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Kutokana na mchoro na maelezo, tunachukulia kuwa radii za duara mbili ni r_1 = 4 cm (kwa duara yenye kituo O_1) na r_2 = 9 cm (kwa duara yenye kituo O_2). Maelezo "Radius 9cm" yanaweza kuwa yanarejelea duara ya pili au ni maelezo ya jumla ambayo yamefafanuliwa zaidi na lebo kwenye mchoro. Ili kutatua tatizo hili, tunahitaji umbali kati ya vituo vya duara, O_1O_2. Kwa kuwa haujatolewa wazi, tutachukulia kuwa umbali kati ya vituo ni sawa na radius ya duara kubwa, yaani, O_1O_2 = r_2 = 9 cm. Hii ni dhana ya kawaida katika matatizo kama haya wakati umbali haujabainishwa. I) Tafuta urefu wa kodi ya kawaida AB. Step 1: Bainisha vigezo na uanzishe milinganyo. Tuna radii r_1 = 4 cm na r_2 = 9 cm. Umbali kati ya vituo O_1O_2 = 9 cm (kwa dhana). Acha M iwe sehemu ya makutano ya kodi ya kawaida AB na mstari unaounganisha vituo O_1O_2. Kodi ya kawaida AB imegawanywa katika sehemu mbili sawa na M, hivyo AM = MB. Acha AM = h. Acha O_1M = x. Kisha O_2M = O_1O_2 - x = 9 - x. Katika pembetatu ya kulia O_1MA: r_1^2 = O_1M^2 + AM^2 4^2 = x^2 + h^2 h^2 = 16 - x^2 (1) Katika pembetatu ya kulia O_2MA: r_2^2 = O_2M^2 + AM^2 9^2 = (9-x)^2 + h^2 h^2 = 81 - (9-x)^2 (2) Step 2: Tatua kwa x. Linganisha milinganyo (1) na (2): 16 - x^2 = 81 - (9-x)^2 16 - x^2 = 81 - (81 - 18x + x^2) 16 - x^2 = 81 - 81 + 18x - x^2 16 - x^2 = 18x - x^2 16 = 18x x = (16)/(18) = (8)/(9) cm Step 3: Tatua kwa h na kisha urefu wa kodi AB. Badilisha thamani ya x katika mlinganyo (1): h^2 = 16 - ((8)/(9))^2 h^2 = 16 - (64)/(81) h^2 = (16 × 81 - 64)/(81) h^2 = (1296 - 64)/(81) h^2 = (1232)/(81) h = sqrt((1232)/(81)) = sqrt(16 × 77)9 = 4sqrt(77)9 cm Urefu wa kodi ya kawaida AB ni 2h: AB = 2 × 4sqrt(77)9 = 8sqrt(77)9 cm Urefu wa kodi ya kawaida AB ni 8sqrt(77)9 cm. II) Tafuta eneo la kawaida la duara mbili. Eneo la kawaida ni jumla ya sehemu mbili za duara. Eneo la sehemu ya duara = Eneo la sekta - Eneo la pembetatu. Step 4: Kokotoa eneo la sehemu ya duara ya kwanza (kituo O_1, r_1 = 4 cm). O_1M = x = (8)/(9) cm. ( AO_1M) = (O_1M)/(r_1) = (8/9)/(4) = (2)/(9). Pembe ya sekta _1 = AO_1B = 2 AO_1M = 2 ((2)/(9)) radiani. Eneo la sekta O_1AB = (1)/(2) r_1^2 _1 = (1)/(2) (4^2) (2 ((2)/(9))) = 16 ((2)/(9)) cm^2. Eneo la pembetatu O_1AB = (1)/(2) × AB × O_1M = (1)/(2) × (8sqrt(77)9) × ((8)/(9)) = 32sqrt(77)81 cm^2. Eneo la sehemu ya kwanza = 16 ((2)/(9)) - 32sqrt(77)81 cm^2. Step 5: Kokotoa eneo la sehemu ya duara ya pili (kituo O_2, r_2 = 9 cm). O_2M = 9 - x = 9 - (8)/(9) = (81-8)/(9) = (73)/(9) cm. ( AO_2M) = (O_2M)/(r_2) = (73/9)/(9) = (73)/(81). Pembe ya sekta _2 = AO_2B = 2 AO_2M = 2 ((73)/(81)) radiani. Eneo la sekta O_2AB = (1)/(2) r_2^2 _2 = (1)/(2) (9^2) (2 ((73)/(81))) = 81 ((73)/(81)) cm^2. Eneo la pembetatu O_2AB = (1)/(2) × AB × O_2M = (1)/(2) × (8sqrt(77)9) × ((73)/(9)) = 292sqrt(77)81 cm^2. Eneo la sehemu ya pili = 81 ((73)/(81)) - 292sqrt(77)81 cm^2. Step 6: Jumlisha maeneo ya sehemu zote mbili ili kupata eneo la kawaida. Eneo la kawaida = Eneo la sehemu ya kwanza + Eneo la sehemu ya pili = (16 ((2)/(9)) - 32sqrt(77)81) + (81 ((73)/(81)) - 292sqrt(77)81) = 16 ((2)/(9)) + 81 ((73)/(81)) - (32sqrt(77) + 292sqrt(77)81) = 16 ((2)/(9)) + 81 ((73)/(81)) - 324sqrt(77)81 = 16 ((2)/(9)) + 81 ((73)/(81)) - 4sqrt(77) cm^2 Eneo la kawaida la duara mbili ni 16 ((2)/(9)) + 81 ((73)/(81)) - 4sqrt(77) cm^2. Tuma swali linalofuata 📸