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3 Find the perimeter of the shape shown.

Mathematics

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Answer

520 m\text{520 m}

Step-by-step solution

4 steps
  1. Tambua vipimo vilivyotolewa.

    Umbo lina sehemu mbili za nusu duara na pande mbili za moja kwa moja.
    Urefu wa upande wa moja kwa moja = 150 m150 \text{ m}
    Kipenyo cha nusu duara = 70 m70 \text{ m}
    Radiasi ya nusu duara, r=70 m2=35 mr = \frac{70 \text{ m}}{2} = 35 \text{ m}

  2. Kokotoa urefu wa arcs za nusu duara.

    Kuna nusu duara mbili, ambazo zikiunganishwa huunda duara kamili.
    Mzingo wa duara kamili hutolewa na fomula C=πdC = \pi d au C=2πrC = 2\pi r.
    Kwa kutumia d=70 md = 70 \text{ m} na π≈227\pi \approx \frac{22}{7}:

    Urefu wa arcs=π×d=227×70 m\text{Urefu wa arcs} = \pi \times d = \frac{22}{7} \times 70 \text{ m} Urefu wa arcs=22×10 m=220 m\text{Urefu wa arcs} = 22 \times 10 \text{ m} = 220 \text{ m}
  3. Kokotoa urefu wa pande mbili za moja kwa moja.

    Kuna pande mbili za moja kwa moja, kila moja ikiwa na urefu wa 150 m150 \text{ m}.

    Urefu wa pande za moja kwa moja=2×150 m=300 m\text{Urefu wa pande za moja kwa moja} = 2 \times 150 \text{ m} = 300 \text{ m}
  4. Tafuta jumla ya perimeta.

    Perimeta ya umbo ni jumla ya urefu wa arcs na urefu wa pande za moja kwa moja.

    Perimeta=Urefu wa arcs+Urefu wa pande za moja kwa moja\text{Perimeta} = \text{Urefu wa arcs} + \text{Urefu wa pande za moja kwa moja} Perimeta=220 m+300 m=520 m\text{Perimeta} = 220 \text{ m} + 300 \text{ m} = 520 \text{ m}

    Perimeta ya umbo ni 520 m\text{520 m}.

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