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 2(a): Bodi ya matangazo ina umbo la pembetatu pacha (isosceles triangle) yenye mzunguko wa 40 m na kila upande sawa ni 15 m. Tunahitaji kupata eneo la bodi hiyo hadi mita ya mraba iliyo karibu zaidi. Step 1: Tafuta urefu wa upande wa tatu (base). Katika pembetatu pacha, pande mbili ni sawa. Tuseme pande hizo ni a = 15 m na b = 15 m. Mzunguko P = a + b + c, ambapo c ni upande wa tatu. 40 m = 15 m + 15 m + c 40 m = 30 m + c c = 40 m - 30 m c = 10 m Hivyo, urefu wa upande wa tatu (base) ni 10 m. Step 2: Tafuta urefu wa kimo (height) cha pembetatu. Katika pembetatu pacha, kimo kinachoshuka kutoka kwenye ncha ya juu hadi kwenye base hugawanya base katika sehemu mbili sawa. Hii inaunda pembetatu mbili za pembe mraba. Hypotenuse ya pembetatu ya pembe mraba ni 15 m (upande sawa wa pembetatu pacha). Upande mmoja wa pembetatu ya pembe mraba ni nusu ya base, yaani 10 m2 = 5 m. Tumia kanuni ya Pythagoras (h^2 + base_half^2 = hypotenuse^2): h^2 + (5 m)^2 = (15 m)^2 h^2 + 25 m^2 = 225 m^2 h^2 = 225 m^2 - 25 m^2 h^2 = 200 m^2 h = sqrt(200) m h = 10sqrt(2) m h ≈ 10 × 1.41421 m h ≈ 14.1421 m Step 3: Hesabu eneo la pembetatu. Eneo la pembetatu A = (1)/(2) × base × height. A = (1)/(2) × 10 m × 10sqrt(2) m A = 5 m × 10sqrt(2) m A = 50sqrt(2) m^2 A ≈ 50 × 1.41421 m^2 A ≈ 70.7105 m^2 Step 4: Zungusha eneo hadi mita ya mraba iliyo karibu zaidi. A ≈ 71 m^2 Eneo la bodi ni 71 m^2. What's next?