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The radius of a wheel is 56cm. Calculate. (a) the circumference of the wheel. (b) the distance moved by the wheel after it makes 10000 revolutions on a flat surface in kilometres.

Mathematics

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Haya, nitakusaidia kutatua maswali haya.

Swali la 29:

a) Radius ya gurudumu ni 56 cm56 \text{ cm}. Tafuta mzingo wa gurudumu.

  1. Tumia formula ya mzingo wa duara.

    Mzingo wa duara (CC) hutolewa na formula C=2πrC = 2\pi r, ambapo rr ni radius na π≈227\pi \approx \frac{22}{7}.

  2. Badilisha thamani za radius na π\pi kwenye formula.

    C=2×227×56 cmC = 2 \times \frac{22}{7} \times 56 \text{ cm}
  3. Rahisisha hesabu.

    C=2×22×567 cmC = 2 \times 22 \times \frac{56}{7} \text{ cm} C=44×8 cmC = 44 \times 8 \text{ cm} C=352 cmC = 352 \text{ cm}

    Mzingo wa gurudumu ni 352 cm\boxed{\text{352 cm}}.

    b) Umbali uliosafirishwa na gurudumu baada ya kufanya mizunguko 10000 kwenye sehemu tambarare kwa kilomita.

  4. Tafuta umbali unaosafirishwa kwa mzunguko mmoja.

    Umbali unaosafirishwa kwa mzunguko mmoja ni sawa na mzingo wa gurudumu, ambao ni 352 cm352 \text{ cm}.

  5. Hesabu jumla ya umbali uliosafirishwa.

    Jumla ya umbali = Mzingo ×\times Idadi ya mizunguko

    Jumla ya umbali=352 cm×10000\text{Jumla ya umbali} = 352 \text{ cm} \times 10000 Jumla ya umbali=3,520,000 cm\text{Jumla ya umbali} = 3,520,000 \text{ cm}
  6. Badilisha umbali kutoka sentimita kwenda kilomita.

    Tunajua kuwa 1 m=100 cm1 \text{ m} = 100 \text{ cm} na 1 km=1000 m1 \text{ km} = 1000 \text{ m}. Hivyo, 1 km=1000×100 cm=100,000 cm1 \text{ km} = 1000 \times 100 \text{ cm} = 100,000 \text{ cm}.

    Jumla ya umbali (km)=3,520,000 cm100,000 cm/km\text{Jumla ya umbali (km)} = \frac{3,520,000 \text{ cm}}{100,000 \text{ cm/km}} Jumla ya umbali (km)=35.2 km\text{Jumla ya umbali (km)} = 35.2 \text{ km}

    Umbali uliosafirishwa ni 35.2 km\boxed{\text{35.2 km}}.

    Swali la 30:
    Urefu wa tao ni 44 cm44 \text{ cm}. Tafuta pembe inayoundwa katikati ya duara na tao ikiwa radius ya duara ni 42 cm42 \text{ cm}.

  7. Tumia formula ya urefu wa tao.

    Urefu wa tao (LL) hutolewa na formula L=θ360∘×2πrL = \frac{\theta}{360^\circ} \times 2\pi r, ambapo θ\theta ni pembe katikati, na rr ni radius. Tumia π=227\pi = \frac{22}{7}.

  8. Badilisha thamani zinazojulikana kwenye formula.

    Tunajua L=44 cmL = 44 \text{ cm} na r=42 cmr = 42 \text{ cm}.

    44=θ360∘×2×227×4244 = \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times 42
  9. Rahisisha na utafute θ\theta.

    44=θ360∘×44×42744 = \frac{\theta}{360^\circ} \times 44 \times \frac{42}{7} 44=θ360∘×44×644 = \frac{\theta}{360^\circ} \times 44 \times 6

    Gawanya pande zote mbili kwa 44:

    1=θ360∘×61 = \frac{\theta}{360^\circ} \times 6 θ=360∘6\theta = \frac{360^\circ}{6} θ=60∘\theta = 60^\circ

    Pembe inayoundwa katikati ni 60∘\boxed{\text{60}^\circ}.

