildi. Qutining hajmi eng katta bo‘ladigan h ni toping. Berilgan: AB=21 BC=16 AK=KM=h KN=16−2h FF1=16−2h h−? Yechish: Shakil to‘g‘ri burchakli parallelepiped shakliga keladi eng katta hajimga ega bo‘lish uchun quyidagi formulani olamiz: Vh=21−2h 16−2hh INTERNATIONAL JOURNAL OF SCIENTIFIC RESEARCHERS ISSN: 3030-332X Impact factor: 8,293 Volume 14, issue 2, November 2025 https://wordlyknowledge.uz/index.php/IJSR worldly knowledge Index: google scholar, research gate, research bib, zenodo, open aire. https://scholar.google.com/scholar?hl=ru&as_sdt=0%2C5&q=wosjournals.com&btnG https://www.researchgate.net/profile/Worldly-Knowledge https://journalseeker.researchbib.com/view/issn/3030-332X 47 bu ifodani soddalashtirsak Vh=336h−74h2+4h3 ko‘rinishga keladi. Ifodadan hosila olamiz va V'h=0tenglamani yechamiz. V'h=336−148h+12h2 12h2−148h+336=0 :4 3h2−74h+84=0 bundan h1=3va h2=91 3yechimlar chiqadi. Masala shartiga ko‘ra h=3bo‘lganda eng katta hajimga ega bo‘ladi. Masala-2. Radiusi Rga teng bo‘lgan yarim aylanaga CDEFto‘g‘ri to‘rtburchak ichki chizilgan SCDEFning eng katta qiymatini toping. Berilgan: R SCDEFmax−? Yechish: EF=b,OF=a 2∆OFE dan R2=b2+a2 4⟺ 4R2=4b2+a2tenglikdan a=2 R2−b2 S=ab=2b R2−b2=2 b2R2−b4 Sb=2 b2R2−b4ifodadan hosila olamiz. S'b=2bR2−4b3 b2R2−b4 S'b=0deb yechamiz 2bR2−4b3=0va b2R2−b4≠0tenglikdan b=R2 a=2 R2−R2 2=2R SCDEF= 2R∙ R2=R2 Javob: R2 Masala-3. Radiusi 6 2ga teng bo‘lgan sharga ichki chizilgan eng katta hajmli konus asosining radiusini toping. Berilgan: AO=62 AD=R−? INTERNATIONAL JOURNAL OF SCIENTIFIC RESEARCHERS ISSN: 3030-332X Impact factor: 8,293 Volume 14, issue 2, November 2025 https://wordlyknowledge.uz/index.php/IJSR worldly knowledge Index: google scholar, research gate, research bib, zenodo, open aire. https://scholar.google.com/scholar?hl=ru&as_sdt=0%2C5&q=wosjournals.com&btnG https://www.researchgate.net/profile/Worldly-Knowledge https://journalseeker.researchbib.com/view/issn/3030-332X 48 Yechish: ∆AODdan AO2=AD2+OD2 6 2 2=R2+H−6 2 2 72=R2+H2−12 2H+72 12 2H−H2=R2 Vk=1 3πR2H=1 3π12 2H−H2H=4 2πH2−1 3πH3 Vk=4 2πH2−1 3πH3ifodadan hosila olamiz. V'H=8 2πH−πH2=πH8 2−H V'H=0danπH≠0vaH=8 2 R2=12 2∙8 2−128=12∙16−128=64 R2=64⟺R=8 Javob:8. Masala-4. Sharga eng katta hajmli silindr ichki chizilgan R−shar radiusi, r−silindr asosi radiusi. a) r Rnisbatni toping. b) Shar hajmining silindr hajmiga nisbatini toping. Berilgan: AO=R−shar radiusi AE=r−silindr asosi radiusi. OE=H 2 r R−?,Vsh Vsil−? Yechish: ∆AEO danR2=r2+H2 4⟺4R2=4r2+H2⟺4R2−H2=4r2 r2=R2−1 4H21 Vsil=πr2H=π R2−1 4H2H=πR2H−πH3 4 INTERNATIONAL JOURNAL OF SCIENTIFIC RESEARCHERS ISSN: 3030-332X Impact factor: 8,293 Volume 14, issue 2, November 2025 https://wordlyknowledge.uz/index.php/IJSR worldly knowledge Index: google scholar, research gate, research bib, zenodo, open aire. https://scholar.google.com/scholar?hl=ru&as_sdt=0%2C5&q=wosjournals.com&btnG https://www.researchgate.net/profile/Worldly-Knowledge https://journalseeker.researchbib.com/view/issn/3030-332X 49 Vsil=πR2H−πH3 4ifodadan hosila olamiz Vsil 'H= πR2H−πH3 4'=πR2−3π 4H2 Vsil 'H=0tenglikdan πR2−3π 4H2=0⇔H=2R 3(2) (2) ifodani (1) ifodaga qo‘yamiz r2=R2−R2 3=2R2 3⇔r= 2R 3=6R 3 a) r R=6R 3:R= 6 3 b) Vsh=4 3πR3Vsil=πr2H=43πR3 9 Vsh Vsil=4 3πR3:4 3πR3 9=4 3πR3∙9 43πR3=33= 3 Javob: a) 6 3b) 3 Masala-5. Apofemasi 3 3 ga teng bo‘lgan muntazam to‘rtburchakli piramida hajmining eng katta qiymatini toping. Berilgan: SABCD−muntazam piramida SE=3 3 Vmax−? Yechish: ∆SEOdan SE2=SO2+OE2⇔3 3 2==H2+a2 4 108=4H2+a2⇔a2=108−4H2 VpH=1 3a2H=1 3108−4H2H=36H−4 3H3 VpH=36H−4 3H3ifodadan hosila olamiz. Vp 'H=36−4H2tenglikdan 36−4H2=0 tenglamadan H2=9⇔H=3 a2=108−36=72 Vp=1 3a2H=1 3∙72∙3=72 Javob: Vmax=72 Xulosa: Geometriya fani o‘sib kelayotgan yosh avlodni kamol toptirishda o‘quv fani sifatida keng imkoniyatlarga ega. U o‘quvchi tafakkurini rivojlantirib, ularni aqlini peshlaydi, uni tartibga soladi, o‘quvchilarga maqsadga yo‘naltirilganlik, mantiqiy fikirlash topqirlik INTERNATIONAL JOURNAL OF SCIENTIFIC RESEARCHERS ISSN: 3030-332X Impact factor: 8,293 Volume 14, issue 2, November 2025 https://wordlyknowledge.uz/index.php/IJSR worldly knowledge Index: google scholar, research gate, research bib, zenodo, open aire. https://scholar.google.com/scholar?hl=ru&as_sdt=0%2C5&q=wosjournals.com&btnG https://www.researchgate.net/profile/Worldly-Knowledge https://journalseeker.researchbib.com/view/issn/3030-332X 50 xislatlarini shakillantirib boradi. Shunday ekan masalalar matematika mutaxassisligida ta’lim olayotgan talabalar va o‘quvchilarning hosila yordamida geometrik masalalarni yechishda qo‘yiladigan talablarni oson o‘rganishga yordam beradi. Foydalanilgan adabiyotlar: 1. I.Isroilov, Z.Pashayev “Geometriya” Toshkent-2010. 2. Sh.Jumayev, B.Ro‘ziyev va boshqalar “Geometriya” Toshkent-2024. 3. 2025-yil BMBA savollar to‘plami.