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A circular pond has a diameter of 21m as shown below. What is the distance around the pond? 6. A metallic rim of a bicycle has a radius of 0.7m. What distance does it cover in 100 revolutions?

Mathematics

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6 parts
5

66 m\text{66 m}

6

440 m\text{440 m}

7

1.4 cm\text{1.4 cm}

1

90 cm2\text{90 cm}^2

2

36 m2\text{36 m}^2

3

12557 cm2\frac{1255}{7}\text{ cm}^2

179.29 cm2\text{179.29 cm}^2

Step-by-step solution

17 steps

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Part 5· 2 steps

Kipenyo cha bwawa ni 21 m. Tunahitaji kupata umbali kuzunguka bwawa, ambayo ni mzingo.
Mzingo wa duara hutolewa na fomula C=πdC = \pi d.

  1. Andika fomula ya mzingo na badilisha thamani zinazojulikana.

    C=πdC = \pi d C=227×21 mC = \frac{22}{7} \times 21\, \text{m}

  2. Rahisisha hesabu.

    C=22×3 mC = 22 \times 3\, \text{m} C=66 mC = 66\, \text{m}

    Umbali kuzunguka bwawa ni 66 m\boxed{\text{66 m}}.

Part 6

440 m\text{440 m}

Mdomo wa baiskeli una rediasi ya 0.7 m. Tunahitaji kujua umbali unaofunikwa katika mizunguko 100.
Kwanza, tafuta mzingo wa mdomo (umbali unaofunikwa katika mzunguko mmoja).
Mzingo wa duara hutolewa na fomula C=2πrC = 2\pi r.

  1. Andika fomula ya mzingo na badilisha thamani zinazojulikana.

    C=2πrC = 2\pi r C=2×227×0.7 mC = 2 \times \frac{22}{7} \times 0.7\, \text{m}

  2. Rahisisha hesabu ya mzingo.

    C=2×227×710 mC = 2 \times \frac{22}{7} \times \frac{7}{10}\, \text{m} C=2×2210 mC = 2 \times \frac{22}{10}\, \text{m} C=4410 mC = \frac{44}{10}\, \text{m} C=4.4 mC = 4.4\, \text{m}

  3. Tafuta umbali wa jumla unaofunikwa katika mizunguko 100.

    Umbali wa jumla=C×100\text{Umbali wa jumla} = C \times 100 Umbali wa jumla=4.4 m×100\text{Umbali wa jumla} = 4.4\, \text{m} \times 100 Umbali wa jumla=440 m\text{Umbali wa jumla} = 440\, \text{m}

    Umbali unaofunikwa katika mizunguko 100 ni 440 m\boxed{\text{440 m}}.

Part 7

1.4 cm\text{1.4 cm}

Mzingo wa pete ya duara ni 8.8 cm. Tunahitaji kupata rediasi yake.
Mzingo wa duara hutolewa na fomula C=2πrC = 2\pi r.

  1. Andika fomula ya mzingo na badilisha thamani zinazojulikana.

    C=2πrC = 2\pi r 8.8 cm=2×227×r8.8\, \text{cm} = 2 \times \frac{22}{7} \times r

  2. Rahisisha upande wa kulia wa mlinganyo.

    8.8 cm=447r8.8\, \text{cm} = \frac{44}{7} r

  3. Tatua kwa rr kwa kuzidisha pande zote mbili kwa 744\frac{7}{44}.

    r=8.8 cm×744r = 8.8\, \text{cm} \times \frac{7}{44} r=8810 cm×744r = \frac{88}{10}\, \text{cm} \times \frac{7}{44} r=210 cm×7r = \frac{2}{10}\, \text{cm} \times 7 r=1410 cmr = \frac{14}{10}\, \text{cm} r=1.4 cmr = 1.4\, \text{cm}

    Rediasi ya pete ni 1.4 cm\boxed{\text{1.4 cm}}.

Part 1

90 cm2\text{90 cm}^2

a (chini ya 3.3 Area)

Tafuta eneo la pembetatu iliyoonyeshwa.
Eneo la pembetatu hutolewa na fomula A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height}.

  1. Tambua urefu na msingi wa pembetatu.

    Msingi = 18 cm
    Urefu = 10 cm

  2. Badilisha thamani kwenye fomula na ukokotoe.

    A=12×18 cm×10 cmA = \frac{1}{2} \times 18\, \text{cm} \times 10\, \text{cm} A=9 cm×10 cmA = 9\, \text{cm} \times 10\, \text{cm} A=90 cm2A = 90\, \text{cm}^2

    Eneo la pembetatu ni 90 cm2\boxed{\text{90 cm}^2}.

Part 2

36 m2\text{36 m}^2

(upande wa kulia)

Sakafu ya Flora ina urefu wa 12 m na upana wa 8 m. Amefunika sakafu kwa zulia, akiacha ukingo wa 1 m pande zote. Tunahitaji kupata eneo la sakafu ambalo halijafunikwa na zulia.