    Swali la 31:
    Hesabu mzingo wa sekta ifuatayo. (Tumia π=3.14\pi = 3.14).
    Sekta ina radius 4 cm4 \text{ cm} na pembe ya 315∘315^\circ.

  10. Tumia formula ya urefu wa tao.

    Urefu wa tao (LL) hutolewa na formula L=θ360∘×2πrL = \frac{\theta}{360^\circ} \times 2\pi r.

  11. Badilisha thamani zinazojulikana kwenye formula.

    Tunajua r=4 cmr = 4 \text{ cm}, θ=315∘\theta = 315^\circ, na π=3.14\pi = 3.14.

    L=315360×2×3.14×4L = \frac{315}{360} \times 2 \times 3.14 \times 4
  12. Rahisisha na hesabu urefu wa tao.

    L=78×8×3.14L = \frac{7}{8} \times 8 \times 3.14 L=7×3.14L = 7 \times 3.14 L=21.98 cmL = 21.98 \text{ cm}
  13. Hesabu mzingo wa sekta.

    Mzingo wa sekta (PP) = Urefu wa tao + 2×2 \times radius.

    P=21.98 cm+(2×4 cm)P = 21.98 \text{ cm} + (2 \times 4 \text{ cm}) P=21.98 cm+8 cmP = 21.98 \text{ cm} + 8 \text{ cm} P=29.98 cmP = 29.98 \text{ cm}

    Mzingo wa sekta ni 29.98 cm\boxed{\text{29.98 cm}}.

    Swali la 32:
    Eneo la mkeka wa duara ni 124.74 m2124.74 \text{ m}^2. Tafuta radius yake.

  14. Tumia formula ya eneo la duara.

    Eneo la duara (AA) hutolewa na formula A=πr2A = \pi r^2, ambapo rr ni radius. Tumia π=227\pi = \frac{22}{7} kwa sababu inatoa jibu safi.

  15. Badilisha thamani zinazojulikana kwenye formula.

    Tunajua A=124.74 m2A = 124.74 \text{ m}^2.

    124.74=227×r2124.74 = \frac{22}{7} \times r^2
  16. Tafuta r2r^2.

    r2=124.74×722r^2 = \frac{124.74 \times 7}{22} r2=873.1822r^2 = \frac{873.18}{22} r2=39.69r^2 = 39.69
  17. Tafuta rr kwa kuhesabu mzizi wa pili.

    r=39.69r = \sqrt{39.69} r=6.3 mr = 6.3 \text{ m}

    Radius ya mkeka wa duara ni 6.3 m\boxed{\text{6.3 m}}.

    Swali la 33:
    Bei ya dizeli ilikuwa ikiuzwa kwa Sh. 140 kwa lita wakati serikali ilitoa ruzuku. Serikali ilipoondoa ruzuku, bei iliongezeka kwa uwiano wa 5:45:4. Hesabu bei mpya ya dizeli bila ruzuku.

  18. Tambua bei ya awali na uwiano wa ongezeko la bei.

    Bei ya awali (na ruzuku) = Sh. 140.
    Uwiano wa bei mpya kwa bei ya awali = 5:45:4. Hii inamaanisha Bei MpyaBei ya Awali=54\frac{\text{Bei Mpya}}{\text{Bei ya Awali}} = \frac{5}{4}.

  19. Hesabu bei mpya.

    Bei Mpya=54×Bei ya Awali\text{Bei Mpya} = \frac{5}{4} \times \text{Bei ya Awali} Bei Mpya=54×140 Sh\text{Bei Mpya} = \frac{5}{4} \times 140 \text{ Sh}
  20. Rahisisha hesabu.

    Bei Mpya=5×1404 Sh\text{Bei Mpya} = 5 \times \frac{140}{4} \text{ Sh} Bei Mpya=5×35 Sh\text{Bei Mpya} = 5 \times 35 \text{ Sh} Bei Mpya=175 Sh\text{Bei Mpya} = 175 \text{ Sh}

    Bei mpya ya dizeli bila ruzuku ni Sh. 175\boxed{\text{Sh. 175}}.

    Swali la 34:
    Tangi la maji la silinda wazi lina kipenyo cha nje cha 2.8 m2.8 \text{ m} na urefu wa 2.4 m2.4 \text{ m}. Hesabu eneo la uso wa tangi la maji.

