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Mathematics

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Ailiyotiwa_kivuli=100(4−π) dm2\mathbf{A_{iliyotiwa\_kivuli} = 100(4 - \pi) \, \text{dm}^2}

Step-by-step solution

5 steps
  1. Tambua maumbo na vipimo vilivyotolewa.

    Mchoro unaonyesha mstatili wenye urefu H=10 dmH = 10 \, \text{dm}. Ndani ya mstatili huu, kuna maumbo mawili yanayofanana kama matao. Tunachukulia kuwa haya ni nusu-duara. Urefu wa kila nusu-duara ni sawa na urefu wa mstatili, hivyo basi rediasi ya kila nusu-duara ni r=H=10 dmr = H = 10 \, \text{dm}. Sehemu zilizotiwa kivuli ni zile sehemu za mstatili ambazo hazijafunikwa na nusu-duara mbili na eneo la mstatili lisilotiwa kivuli katikati yao.

  2. Kokotoa eneo la nusu-duara mbili.

    Eneo la nusu-duara moja ni 12πr2\frac{1}{2} \pi r^2.
    Eneo la nusu-duara mbili ni Anusu−duara=2×12πr2=πr2A_{nusu-duara} = 2 \times \frac{1}{2} \pi r^2 = \pi r^2.
    Badilisha r=10 dmr = 10 \, \text{dm}:

    Anusu−duara=π(10 dm)2=100π dm2A_{nusu-duara} = \pi (10 \, \text{dm})^2 = 100\pi \, \text{dm}^2
  3. Bainisha eneo lote la mstatili.

    Acha upana wa mstatili uwe WW.
    Upana wa kila nusu-duara ni kipenyo chake, D=2r=2×10 dm=20 dmD = 2r = 2 \times 10 \, \text{dm} = 20 \, \text{dm}.
    Acha upana wa eneo la mstatili lisilotiwa kivuli katikati ya nusu-duara mbili uwe xx.
    Upana wote wa mstatili utakuwa W=D+x+D=20 dm+x+20 dm=(40+x) dmW = D + x + D = 20 \, \text{dm} + x + 20 \, \text{dm} = (40 + x) \, \text{dm}.
    Urefu wa mstatili ni H=10 dmH = 10 \, \text{dm}.
    Eneo lote la mstatili ni Amstatili=W×H=(40+x) dm×10 dm=(400+10x) dm2A_{mstatili} = W \times H = (40 + x) \, \text{dm} \times 10 \, \text{dm} = (400 + 10x) \, \text{dm}^2.

  4. Kokotoa eneo la sehemu ya mstatili isiyotiwa kivuli katikati.

    Upana wa eneo hili ni xx na urefu wake ni H=10 dmH = 10 \, \text{dm}.

    Akati_isiyotiwa_kivuli=x×10 dm=10x dm2A_{kati\_isiyotiwa\_kivuli} = x \times 10 \, \text{dm} = 10x \, \text{dm}^2
  5. Kokotoa eneo la sehemu iliyotiwa kivuli.

    Eneo lililotiwa kivuli ni eneo lote la mstatili ukitoa eneo la nusu-duara mbili na eneo la mstatili lisilotiwa kivuli katikati.

    Ailiyotiwa_kivuli=Amstatili−Anusu−duara−Akati_isiyotiwa_kivuliA_{iliyotiwa\_kivuli} = A_{mstatili} - A_{nusu-duara} - A_{kati\_isiyotiwa\_kivuli} Ailiyotiwa_kivuli=(400+10x) dm2−100π dm2−10x dm2A_{iliyotiwa\_kivuli} = (400 + 10x) \, \text{dm}^2 - 100\pi \, \text{dm}^2 - 10x \, \text{dm}^2 Ailiyotiwa_kivuli=400+10x−100π−10x dm2A_{iliyotiwa\_kivuli} = 400 + 10x - 100\pi - 10x \, \text{dm}^2 Ailiyotiwa_kivuli=(400−100π) dm2A_{iliyotiwa\_kivuli} = (400 - 100\pi) \, \text{dm}^2

    Eneo lililotiwa kivuli linaweza pia kuandikwa kama:

    Ailiyotiwa_kivuli=100(4−π) dm2\mathbf{A_{iliyotiwa\_kivuli} = 100(4 - \pi) \, \text{dm}^2}
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