uchburchaklar va aylana

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1662926439.doc g b a sin 2 sin 2 sin 2 c в a r = = = в в 2 c p s r d = с в а h h h r 1 1 1 1 + + = 2 1 в c в a h h h , , в в d = s авс r 4 в 12 12 s авс r 4 = 2 3 x 3 2 1 4 3 2 x 2 1 2 1 3 4 12 4 3 2 x 2 x 3 4 12 3 4 3 / 4 4 3 2 x 3 3 2 2 сd аd + 64 36 + 2 1 2 1 сd c к g b a sin sin sin c в a = = 2 1 в 2 1 в 2 a 2 c 2 a a a g 2 sin 2 ) 2 …

hizilgan aylana diametri 16 sm. uchburchak tomonlari uzunliklari ko‘paytmasini toping. yechish: a, , c – uchburchak tomonlari; r – tashqi chizilgan aylana radiusi. shartga ko‘ra 2r = 16 sm, sδ = 10 sm2. formuladan a c = 4sδr = 320 cm3 javob: 320 cm3 2 – misol. 1200 li teng yonli uchburchak yuzini toping. bu yerda ichki chizilgan aylana radiusi yechish: shartga ko‘ra < acb = 1200, od = r = ac=bc=x deb olsak, ma’lumki <acd = <bcd=600, ad=ac sin600 = demak ac = x . avval x ni topamiz. abc uchburchakni yuzini topishni quyi- dagi ikki yuza formulasidan: sδ = ca · cb sin1200 = ; sδ =p · r = (2ac +ab) · r = (2x+ +x ) . ushbu tenglamadan = (2+ ) · noma’lum x ni topamiz: x= 2(2+ ) · . u holda sδ = = 2 (7+4 ) cm2. javob: 2 (7 + 4 …

chakka ichki chizilgan aylana radiusi 4 sm. tomonlaridan biri aylanaga o‘rinish nuqtasida 6 sm, 8 sm, bo‘laklarga bo‘lingan. uchburchakni qolgan ikki tomonini toping. yechish. ad= 6 sm, cd= 8 sm bo‘lsin. abc uchburchakni ac va bc tomonlarini aniqlash uchun eb=bf =x, shuningdek abe=ad= 6sm, cf=cd=8 sm ekanidan foydalanish mumkin. buning uchun uchburchakni yuzini topishishining quyidagi ikki formulasidan foydalanamiz: sδ = p ∙ r va sδ = , r – uchburchakni yarim perimetri, shunga ko‘ra p = (ae + ad + dc + cf + fb + be) = = (28 + 2x) = 14 + x. navbatdagi tenglamani hosil qilamiz: 4 (14 +x) = bu yerdan x = 7 sm, u holda ab=abe +x=13sm, bc=cf +x = 15 sm. javob: 13sm, 15 sm. 7 – misol. tomoni a ga teng, teng tomonli uchburchakka doira ichki chizilgan. bu doiraga va berilgan uchburchak tomonlariga o‘rinuvchi yana uchta doira ichki chchizilgan va bu …

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1662926439.doc g b a sin 2 sin 2 sin 2 c в a r = = = в в 2 c p s r d = с в а h h h r 1 1 1 1 + + = 2 1 в c в a h h h , , в в d = s авс r 4 в 12 12 s авс r 4 = 2 3 x 3 2 1 4 3 2 x 2 1 2 1 3 4 12 4 3 2 x 2 x 3 4 12 3 4 3 / 4 4 3 2 x 3 3 2 2 сd аd + 64 36 + 2 1 2 1 сd c к g …

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