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The trough is open at the top. Calculate the Total surface area of feeding trough. Square based. Calculate the Surface area of pyramid.

Mathematics

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2 parts
1

8.96 m28.96 \text{ m}^2

2

333 cm2333 \text{ cm}^2

Step-by-step solution

8 steps

Haya, twende na maswali haya mawili.

Part 1· 4 steps

Eneo la uso la hori ya kulishia

  1. Tambua vipimo vya hori.

    Hori ina urefu (LL) wa 2 m2 \text{ m}, upana (WW) wa 1.8 m1.8 \text{ m}, na urefu wa jumla (HH) wa 1 m1 \text{ 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 m1.8 \text{ m} ni kipenyo cha nusu-duara chini. Hivyo, rediasi (rr) ya nusu-duara ni 1.8 m2=0.9 m\frac{1.8 \text{ m}}{2} = 0.9 \text{ m}.
    Urefu wa sehemu ya mstatili juu ya nusu-duara ni 1 m−0.9 m=0.1 m1 \text{ m} - 0.9 \text{ m} = 0.1 \text{ m}.

  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. Achini=12×(2πrL)=πrLA_{chini} = \frac{1}{2} \times (2 \pi r L) = \pi r L Achini=π×0.9 m×2 m=1.8π m2A_{chini} = \pi \times 0.9 \text{ m} \times 2 \text{ m} = 1.8 \pi \text{ m}^2
    • Eneo la kuta mbili za pembeni (mstatili): Kila ukuta una urefu wa L=2 mL = 2 \text{ m} na urefu wa sehemu ya mstatili h=0.1 mh = 0.1 \text{ m}. Akuta=2×(L×h)=2×(2 m×0.1 m)=2×0.2 m2=0.4 m2A_{kuta} = 2 \times (L \times h) = 2 \times (2 \text{ m} \times 0.1 \text{ m}) = 2 \times 0.2 \text{ m}^2 = 0.4 \text{ 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 m21.8 \text{ m} \times 0.1 \text{ m} = 0.18 \text{ m}^2. Eneo la nusu-duara = 12πr2=12π(0.9 m)2=12π(0.81) m2=0.405π m2\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (0.9 \text{ m})^2 = \frac{1}{2} \pi (0.81) \text{ m}^2 = 0.405 \pi \text{ m}^2. Eneo la uso mmoja wa mwisho = 0.18 m2+0.405π m20.18 \text{ m}^2 + 0.405 \pi \text{ m}^2. Eneo la nyuso mbili za mwisho = 2×(0.18 m2+0.405π m2)=0.36 m2+0.81π m22 \times (0.18 \text{ m}^2 + 0.405 \pi \text{ m}^2) = 0.36 \text{ m}^2 + 0.81 \pi \text{ m}^2.
  3. Kokotoa jumla ya eneo la uso.

    Jumla ya eneo la uso (AjumlaA_{jumla}) ni jumla ya maeneo yote yaliyokokotolewa, kwani hori iko wazi juu.

    Ajumla=Achini+Akuta+Anyuso_za_mwishoA_{jumla} = A_{chini} + A_{kuta} + A_{nyuso\_za\_mwisho} Ajumla=1.8π m2+0.4 m2+(0.36 m2+0.81π m2)A_{jumla} = 1.8 \pi \text{ m}^2 + 0.4 \text{ m}^2 + (0.36 \text{ m}^2 + 0.81 \pi \text{ m}^2) Ajumla=(1.8π+0.81π) m2+(0.4+0.36) m2A_{jumla} = (1.8 \pi + 0.81 \pi) \text{ m}^2 + (0.4 + 0.36) \text{ m}^2 Ajumla=2.61π m2+0.76 m2A_{jumla} = 2.61 \pi \text{ m}^2 + 0.76 \text{ m}^2
  4. Badilisha thamani ya π\pi na fanya hesabu.

    Tutatumia π≈3.14159\pi \approx 3.14159.

    Ajumla=(2.61×3.14159) m2+0.76 m2A_{jumla} = (2.61 \times 3.14159) \text{ m}^2 + 0.76 \text{ m}^2 Ajumla=8.2005099 m2+0.76 m2A_{jumla} = 8.2005099 \text{ m}^2 + 0.76 \text{ m}^2 Ajumla=8.9605099 m2A_{jumla} = 8.9605099 \text{ m}^2

    Kuzungusha hadi sehemu mbili za desimali:

    8.96 m2\boxed{8.96 \text{ m}^2}

Part 2

333 cm2333 \text{ cm}^2

Eneo la uso la piramidi yenye msingi wa mraba

  1. Tambua vipimo vya piramidi.

    Piramidi ina msingi wa mraba wenye urefu wa upande (ss) wa 9 cm9 \text{ cm}.
    Urefu mteremko (ll) wa nyuso za pembeni ni 14 cm14 \text{ cm}.

  2. Kokotoa eneo la msingi.

    Msingi ni mraba.

    Amsingi=s2=(9 cm)2=81 cm2A_{msingi} = s^2 = (9 \text{ cm})^2 = 81 \text{ cm}^2
  3. Kokotoa eneo la nyuso nne za pembeni (triangles).

    Kila uso wa pembeni ni pembetatu yenye msingi s=9 cms = 9 \text{ cm} na urefu mteremko l=14 cml = 14 \text{ cm}.
    Eneo la pembetatu moja = 12×msingi×urefu mteremko\frac{1}{2} \times \text{msingi} \times \text{urefu mteremko}

    Apembetatu=12×9 cm×14 cm=9 cm×7 cm=63 cm2A_{pembetatu} = \frac{1}{2} \times 9 \text{ cm} \times 14 \text{ cm} = 9 \text{ cm} \times 7 \text{ cm} = 63 \text{ cm}^2

    Eneo la nyuso nne za pembeni = 4×Apembetatu4 \times A_{pembetatu}

    Apembeni=4×63 cm2=252 cm2A_{pembeni} = 4 \times 63 \text{ cm}^2 = 252 \text{ cm}^2
  4. Kokotoa jumla ya eneo la uso la piramidi.

    Jumla ya eneo la uso (AjumlaA_{jumla}) ni jumla ya eneo la msingi na eneo la nyuso nne za pembeni.

    Ajumla=Amsingi+ApembeniA_{jumla} = A_{msingi} + A_{pembeni} Ajumla=81 cm2+252 cm2A_{jumla} = 81 \text{ cm}^2 + 252 \text{ cm}^2 Ajumla=333 cm2A_{jumla} = 333 \text{ cm}^2 333 cm2\boxed{333 \text{ cm}^2}
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