  1. Kokotoa eneo la sakafu nzima.

    Eneo la sakafu=urefu×upana\text{Eneo la sakafu} = \text{urefu} \times \text{upana} Eneo la sakafu=12 m×8 m\text{Eneo la sakafu} = 12\, \text{m} \times 8\, \text{m} Eneo la sakafu=96 m2\text{Eneo la sakafu} = 96\, \text{m}^2

  2. Kokotoa vipimo vya zulia.

    Kwa kuwa kuna ukingo wa 1 m pande zote, urefu na upana wa zulia utapungua kwa 1+1=21+1=2 m.

    Urefu wa zulia=12 m−1 m−1 m=10 m\text{Urefu wa zulia} = 12\, \text{m} - 1\, \text{m} - 1\, \text{m} = 10\, \text{m} Upana wa zulia=8 m−1 m−1 m=6 m\text{Upana wa zulia} = 8\, \text{m} - 1\, \text{m} - 1\, \text{m} = 6\, \text{m}

  3. Kokotoa eneo la zulia.

    Eneo la zulia=urefu wa zulia×upana wa zulia\text{Eneo la zulia} = \text{urefu wa zulia} \times \text{upana wa zulia} Eneo la zulia=10 m×6 m\text{Eneo la zulia} = 10\, \text{m} \times 6\, \text{m} Eneo la zulia=60 m2\text{Eneo la zulia} = 60\, \text{m}^2

  4. Tafuta eneo la sakafu ambalo halijafunikwa na zulia.

    Eneo lisilofunikwa=Eneo la sakafu−Eneo la zulia\text{Eneo lisilofunikwa} = \text{Eneo la sakafu} - \text{Eneo la zulia} Eneo lisilofunikwa=96 m2−60 m2\text{Eneo lisilofunikwa} = 96\, \text{m}^2 - 60\, \text{m}^2 Eneo lisilofunikwa=36 m2\text{Eneo lisilofunikwa} = 36\, \text{m}^2

    Eneo la sakafu ambalo halijafunikwa na zulia ni 36 m2\boxed{\text{36 m}^2}.

Part 3

12557 cm2\frac{1255}{7}\text{ cm}^2

(upande wa kulia)

Tafuta eneo la umbo lililoonyeshwa. Umbo hili lina mstatili na nusu duara.

  1. Kokotoa eneo la mstatili.

    Urefu wa mstatili = 14 cm
    Upana wa mstatili = 10 cm

    Eneo la mstatili=urefu×upana\text{Eneo la mstatili} = \text{urefu} \times \text{upana} Eneo la mstatili=14 cm×10 cm\text{Eneo la mstatili} = 14\, \text{cm} \times 10\, \text{cm} Eneo la mstatili=140 cm2\text{Eneo la mstatili} = 140\, \text{cm}^2

  2. Kokotoa eneo la nusu duara.

    Kipenyo cha nusu duara ni 10 cm, hivyo rediasi yake ni r=102=5r = \frac{10}{2} = 5 cm.
    Eneo la duara ni πr2\pi r^2, kwa hivyo eneo la nusu duara ni 12πr2\frac{1}{2} \pi r^2.

    Eneo la nusu duara=12×227×(5 cm)2\text{Eneo la nusu duara} = \frac{1}{2} \times \frac{22}{7} \times (5\, \text{cm})^2 Eneo la nusu duara=12×227×25 cm2\text{Eneo la nusu duara} = \frac{1}{2} \times \frac{22}{7} \times 25\, \text{cm}^2 Eneo la nusu duara=11×257 cm2\text{Eneo la nusu duara} = \frac{11 \times 25}{7}\, \text{cm}^2 Eneo la nusu duara=2757 cm2\text{Eneo la nusu duara} = \frac{275}{7}\, \text{cm}^2

  3. Tafuta eneo la jumla la umbo.

    Eneo la jumla=Eneo la mstatili+Eneo la nusu duara\text{Eneo la jumla} = \text{Eneo la mstatili} + \text{Eneo la nusu duara} Eneo la jumla=140 cm2+2757 cm2\text{Eneo la jumla} = 140\, \text{cm}^2 + \frac{275}{7}\, \text{cm}^2

    Ili kuongeza, tunahitaji kupata denominator ya kawaida.

    140=140×77=9807140 = \frac{140 \times 7}{7} = \frac{980}{7} Eneo la jumla=9807 cm2+2757 cm2\text{Eneo la jumla} = \frac{980}{7}\, \text{cm}^2 + \frac{275}{7}\, \text{cm}^2 Eneo la jumla=980+2757 cm2\text{Eneo la jumla} = \frac{980 + 275}{7}\, \text{cm}^2 Eneo la jumla=12557 cm2\text{Eneo la jumla} = \frac{1255}{7}\, \text{cm}^2

    Kama desimali (takriban):

    Eneo la jumla≈179.29 cm2\text{Eneo la jumla} \approx 179.29\, \text{cm}^2

    Eneo la umbo ni 12557 cm2\boxed{\frac{1255}{7}\text{ cm}^2} au takriban 179.29 cm2\boxed{\text{179.29 cm}^2}.

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