This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Sawa Victor, tuendelee na maswali haya. 2. One of the interior angles of a polygon is 84°. If each of the other angles is 147°, how many sides does the polygon have? Tunahitaji kutumia formula ya jumla ya pembe za ndani za poligoni. Step 1: Andika formula ya jumla ya pembe za ndani za poligoni. Jumla ya pembe za ndani za poligoni yenye n pande ni (n-2) × 180^. Step 2: Unda mlinganyo kwa kutumia taarifa iliyotolewa. Poligoni ina pembe moja ya 84^ na pembe zingine (n-1) kila moja ikiwa 147^. Hivyo, jumla ya pembe ni 84^ + (n-1)147^. 84 + (n-1)147 = (n-2)180 Step 3: Rahisisha na tatua mlinganyo kwa n. 84 + 147n - 147 = 180n - 360 147n - 63 = 180n - 360 Kusanya n upande mmoja na nambari upande mwingine: 360 - 63 = 180n - 147n 297 = 33n n = (297)/(33) n = 9 Poligoni ina pande 9. Jibu la mwisho ni 9 pande. 3. Two circles of equal radius have their centres at O (0,0) and A (5,0) respectively as shown below. Given that angle PAQ = 120°. Calculate the area of the shaded region. Eneo lililotiwa kivuli ni eneo la makutano ya duru hizo mbili. Step 1: Tambua taarifa muhimu na uhusiano kati ya vitu. Vituo vya duara ni O(0,0) na A(5,0). Umbali kati ya vituo ni OA = 5. Duru zina radius sawa, tuseme r. P na Q ni sehemu za makutano ya duru hizo mbili. Pembe PAQ = 120^. Step 2: Tafuta radius r. Katika duara iliyo na kituo A, AP = AQ = r. Kwa kuwa PAQ = 120^, na APQ ni pembetatu pacha, tunaweza kugawanya PAQ kwa nusu. Acha M iwe katikati ya PQ. Basi AM PQ. Katika pembetatu ya kulia AMP, PAM = (120^)/(2) = 60^. Tunajua AM = AP ( PAM) = r (60^) = r × (1)/(2) = (r)/(2). Kwa ulinganifu, katika duara iliyo na kituo O, POQ pia ni 120^. Hivyo, OM = OP ( POM) = r (60^) = r × (1)/(2) = (r)/(2). Umbali kati ya vituo OA = OM + AM. 5 = (r)/(2) + (r)/(2) 5 = r Hivyo, radius ya kila duara ni r=5. Step 3: Kokotoa eneo la sehemu ya duara (circular segment) kwa kila duara. Eneo la sehemu ya duara = Eneo la sekta - Eneo la pembetatu. Kwa duara iliyo na kituo A, sekta APQ ina pembe ya kati = 120^. Eneo la sekta APQ = ()/(360^) r^2 = (120)/(360) (5^2) = (1)/(3) (25) = (25)/(3). Eneo la pembetatu APQ = (1)/(2) r^2 () = (1)/(2) (5^2) (120^) = (1)/(2) (25) sqrt(3)2 = 25sqrt(3)4. Eneo la sehemu ya duara kutoka duara A = (25)/(3) - 25sqrt(3)4. Step 4: Kokotoa eneo la jumla lililotiwa kivuli. Eneo lililotiwa kivuli ni jumla ya sehemu mbili za duara, moja kutoka kila duara. Kwa kuwa duru ni sawa na pembe za kati ni sawa, sehemu hizi mbili za duara ni sawa. Eneo lililotiwa kivuli = 2 × ( (25)/(3) - 25sqrt(3)4 ) = (50)/(3) - 50sqrt(3)4 = (50)/(3) - 25sqrt(3)2 Jibu la mwisho ni (50)/(3) - 25sqrt(3)2. That's 2 down. 3 left today — send the next one.