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Hannah drew two intersecting circles. The length OA is 10.5cm, OB is 14cm, and angle AOB is 63.62 degrees and angle AO,B is 60 degrees. Calculate the Area of the shaded part.

Mathematics

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Answer

43.46 cm2\text{43.46 cm}^2

Step-by-step solution

8 steps

Huu hapa ni ukokotozi wa eneo lililotiwa kivuli:

  1. Tambua taarifa ulizopewa.

    • Radius ya duara ya kwanza (r1r_1): 10.5 cm
    • Pembe ya katikati ya duara ya kwanza (θ1\theta_1): 83.62∘83.62^\circ
    • Radius ya duara ya pili (r2r_2): 14 cm
    • Pembe ya katikati ya duara ya pili (θ2\theta_2): 60∘60^\circ Eneo lililotiwa kivuli ni jumla ya sehemu mbili za duara (segments) zinazoundwa na kodi AB.
  2. Kokotoa eneo la sekta O1ABO_1AB.

    Eneo la sekta hutolewa na fomula Asekta=θ360∘πr2A_{sekta} = \frac{\theta}{360^\circ} \pi r^2.

    Asekta1=83.62360×π×(10.5)2A_{sekta1} = \frac{83.62}{360} \times \pi \times (10.5)^2 Asekta1=83.62360×π×110.25A_{sekta1} = \frac{83.62}{360} \times \pi \times 110.25 Asekta1≈80.48 cm2A_{sekta1} \approx 80.48 \text{ cm}^2
  3. Kokotoa eneo la pembetatu O1ABO_1AB.

    Eneo la pembetatu ya isosceles hutolewa na fomula Apembetatu=12r2sin⁡(θ)A_{pembetatu} = \frac{1}{2} r^2 \sin(\theta).

    Apembetatu1=12×(10.5)2×sin⁡(83.62∘)A_{pembetatu1} = \frac{1}{2} \times (10.5)^2 \times \sin(83.62^\circ) Apembetatu1=12×110.25×0.99373A_{pembetatu1} = \frac{1}{2} \times 110.25 \times 0.99373 Apembetatu1≈54.78 cm2A_{pembetatu1} \approx 54.78 \text{ cm}^2
  4. Kokotoa eneo la sehemu ya duara (segment) ya kwanza.

    Eneo la sehemu ya duara ni tofauti kati ya eneo la sekta na eneo la pembetatu.

    Asehemu1=Asekta1−Apembetatu1A_{sehemu1} = A_{sekta1} - A_{pembetatu1} Asehemu1=80.48−54.78=25.70 cm2A_{sehemu1} = 80.48 - 54.78 = 25.70 \text{ cm}^2
  5. Kokotoa eneo la sekta O2ABO_2AB.

    Asekta2=60360×π×(14)2A_{sekta2} = \frac{60}{360} \times \pi \times (14)^2 Asekta2=16×π×196A_{sekta2} = \frac{1}{6} \times \pi \times 196 Asekta2≈102.63 cm2A_{sekta2} \approx 102.63 \text{ cm}^2
  6. Kokotoa eneo la pembetatu O2ABO_2AB.

    Apembetatu2=12×(14)2×sin⁡(60∘)A_{pembetatu2} = \frac{1}{2} \times (14)^2 \times \sin(60^\circ) Apembetatu2=12×196×32A_{pembetatu2} = \frac{1}{2} \times 196 \times \frac{\sqrt{3}}{2} Apembetatu2=493≈84.87 cm2A_{pembetatu2} = 49\sqrt{3} \approx 84.87 \text{ cm}^2
  7. Kokotoa eneo la sehemu ya duara (segment) ya pili.

    Asehemu2=Asekta2−Apembetatu2A_{sehemu2} = A_{sekta2} - A_{pembetatu2} Asehemu2=102.63−84.87=17.76 cm2A_{sehemu2} = 102.63 - 84.87 = 17.76 \text{ cm}^2
  8. Kokotoa jumla ya eneo lililotiwa kivuli.

    Eneo lililotiwa kivuli ni jumla ya maeneo ya sehemu mbili za duara.

    Ajumla=Asehemu1+Asehemu2A_{jumla} = A_{sehemu1} + A_{sehemu2} Ajumla=25.70+17.76=43.46 cm2A_{jumla} = 25.70 + 17.76 = 43.46 \text{ cm}^2

    Eneo lililotiwa kivuli ni 43.46 cm2\text{43.46 cm}^2.

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