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Find the area of the following polygon with centre O

Mathematics

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Answer

25.524 cm225.524 \text{ cm}^2

Step-by-step solution

6 steps

Hapa kuna suluhisho la swali la 1):

1) Tafuta eneo la poligoni ifuatayo yenye kituo O.

Poligoni hii imeundwa na sehemu mbili: heksagoni ya kawaida (regular hexagon) na pembetatu ya juu (isosceles triangle). Kituo O ni kituo cha heksagoni.

  1. Tambua vipimo vilivyotolewa.

    Apothem ya heksagoni (umbali kutoka kituo O hadi katikati ya upande mmoja, perpendicular) ni a=2.55 cma = 2.55 \text{ cm}.

  2. Kokotoa urefu wa upande (ss) wa heksagoni.

    Kwa heksagoni ya kawaida, apothem aa inahusiana na urefu wa upande ss kwa fomula a=32sa = \frac{\sqrt{3}}{2}s.
    Hivyo, tunaweza kupata ss:

    s=2a3s = \frac{2a}{\sqrt{3}}

    Weka thamani ya aa:

    s=2×2.553=5.13 cms = \frac{2 \times 2.55}{\sqrt{3}} = \frac{5.1}{\sqrt{3}} \text{ cm}
  3. Kokotoa eneo la heksagoni.

    Eneo la heksagoni ya kawaida ni Aheksagoni=12PaA_{heksagoni} = \frac{1}{2} P a, ambapo PP ni mzingo (P=6sP=6s).

    Aheksagoni=12(6s)a=3saA_{heksagoni} = \frac{1}{2} (6s) a = 3sa

    Weka thamani za ss na aa:

    Aheksagoni=3×5.13×2.55A_{heksagoni} = 3 \times \frac{5.1}{\sqrt{3}} \times 2.55 Aheksagoni=3×5.1×2.553=39.0153 cm2A_{heksagoni} = \frac{3 \times 5.1 \times 2.55}{\sqrt{3}} = \frac{39.015}{\sqrt{3}} \text{ cm}^2
  4. Bainisha urefu wa pembetatu ya juu (hth_t).

    Msingi wa pembetatu ya juu ni urefu wa upande ss wa heksagoni. Urefu wa pembetatu ya juu (hth_t) haujatolewa wazi, lakini tunaweza kuukadiria kutoka kwenye mchoro kwa kutumia gridi.
    Kutoka kwenye mchoro, apothem a=2.55 cma = 2.55 \text{ cm} inalingana na takriban vitengo 2.5 vya gridi.
    Urefu wa pembetatu ya juu unaonekana kuwa takriban vitengo 2 vya gridi.
    Kwa hiyo, tunakadiria hth_t kwa kutumia uwiano:

    hta≈2 vitengo vya gridi2.5 vitengo vya gridi\frac{h_t}{a} \approx \frac{2 \text{ vitengo vya gridi}}{2.5 \text{ vitengo vya gridi}} ht≈22.5×a=45×2.55=0.8×2.55=2.04 cmh_t \approx \frac{2}{2.5} \times a = \frac{4}{5} \times 2.55 = 0.8 \times 2.55 = 2.04 \text{ cm}
  5. Kokotoa eneo la pembetatu ya juu.

    Eneo la pembetatu ni Apembetatu=12×msingi×urefuA_{pembetatu} = \frac{1}{2} \times \text{msingi} \times \text{urefu}.

    Apembetatu_ya_juu=12×s×htA_{pembetatu\_ya\_juu} = \frac{1}{2} \times s \times h_t

    Weka thamani za ss na hth_t:

    Apembetatu_ya_juu=12×5.13×2.04A_{pembetatu\_ya\_juu} = \frac{1}{2} \times \frac{5.1}{\sqrt{3}} \times 2.04 Apembetatu_ya_juu=5.1×1.023=5.2023 cm2A_{pembetatu\_ya\_juu} = \frac{5.1 \times 1.02}{\sqrt{3}} = \frac{5.202}{\sqrt{3}} \text{ cm}^2
  6. Kokotoa eneo lote la poligoni.

    Eneo lote ni jumla ya eneo la heksagoni na eneo la pembetatu ya juu.

    Ajumla=Aheksagoni+Apembetatu_ya_juuA_{jumla} = A_{heksagoni} + A_{pembetatu\_ya\_juu} Ajumla=39.0153+5.2023=39.015+5.2023=44.2173 cm2A_{jumla} = \frac{39.015}{\sqrt{3}} + \frac{5.202}{\sqrt{3}} = \frac{39.015 + 5.202}{\sqrt{3}} = \frac{44.217}{\sqrt{3}} \text{ cm}^2

    Ili kupata thamani ya namba, tunazidisha na 33\frac{\sqrt{3}}{\sqrt{3}} au tunatumia 3≈1.73205\sqrt{3} \approx 1.73205:

    Ajumla=44.21733=14.7393 cm2A_{jumla} = \frac{44.217 \sqrt{3}}{3} = 14.739 \sqrt{3} \text{ cm}^2 Ajumla≈14.739×1.73205≈25.524 cm2A_{jumla} \approx 14.739 \times 1.73205 \approx 25.524 \text{ cm}^2

    Eneo la mwisho ni:

    25.524 cm225.524 \text{ cm}^2
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