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Two circles with equal radii, r = 9 ext cm. Centers A and B. Angle CAD = 80^. i) Calculate the area of Sector CAD. ii) Calculate the area of Triangle CAD.

Mathematics

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1) Geometry Problem

Given:

  • Two circles with equal radii, r=9 cmr = 9 \text{ cm}.

  • Centers A and B.

  • Angle ∠CAD=80∘\angle CAD = 80^\circ.

  • i) Calculate the area of Sector CAD.

  1. Identify the formula for the area of a sector.

    The area of a sector is given by the formula:

    Asector=θ360∘πr2A_{\text{sector}} = \frac{\theta}{360^\circ} \pi r^2

    where θ\theta is the central angle in degrees and rr is the radius.

  2. Substitute the given values into the formula.

    Given θ=80∘\theta = 80^\circ and r=9 cmr = 9 \text{ cm}.

    Asector=80∘360∘×π×(9 cm)2A_{\text{sector}} = \frac{80^\circ}{360^\circ} \times \pi \times (9 \text{ cm})^2
  3. Simplify the expression.

    Asector=29×π×81 cm2A_{\text{sector}} = \frac{2}{9} \times \pi \times 81 \text{ cm}^2 Asector=2×9×π cm2A_{\text{sector}} = 2 \times 9 \times \pi \text{ cm}^2 Asector=18π cm2A_{\text{sector}} = 18\pi \text{ cm}^2

    Using π≈3.14159\pi \approx 3.14159:

    Asector≈18×3.14159 cm2A_{\text{sector}} \approx 18 \times 3.14159 \text{ cm}^2 Asector≈56.55 cm2A_{\text{sector}} \approx 56.55 \text{ cm}^2

    The area of Sector CAD is 18π cm2≈56.55 cm2\boxed{18\pi \text{ cm}^2 \approx 56.55 \text{ cm}^2}.

    • ii) Calculate the area of Triangle CAD.
  4. Identify the formula for the area of a triangle given two sides and the included angle.

    Triangle CAD is an isosceles triangle with sides AC=AD=r=9 cmAC = AD = r = 9 \text{ cm} and the included angle ∠CAD=80∘\angle CAD = 80^\circ.
    The area of a triangle is given by the formula:

    Atriangle=12absin⁡CA_{\text{triangle}} = \frac{1}{2}ab\sin C

    where aa and bb are the lengths of two sides and CC is the included angle.

  5. Substitute the given values into the formula.

    Given a=9 cma = 9 \text{ cm}, b=9 cmb = 9 \text{ cm}, and C=80∘C = 80^\circ.

    Atriangle=12×9 cm×9 cm×sin⁡(80∘)A_{\text{triangle}} = \frac{1}{2} \times 9 \text{ cm} \times 9 \text{ cm} \times \sin(80^\circ)
  6. Simplify the expression.

    Atriangle=812sin⁡(80∘) cm2A_{\text{triangle}} = \frac{81}{2} \sin(80^\circ) \text{ cm}^2 Atriangle=40.5sin⁡(80∘) cm2A_{\text{triangle}} = 40.5 \sin(80^\circ) \text{ cm}^2

    Using sin⁡(80∘)≈0.9848\sin(80^\circ) \approx 0.9848:

    Atriangle≈40.5×0.9848 cm2A_{\text{triangle}} \approx 40.5 \times 0.9848 \text{ cm}^2 Atriangle≈39.88 cm2A_{\text{triangle}} \approx 39.88 \text{ cm}^2

    The area of Triangle CAD is 40.5sin⁡(80∘) cm2≈39.88 cm2\boxed{40.5 \sin(80^\circ) \text{ cm}^2 \approx 39.88 \text{ cm}^2}.

    • iii) Calculate the area of the shaded region.
  7. Understand the shaded region.

    The shaded region is the area of intersection of the two circles (the lens shape). This area is composed of two identical segments, one from each circle. The area of one segment (e.g., from circle A, formed by chord CD) is the area of Sector CAD minus the area of Triangle CAD.

  8. Calculate the area of one segment.

    Asegment=Asector−AtriangleA_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} Asegment=18π cm2−40.5sin⁡(80∘) cm2A_{\text{segment}} = 18\pi \text{ cm}^2 - 40.5 \sin(80^\circ) \text{ cm}^2 Asegment≈56.55 cm2−39.88 cm2A_{\text{segment}} \approx 56.55 \text{ cm}^2 - 39.88 \text{ cm}^2 Asegment≈16.67 cm2A_{\text{segment}} \approx 16.67 \text{ cm}^2
  9. Calculate the total shaded area.

    Since the two circles are identical and the intersection is symmetrical, the total shaded area is twice the area of one segment.

    Ashaded=2×AsegmentA_{\text{shaded}} = 2 \times A_{\text{segment}} Ashaded=2×(18π−40.5sin⁡(80∘)) cm2A_{\text{shaded}} = 2 \times (18\pi - 40.5 \sin(80^\circ)) \text{ cm}^2 Ashaded=(36π−81sin⁡(80∘)) cm2A_{\text{shaded}} = (36\pi - 81 \sin(80^\circ)) \text{ cm}^2

    Using the approximate values:

    Ashaded≈2×16.67 cm2A_{\text{shaded}} \approx 2 \times 16.67 \text{ cm}^2 Ashaded≈33.34 cm2A_{\text{shaded}} \approx 33.34 \text{ cm}^2

    The area of the shaded region is (36π−81sin⁡(80∘)) cm2≈33.34 cm2\boxed{(36\pi - 81 \sin(80^\circ)) \text{ cm}^2 \approx 33.34 \text{ cm}^2}.

    2) Second Problem

    The second problem is incomplete in the image provided. Please send the full question and diagram for "2) The figure shows a circle ∠QOT\angle QOT is the diameter and ∠SRT=37∘\angle SRT = 37^\circ".

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