Math worksheet on arc length and sector area featuring three problems with diagrams and instructions.

Problem 1: Circle with Centre \( O \)

#### Given:
- Radius \( r = 10 \) cm
- Central angle \( \angle AOB = 60^\circ \)
- Use \( \pi = 3.14 \)

#### Task:
1. Calculate the length of the minor arc \( AB \).
2. Calculate the area of the shaded sector \( AOB \).

---

#### Part (a): Length of the Minor Arc \( AB \)

The formula for the length of an arc is:
\[
\text{Arc length} = \frac{\theta}{360^\circ} \times 2\pi r
\]
where:
- \( \theta \) is the central angle in degrees,
- \( r \) is the radius of the circle.

Substitute the given values:
\[
\theta = 60^\circ, \quad r = 10 \, \text{cm}, \quad \pi = 3.14
\]

\[
\text{Arc length} = \frac{60^\circ}{360^\circ} \times 2 \times 3.14 \times 10
\]

Simplify the fraction:
\[
\frac{60^\circ}{360^\circ} = \frac{1}{6}
\]

Now calculate:
\[
\text{Arc length} = \frac{1}{6} \times 2 \times 3.14 \times 10 = \frac{1}{6} \times 62.8 = 10.47 \, \text{cm}
\]

Thus, the length of the minor arc \( AB \) is:
\[
\boxed{10.47}
\]

---

#### Part (b): Area of the Shaded Sector \( AOB \)

The formula for the area of a sector is:
\[
\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2
\]

Substitute the given values:
\[
\theta = 60^\circ, \quad r = 10 \, \text{cm}, \quad \pi = 3.14
\]

\[
\text{Area of sector} = \frac{60^\circ}{360^\circ} \times 3.14 \times 10^2
\]

Simplify the fraction:
\[
\frac{60^\circ}{360^\circ} = \frac{1}{6}
\]

Now calculate:
\[
\text{Area of sector} = \frac{1}{6} \times 3.14 \times 100 = \frac{1}{6} \times 314 = 52.33 \, \text{cm}^2
\]

Thus, the area of the shaded sector \( AOB \) is:
\[
\boxed{52.33}
\]

---

Problem 2: Lighthouse Beam

#### Given:
- Radius \( r = 8 \) km
- Central angle \( \theta = 35^\circ \)
- Use \( \pi = 3.14 \)

#### Task:
Calculate the area covered by the beam from the lighthouse.

---

#### Solution:

The formula for the area of a sector is:
\[
\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2
\]

Substitute the given values:
\[
\theta = 35^\circ, \quad r = 8 \, \text{km}, \quad \pi = 3.14
\]

\[
\text{Area of sector} = \frac{35^\circ}{360^\circ} \times 3.14 \times 8^2
\]

Simplify the fraction:
\[
\frac{35^\circ}{360^\circ} = \frac{7}{72}
\]

Now calculate:
\[
\text{Area of sector} = \frac{7}{72} \times 3.14 \times 64
\]

First, calculate \( 3.14 \times 64 \):
\[
3.14 \times 64 = 200.96
\]

Now multiply by \( \frac{7}{72} \):
\[
\text{Area of sector} = \frac{7}{72} \times 200.96 = \frac{1406.72}{72} \approx 19.54 \, \text{km}^2
\]

Thus, the area covered by the beam is:
\[
\boxed{19.5}
\]

---

Problem 3: Arc Length and Radius

#### Given:
- Arc length \( PS = 9.42 \) cm
- Central angle \( \angle POS = 90^\circ \)
- Use \( \pi = 3.14 \)

#### Task:
Find the radius \( r \).

---

#### Solution:

The formula for the arc length is:
\[
\text{Arc length} = \frac{\theta}{360^\circ} \times 2\pi r
\]

Substitute the given values:
\[
\text{Arc length} = 9.42 \, \text{cm}, \quad \theta = 90^\circ, \quad \pi = 3.14
\]

\[
9.42 = \frac{90^\circ}{360^\circ} \times 2 \times 3.14 \times r
\]

Simplify the fraction:
\[
\frac{90^\circ}{360^\circ} = \frac{1}{4}
\]

Now substitute:
\[
9.42 = \frac{1}{4} \times 2 \times 3.14 \times r
\]

Simplify further:
\[
9.42 = \frac{1}{2} \times 3.14 \times r
\]

\[
9.42 = 1.57 \times r
\]

Solve for \( r \):
\[
r = \frac{9.42}{1.57} = 6 \, \text{cm}
\]

Thus, the radius is:
\[
\boxed{6}
\]

---

Final Answers:

1. (a) Length of minor arc \( AB \): \(\boxed{10.47}\)
(b) Area of shaded sector \( AOB \): \(\boxed{52.33}\)
2. Area covered by the beam: \(\boxed{19.5}\)
3. Radius: \(\boxed{6}\)