This worksheet provides nine practice problems for calculating arc length and sector area, featuring both diagrams and word problems.

Let's solve each problem step by step using the formulas for arc length and sector area.

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🔷 Key Formulas:

- Arc Length:
$$
\text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r
$$

- Sector Area:
$$
\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2
$$

Where:
- $ \theta $ is the central angle in degrees,
- $ r $ is the radius of the circle.

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## ✔ Problem 1: Find the length of the bolded arc.
- Radius $ r = 3 $ mm
- Central angle $ \theta = 108^\circ $

$$
\text{Arc Length} = \frac{108}{360} \times 2\pi \times 3 = \frac{3}{10} \times 6\pi = 1.8\pi \approx 5.65 \text{ mm}
$$

✔ Answer: $ 1.8\pi $ mm or approximately 5.65 mm

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## ✔ Problem 2: Find the area of the shaded sector.
- Radius $ r = 8 $ ft
- Central angle $ \theta = 32^\circ $

$$
\text{Area} = \frac{32}{360} \times \pi \times 8^2 = \frac{8}{90} \times 64\pi = \frac{512}{90}\pi = \frac{256}{45}\pi \approx 18.07 \text{ ft}^2
$$

✔ Answer: $ \frac{256}{45}\pi $ ft² or approximately 18.07 ft²

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## ✔ Problem 3: Find the length of the bolded arc.
- Radius $ r = 6 $ in
- Central angle $ \theta = 143^\circ $

$$
\text{Arc Length} = \frac{143}{360} \times 2\pi \times 6 = \frac{143}{360} \times 12\pi = \frac{1716}{360}\pi = \frac{143}{30}\pi \approx 14.96 \text{ in}
$$

✔ Answer: $ \frac{143}{30}\pi $ in or approximately 14.96 in

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## ✔ Problem 4: Find the area of the shaded sector.
- Radius $ r = 12 $ cm
- Central angle $ \theta = 204^\circ $

$$
\text{Area} = \frac{204}{360} \times \pi \times 12^2 = \frac{17}{30} \times 144\pi = \frac{2448}{30}\pi = 81.6\pi \approx 256.36 \text{ cm}^2
$$

✔ Answer: $ 81.6\pi $ cm² or approximately 256.36 cm²

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## ✔ Problem 5: Find the length of the bolded arc.
- Diameter $ d = 30 $ m → Radius $ r = 15 $ m
- Central angle $ \theta = 47^\circ $

$$
\text{Arc Length} = \frac{47}{360} \times 2\pi \times 15 = \frac{47}{360} \times 30\pi = \frac{1410}{360}\pi = \frac{47}{12}\pi \approx 12.27 \text{ m}
$$

✔ Answer: $ \frac{47}{12}\pi $ m or approximately 12.27 m

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## ✔ Problem 6: Find the area of the shaded sector.
- Diameter $ d = 17 $ mi → Radius $ r = 8.5 $ mi
- Central angle $ \theta = 126^\circ $

$$
\text{Area} = \frac{126}{360} \times \pi \times (8.5)^2 = \frac{7}{20} \times \pi \times 72.25 = \frac{505.75}{20}\pi = 25.2875\pi \approx 79.43 \text{ mi}^2
$$

✔ Answer: $ 25.2875\pi $ mi² or approximately 79.43 mi²

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## ✔ Problem 7: Arc length with $ \theta = 70^\circ $, $ r = 10 $ cm

$$
\text{Arc Length} = \frac{70}{360} \times 2\pi \times 10 = \frac{7}{36} \times 20\pi = \frac{140}{36}\pi = \frac{35}{9}\pi \approx 12.22 \text{ cm}
$$

✔ Answer: $ \frac{35}{9}\pi $ cm or approximately 12.22 cm

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## ✔ Problem 8: Sector area with $ \theta = 105^\circ $, $ r = 5 $ in

$$
\text{Area} = \frac{105}{360} \times \pi \times 5^2 = \frac{7}{24} \times 25\pi = \frac{175}{24}\pi \approx 23.04 \text{ in}^2
$$

✔ Answer: $ \frac{175}{24}\pi $ in² or approximately 23.04 in²

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## ✔ Problem 9: Arc length with $ \theta = 280^\circ $, diameter = 100 m → $ r = 50 $ m

$$
\text{Arc Length} = \frac{280}{360} \times 2\pi \times 50 = \frac{7}{9} \times 100\pi = \frac{700}{9}\pi \approx 244.35 \text{ m}
$$

✔ Answer: $ \frac{700}{9}\pi $ m or approximately 244.35 m

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📌 Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ 1.8\pi $ mm ≈ 5.65 mm |
| 2 | $ \frac{256}{45}\pi $ ft² ≈ 18.07 ft² |
| 3 | $ \frac{143}{30}\pi $ in ≈ 14.96 in |
| 4 | $ 81.6\pi $ cm² ≈ 256.36 cm² |
| 5 | $ \frac{47}{12}\pi $ m ≈ 12.27 m |
| 6 | $ 25.2875\pi $ mi² ≈ 79.43 mi² |
| 7 | $ \frac{35}{9}\pi $ cm ≈ 12.22 cm |
| 8 | $ \frac{175}{24}\pi $ in² ≈ 23.04 in² |
| 9 | $ \frac{700}{9}\pi $ m ≈ 244.35 m |

> ✔ All answers are exact in terms of $ \pi $ and approximate values rounded to two decimal places.

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