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The diagram below shows a wedge. Determine its total surface area.

Mathematics

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Answer

264 cm2\text{264 cm}^2

Step-by-step solution

4 steps

Huu hapa ni ufumbuzi wa swali la 31:

Ili kupata eneo lote la uso wa 'wedge' (prismu ya pembetatu), tunahitaji kuhesabu eneo la nyuso zake zote: nyuso mbili za pembetatu na nyuso tatu za mstatili.

  1. Tafuta urefu wa hypotenuse ya msingi wa pembetatu.

    Msingi wa pembetatu ni pembetatu ya pembe mraba yenye pande za 6 cm na 8 cm.
    Tunatumia theorem ya Pythagoras: c2=a2+b2c^2 = a^2 + b^2.

    c=62+82c = \sqrt{6^2 + 8^2} c=36+64c = \sqrt{36 + 64} c=100c = \sqrt{100} c=10 cmc = 10 \text{ cm}
  2. Kokotoa eneo la nyuso mbili za pembetatu.

    Eneo la pembetatu moja ni 12×base×height\frac{1}{2} \times \text{base} \times \text{height}.

    Eneo la pembetatu moja=12×8 cm×6 cm=24 cm2\text{Eneo la pembetatu moja} = \frac{1}{2} \times 8 \text{ cm} \times 6 \text{ cm} = 24 \text{ cm}^2 Eneo la pembetatu mbili=2×24 cm2=48 cm2\text{Eneo la pembetatu mbili} = 2 \times 24 \text{ cm}^2 = 48 \text{ cm}^2
  3. Kokotoa eneo la nyuso tatu za mstatili.

    Urefu wa 'wedge' ni 9 cm.

    • Eneo la mstatili wa chini: 8 cm×9 cm=72 cm28 \text{ cm} \times 9 \text{ cm} = 72 \text{ cm}^2
    • Eneo la mstatili wa upande: 6 cm×9 cm=54 cm26 \text{ cm} \times 9 \text{ cm} = 54 \text{ cm}^2
    • Eneo la mstatili wa juu uliopinda (hypotenuse ×\times urefu): 10 cm×9 cm=90 cm210 \text{ cm} \times 9 \text{ cm} = 90 \text{ cm}^2
  4. Jumlisha maeneo yote ili kupata eneo lote la uso.

    Eneo lote la uso=48 cm2+72 cm2+54 cm2+90 cm2\text{Eneo lote la uso} = 48 \text{ cm}^2 + 72 \text{ cm}^2 + 54 \text{ cm}^2 + 90 \text{ cm}^2 Eneo lote la uso=264 cm2\text{Eneo lote la uso} = 264 \text{ cm}^2

    Eneo lote la uso wa 'wedge' ni 264 cm2\text{264 cm}^2.

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