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Find the largest possible size of square tiles that can be used to tile a room of 270 cm by 150 cm without any gaps, and calculate the number of tiles needed.

Mathematics

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Habari! Hakika, wacha tuendelee na maswali ya jana. Haya hapa majibu ya maswali uliyotuma awali:

Part 1· 6 steps

  1. Tafuta urefu na upana wa chumba.

    Urefu wa chumba =270 cm= 270 \text{ cm}
    Upana wa chumba =150 cm= 150 \text{ cm}

  2. Ili kupata ukubwa mkubwa zaidi wa vigae vya mraba ambavyo vinaweza kutumika bila kuacha nafasi yoyote, tunahitaji kutafuta Kigawo Kikubwa cha Shirika (GCD) cha urefu na upana wa chumba.

    Ufafanuzi wa nambari tasa wa 270:

    270=2×33×5270 = 2 \times 3^3 \times 5

    Ufafanuzi wa nambari tasa wa 150:

    150=2×3×52150 = 2 \times 3 \times 5^2

  3. Kokotoa GCD ya 270 na 150.

    GCD(270,150)=21×31×51=2×3×5=30\text{GCD}(270, 150) = 2^1 \times 3^1 \times 5^1 = 2 \times 3 \times 5 = 30

    Hivyo, ukubwa mkubwa zaidi wa kigae cha mraba ni 30 cm×30 cm30 \text{ cm} \times 30 \text{ cm}.

  4. Kokotoa eneo la chumba.

    Eneo la chumba =urefu×upana= \text{urefu} \times \text{upana}

    =270 cm×150 cm=40500 cm2= 270 \text{ cm} \times 150 \text{ cm} = 40500 \text{ cm}^2

  5. Kokotoa eneo la kigae kimoja.

    Eneo la kigae kimoja =upande×upande= \text{upande} \times \text{upande}

    =30 cm×30 cm=900 cm2= 30 \text{ cm} \times 30 \text{ cm} = 900 \text{ cm}^2

  6. Kokotoa idadi ya vigae vinavyohitajika.

    Idadi ya vigae =Eneo la chumbaEneo la kigae kimoja= \frac{\text{Eneo la chumba}}{\text{Eneo la kigae kimoja}}

    =40500 cm2900 cm2=45= \frac{40500 \text{ cm}^2}{900 \text{ cm}^2} = 45

    Ukubwa mkubwa zaidi wa kigae ni 30 cm, na vigae 45 vitatumika kufunika sakafu.

Part 2· 3 steps

  1. Badilisha kiasi cha kitambaa kinachohitajika kwa sketi moja na jumla ya kitambaa kuwa sehemu zisizofaa.

    Kitambaa kwa sketi moja =134 m=1×4+34 m=74 m= 1 \frac{3}{4} \text{ m} = \frac{1 \times 4 + 3}{4} \text{ m} = \frac{7}{4} \text{ m}
    Jumla ya kitambaa =1012 m=10×2+12 m=212 m= 10 \frac{1}{2} \text{ m} = \frac{10 \times 2 + 1}{2} \text{ m} = \frac{21}{2} \text{ m}

  2. Gawanya jumla ya kitambaa kwa kiasi cha kitambaa kinachohitajika kwa sketi moja ili kupata idadi ya sketi zinazoweza kutengenezwa.

    Idadi ya sketi =Jumla ya kitambaaKitambaa kwa sketi moja= \frac{\text{Jumla ya kitambaa}}{\text{Kitambaa kwa sketi moja}}

    =212÷74= \frac{21}{2} \div \frac{7}{4}

  3. Ili kugawanya sehemu, zidisha sehemu ya kwanza na kinyume cha sehemu ya pili.

    =212×47= \frac{21}{2} \times \frac{4}{7} =21×42×7= \frac{21 \times 4}{2 \times 7} =8414= \frac{84}{14} =6= 6

    Sketi 6 zinaweza kutengenezwa kutoka 1012 m10 \frac{1}{2} \text{ m} ya kitambaa.

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