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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Hapa kuna suluhisho la swali la 1): 1) Tafuta eneo la poligoni ifuatayo yenye kituo O. Poligoni hii imeundwa na sehemu mbili: heksagoni ya kawaida (regular hexagon) na pembetatu ya juu (isosceles triangle). Kituo O ni kituo cha heksagoni. Step 1: Tambua vipimo vilivyotolewa. Apothem ya heksagoni (umbali kutoka kituo O hadi katikati ya upande mmoja, perpendicular) ni a = 2.55 cm. Step 2: Kokotoa urefu wa upande (s) wa heksagoni. Kwa heksagoni ya kawaida, apothem a inahusiana na urefu wa upande s kwa fomula a = sqrt(3)2s. Hivyo, tunaweza kupata s: s = (2a)/(sqrt(3)) Weka thamani ya a: s = (2 × 2.55)/(sqrt(3)) = (5.1)/(sqrt(3)) cm Step 3: Kokotoa eneo la heksagoni. Eneo la heksagoni ya kawaida ni A_heksagoni = (1)/(2) P a, ambapo P ni mzingo (P=6s). A_heksagoni = (1)/(2) (6s) a = 3sa Weka thamani za s na a: A_heksagoni = 3 × (5.1)/(sqrt(3)) × 2.55 A_heksagoni = (3 × 5.1 × 2.55)/(sqrt(3)) = (39.015)/(sqrt(3)) cm^2 Step 4: Bainisha urefu wa pembetatu ya juu (h_t). Msingi wa pembetatu ya juu ni urefu wa upande s wa heksagoni. Urefu wa pembetatu ya juu (h_t) haujatolewa wazi, lakini tunaweza kuukadiria kutoka kwenye mchoro kwa kutumia gridi. Kutoka kwenye mchoro, apothem a = 2.55 cm inalingana na takriban vitengo 2.5 vya gridi. Urefu wa pembetatu ya juu unaonekana kuwa takriban vitengo 2 vya gridi. Kwa hiyo, tunakadiria h_t kwa kutumia uwiano: (h_t)/(a) ≈ 2 vitengo vya gridi2.5 vitengo vya gridi h_t ≈ (2)/(2.5) × a = (4)/(5) × 2.55 = 0.8 × 2.55 = 2.04 cm Step 5: Kokotoa eneo la pembetatu ya juu. Eneo la pembetatu ni A_pembetatu = (1)/(2) × msingi × urefu. A_pembetatu\_ya\_juu = (1)/(2) × s × h_t Weka thamani za s na h_t: A_pembetatu\_ya\_juu = (1)/(2) × (5.1)/(sqrt(3)) × 2.04 A_pembetatu\_ya\_juu = (5.1 × 1.02)/(sqrt(3)) = (5.202)/(sqrt(3)) cm^2 Step 6: Kokotoa eneo lote la poligoni. Eneo lote ni jumla ya eneo la heksagoni na eneo la pembetatu ya juu. A_jumla = A_heksagoni + A_pembetatu\_ya\_juu A_jumla = (39.015)/(sqrt(3)) + (5.202)/(sqrt(3)) = (39.015 + 5.202)/(sqrt(3)) = (44.217)/(sqrt(3)) cm^2 Ili kupata thamani ya namba, tunazidisha na sqrt(3)sqrt(3) au tunatumia sqrt(3) ≈ 1.73205: A_jumla = 44.217 sqrt(3)3 = 14.739 sqrt(3) cm^2 A_jumla ≈ 14.739 × 1.73205 ≈ 25.524 cm^2 Eneo la mwisho ni: 25.524 cm^2 Last free one today — make it count tomorrow, or type /upgrade for unlimited.