  21. Tambua vipimo na formula za eneo la uso.

    Kipenyo (DD) = 2.8 m2.8 \text{ m}, hivyo radius (rr) = 2.82=1.4 m\frac{2.8}{2} = 1.4 \text{ m}.
    Urefu (hh) = 2.4 m2.4 \text{ m}.
    Kwa kuwa tangi ni wazi (haina kifuniko), eneo la uso ni jumla ya eneo la msingi na eneo la uso wa kando.
    Eneo la uso (AA) = Eneo la msingi + Eneo la uso wa kando
    A=πr2+2πrhA = \pi r^2 + 2\pi rh. Tumia π=227\pi = \frac{22}{7}.

  22. Badilisha thamani kwenye formula.

    A=(227×(1.4)2)+(2×227×1.4×2.4)A = \left(\frac{22}{7} \times (1.4)^2\right) + \left(2 \times \frac{22}{7} \times 1.4 \times 2.4\right)
  23. Hesabu eneo la msingi.

    Eneo la msingi=227×1.96=22×0.28=6.16 m2\text{Eneo la msingi} = \frac{22}{7} \times 1.96 = 22 \times 0.28 = 6.16 \text{ m}^2
  24. Hesabu eneo la uso wa kando.

    Eneo la uso wa kando=2×227×1.4×2.4=2×22×0.2×2.4=4.4×2.4=21.12 m2\text{Eneo la uso wa kando} = 2 \times \frac{22}{7} \times 1.4 \times 2.4 = 2 \times 22 \times 0.2 \times 2.4 = 4.4 \times 2.4 = 21.12 \text{ m}^2
  25. Jumlisha maeneo kupata eneo la uso lote.

    A=6.16 m2+21.12 m2A = 6.16 \text{ m}^2 + 21.12 \text{ m}^2 A=27.28 m2A = 27.28 \text{ m}^2

    Eneo la uso wa tangi la maji ni 27.28 m2\boxed{\text{27.28 m}^2}.

    Swali la 35:
    Eneo la uwanja wa michezo wa mraba ni 3287 m23287 \text{ m}^2. Wakati wa somo la P.E, wanafunzi wa Darasa la 8 walikimbia kuzunguka uwanja mara mbili. Tambua jumla ya umbali uliosafirishwa na wanafunzi. Tumia jedwali la Hisabati.

  26. Tafuta urefu wa upande wa uwanja wa mraba.

    Eneo la mraba (AA) = s2s^2, ambapo ss ni urefu wa upande.

    s2=3287 m2s^2 = 3287 \text{ m}^2 s=3287 ms = \sqrt{3287} \text{ m}

    Kutumia kikokotoo (kama mbadala wa jedwali la Hisabati):

    s≈57.332 ms \approx 57.332 \text{ m}
  27. Hesabu mzingo wa uwanja wa mraba.

    Mzingo wa mraba (PP) = 4s4s.

    P=4×57.332 mP = 4 \times 57.332 \text{ m} P≈229.328 mP \approx 229.328 \text{ m}
  28. Hesabu jumla ya umbali uliosafirishwa.

    Wanafunzi walikimbia mara mbili.
    Jumla ya umbali = Mzingo ×\times Idadi ya mizunguko

    Jumla ya umbali=229.328 m×2\text{Jumla ya umbali} = 229.328 \text{ m} \times 2 Jumla ya umbali=458.656 m\text{Jumla ya umbali} = 458.656 \text{ m}

    Jumla ya umbali uliosafirishwa na wanafunzi ni takriban 458.66 m\boxed{\text{458.66 m}}.

    Swali la 36:
    Tumia kikokotoo kutatua matatizo yafuatayo:

    a) 0.08520.085^2

  29. Hesabu mraba wa nambari.

    0.0852=0.085×0.0850.085^2 = 0.085 \times 0.085 0.0852=0.0072250.085^2 = 0.007225

    Jibu ni 0.007225\boxed{\text{0.007225}}.

    b) 8752875^2

  30. Hesabu mraba wa nambari.

    8752=875×875875^2 = 875 \times 875 8752=765625875^2 = 765625

    Jibu ni 765625\boxed{\text{765625}}.

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