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A shot put throwing area is a sector of an annulus with R=10m, r=2m, and angle theta=40 degrees. Calculate arc lengths EF and DG.

Mathematics

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Answers

3 parts
a

6.982 m\text{6.982 m}

b

1.396 m\text{1.396 m}

c

24.379 m\text{24.379 m}

Step-by-step solution

6 steps

The figure shows a shot put throwing area, which is a sector of an annulus.
The common center is C.
The radius of the larger circle is R=10R = 10 m.
The radius of the smaller circle is r=2r = 2 m.
The angle θ=∠ECF=40∘\theta = \angle ECF = 40^\circ.
We will use π=3.142\pi = 3.142.

The formula for the arc length LL of a sector with radius RR and angle θ\theta (in degrees) is:

L=θ360∘×2πRL = \frac{\theta}{360^\circ} \times 2\pi R

Part a· 2 steps

Calculate the arc length EF.

  1. Identify the radius and angle for arc EF.

    Arc EF is part of the larger circle with radius R=10R = 10 m and angle θ=40∘\theta = 40^\circ.

  2. Apply the arc length formula and calculate.

    LEF=40∘360∘×2×3.142×10L_{EF} = \frac{40^\circ}{360^\circ} \times 2 \times 3.142 \times 10 LEF=19×20×3.142L_{EF} = \frac{1}{9} \times 20 \times 3.142 LEF=62.849L_{EF} = \frac{62.84}{9} LEF≈6.9822L_{EF} \approx 6.9822

    Rounding to three decimal places:

    LEF≈6.982 mL_{EF} \approx 6.982 \text{ m}

    The arc length EF is 6.982 m\boxed{\text{6.982 m}}.

Part b

1.396 m\text{1.396 m}

Calculate the arc length DG.

  1. Identify the radius and angle for arc DG.

    Arc DG is part of the smaller circle with radius r=2r = 2 m and angle θ=40∘\theta = 40^\circ.

  2. Apply the arc length formula and calculate.

    LDG=40∘360∘×2×3.142×2L_{DG} = \frac{40^\circ}{360^\circ} \times 2 \times 3.142 \times 2 LDG=19×4×3.142L_{DG} = \frac{1}{9} \times 4 \times 3.142 LDG=12.5689L_{DG} = \frac{12.568}{9} LDG≈1.3964L_{DG} \approx 1.3964

    Rounding to three decimal places:

    LDG≈1.396 mL_{DG} \approx 1.396 \text{ m}

    The arc length DG is 1.396 m\boxed{\text{1.396 m}}.

Part c

24.379 m\text{24.379 m}

Calculate the perimeter of the shaded region DEFG.

  1. Identify the lengths of the straight segments DE and GF.

    The straight segments DE and GF are the difference between the radii of the two circles.

    DE=GF=R−r=10 m−2 m=8 mDE = GF = R - r = 10 \text{ m} - 2 \text{ m} = 8 \text{ m}

  2. Sum all the lengths to find the perimeter.

    The perimeter PP of the shaded region DEFG is the sum of arc length EF, arc length DG, and the two straight segments DE and GF.

    P=LEF+LDG+DE+GFP = L_{EF} + L_{DG} + DE + GF P=6.9822...+1.3964...+8+8P = 6.9822... + 1.3964... + 8 + 8 P=8.3786...+16P = 8.3786... + 16 P=24.3786...P = 24.3786...

    Rounding to three decimal places:

    P≈24.379 mP \approx 24.379 \text{ m}

    The perimeter of the shaded region DEFG is 24.379 m\boxed{\text{24.379 m}}.

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