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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Haya, twende na maswali haya mawili. Swali la 1: Eneo la uso la hori ya kulishia Step 1: Tambua vipimo vya hori. Hori ina urefu (L) wa 2 m, upana (W) wa 1.8 m, na urefu wa jumla (H) wa 1 m. Imeelezwa kuwa hori iko wazi juu. Umbo la hori ni mchemraba wenye sehemu ya chini ya nusu-silinda. Kutoka kwenye mchoro, upana wa 1.8 m ni kipenyo cha nusu-duara chini. Hivyo, rediasi (r) ya nusu-duara ni 1.8 m2 = 0.9 m. Urefu wa sehemu ya mstatili juu ya nusu-duara ni 1 m - 0.9 m = 0.1 m. Step 2: Kokotoa eneo la nyuso mbalimbali. Hori ina nyuso zifuatazo: • Eneo la chini lililopinda (nusu-silinda): Hili ni eneo la uso wa pembeni wa nusu-silinda. A_chini = (1)/(2) × (2 r L) = r L A_chini = × 0.9 m × 2 m = 1.8 m^2 • Eneo la kuta mbili za pembeni (mstatili): Kila ukuta una urefu wa L = 2 m na urefu wa sehemu ya mstatili h = 0.1 m. A_kuta = 2 × (L × h) = 2 × (2 m × 0.1 m) = 2 × 0.2 m^2 = 0.4 m^2 • Eneo la nyuso mbili za mwisho (cross-sections): Kila uso wa mwisho una mstatili na nusu-duara. Eneo la mstatili = 1.8 m × 0.1 m = 0.18 m^2. Eneo la nusu-duara = (1)/(2) r^2 = (1)/(2) (0.9 m)^2 = (1)/(2) (0.81) m^2 = 0.405 m^2. Eneo la uso mmoja wa mwisho = 0.18 m^2 + 0.405 m^2. Eneo la nyuso mbili za mwisho = 2 × (0.18 m^2 + 0.405 m^2) = 0.36 m^2 + 0.81 m^2. Step 3: Kokotoa jumla ya eneo la uso. Jumla ya eneo la uso (A_jumla) ni jumla ya maeneo yote yaliyokokotolewa, kwani hori iko wazi juu. A_jumla = A_chini + A_kuta + A_nyuso\_za\_mwisho A_jumla = 1.8 m^2 + 0.4 m^2 + (0.36 m^2 + 0.81 m^2) A_jumla = (1.8 + 0.81 ) m^2 + (0.4 + 0.36) m^2 A_jumla = 2.61 m^2 + 0.76 m^2 Step 4: Badilisha thamani ya na fanya hesabu. Tutatumia ≈ 3.14159. A_jumla = (2.61 × 3.14159) m^2 + 0.76 m^2 A_jumla = 8.2005099 m^2 + 0.76 m^2 A_jumla = 8.9605099 m^2 Kuzungusha hadi sehemu mbili za desimali: 8.96 m^2 Swali la 2: Eneo la uso la piramidi yenye msingi wa mraba Step 1: Tambua vipimo vya piramidi. Piramidi ina msingi wa mraba wenye urefu wa upande (s) wa 9 cm. Urefu mteremko (l) wa nyuso za pembeni ni 14 cm. Step 2: Kokotoa eneo la msingi. Msingi ni mraba. A_msingi = s^2 = (9 cm)^2 = 81 cm^2 Step 3: Kokotoa eneo la nyuso nne za pembeni (triangles). Kila uso wa pembeni ni pembetatu yenye msingi s = 9 cm na urefu mteremko l = 14 cm. Eneo la pembetatu moja = (1)/(2) × msingi × urefu mteremko A_pembetatu = (1)/(2) × 9 cm × 14 cm = 9 cm × 7 cm = 63 cm^2 Eneo la nyuso nne za pembeni = 4 × A_pembetatu A_pembeni = 4 × 63 cm^2 = 252 cm^2 Step 4: Kokotoa jumla ya eneo la uso la piramidi. Jumla ya eneo la uso (A_jumla) ni jumla ya eneo la msingi na eneo la nyuso nne za pembeni. A_jumla = A_msingi + A_pembeni A_jumla = 81 cm^2 + 252 cm^2 A_jumla = 333 cm^2 333 cm^2 That's 2 down. 3 left today — send the next one.