Student task worksheet on arcs and sectors with problems and diagrams.

Problem Analysis and Solution

The task involves solving problems related to arcs and sectors of circles. We will solve each part step by step using the relevant formulas for arc length, perimeter, area, and relationships between these quantities.

#### Formula Recap:
1. Arc Length (\( L \)):
\[
L = \frac{\theta}{360^\circ} \times 2\pi r
\]
where \( \theta \) is the central angle in degrees and \( r \) is the radius.

2. Perimeter of a Sector:
\[
\text{Perimeter} = L + 2r
\]
where \( L \) is the arc length and \( r \) is the radius.

3. Area of a Sector:
\[
\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2
\]

4. Angle from Arc Length:
\[
\theta = \frac{L \times 360^\circ}{2\pi r}
\]

5. Radius from Arc Length:
\[
r = \frac{L \times 360^\circ}{2\pi \theta}
\]

6. Radius from Area:
\[
r = \sqrt{\frac{\text{Area} \times 360^\circ}{\pi \theta}}
\]

7. Diameter:
\[
\text{Diameter} = 2r
\]

---

Problem 1:

The sector has an angle of \( 160^\circ \) and a radius of \( 9 \) cm.

#### a. Find the length of the arc.
Using the arc length formula:
\[
L = \frac{\theta}{360^\circ} \times 2\pi r
\]
Substitute \( \theta = 160^\circ \) and \( r = 9 \):
\[
L = \frac{160}{360} \times 2\pi \times 9
\]
Simplify:
\[
L = \frac{4}{9} \times 18\pi = 8\pi
\]
Approximate \( \pi \approx 3.14159 \):
\[
L \approx 8 \times 3.14159 = 25.13272
\]
Round to 1 decimal place:
\[
L \approx 25.1 \, \text{cm}
\]

#### b. What is the perimeter of the sector?
The perimeter of the sector is the sum of the arc length and the two radii:
\[
\text{Perimeter} = L + 2r
\]
Substitute \( L \approx 25.1 \) and \( r = 9 \):
\[
\text{Perimeter} = 25.1 + 2 \times 9 = 25.1 + 18 = 43.1 \, \text{cm}
\]

#### c. Find the area of the sector.
Using the area formula:
\[
\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2
\]
Substitute \( \theta = 160^\circ \) and \( r = 9 \):
\[
\text{Area} = \frac{160}{360} \times \pi \times 9^2
\]
Simplify:
\[
\text{Area} = \frac{4}{9} \times \pi \times 81 = 36\pi
\]
Approximate \( \pi \approx 3.14159 \):
\[
\text{Area} \approx 36 \times 3.14159 = 113.09724
\]
Round to 1 decimal place:
\[
\text{Area} \approx 113.1 \, \text{cm}^2
\]

---

Problem 2:

Find the angle of the sector when the arc length is \( 12 \) cm and the radius is \( 5 \) cm.

Using the formula for the angle:
\[
\theta = \frac{L \times 360^\circ}{2\pi r}
\]
Substitute \( L = 12 \) and \( r = 5 \):
\[
\theta = \frac{12 \times 360^\circ}{2\pi \times 5}
\]
Simplify:
\[
\theta = \frac{4320^\circ}{10\pi} = \frac{432^\circ}{\pi}
\]
Approximate \( \pi \approx 3.14159 \):
\[
\theta \approx \frac{432}{3.14159} \approx 137.577
\]
Round to 1 decimal place:
\[
\theta \approx 137.6^\circ
\]

---

Problem 3:

What angle will the sector have when the radius is \( 8 \) cm and the arc length is \( 13 \) cm?

Using the formula for the angle:
\[
\theta = \frac{L \times 360^\circ}{2\pi r}
\]
Substitute \( L = 13 \) and \( r = 8 \):
\[
\theta = \frac{13 \times 360^\circ}{2\pi \times 8}
\]
Simplify:
\[
\theta = \frac{4680^\circ}{16\pi} = \frac{292.5^\circ}{\pi}
\]
Approximate \( \pi \approx 3.14159 \):
\[
\theta \approx \frac{292.5}{3.14159} \approx 93.11
\]
Round to 2 significant figures:
\[
\theta \approx 93^\circ
\]

---

Problem 4:

Find the diameter of the circle when the sector with an angle of \( 80^\circ \) has an arc length of \( 7.5 \) cm.

First, find the radius using the arc length formula:
\[
L = \frac{\theta}{360^\circ} \times 2\pi r
\]
Rearrange to solve for \( r \):
\[
r = \frac{L \times 360^\circ}{2\pi \theta}
\]
Substitute \( L = 7.5 \), \( \theta = 80^\circ \):
\[
r = \frac{7.5 \times 360^\circ}{2\pi \times 80}
\]
Simplify:
\[
r = \frac{2700}{160\pi} = \frac{135}{8\pi}
\]
Approximate \( \pi \approx 3.14159 \):
\[
r \approx \frac{135}{8 \times 3.14159} \approx \frac{135}{25.13272} \approx 5.37
\]
Now, find the diameter:
\[
\text{Diameter} = 2r \approx 2 \times 5.37 = 10.74
\]
Round to 2 decimal places:
\[
\text{Diameter} \approx 10.7 \, \text{cm}
\]

---

Problem 5:

Find the radius of the circle when the sector has an arc length of \( 12 \) cm and an angle of \( 134^\circ \).

Using the formula for the radius:
\[
r = \frac{L \times 360^\circ}{2\pi \theta}
\]
Substitute \( L = 12 \) and \( \theta = 134^\circ \):
\[
r = \frac{12 \times 360^\circ}{2\pi \times 134}
\]
Simplify:
\[
r = \frac{4320}{268\pi} = \frac{2160}{134\pi}
\]
Approximate \( \pi \approx 3.14159 \):
\[
r \approx \frac{2160}{134 \times 3.14159} \approx \frac{2160}{420.97} \approx 5.13
\]
Round to 1 decimal place:
\[
r \approx 5.1 \, \text{cm}
\]

---

Problem 6:

A sector with a radius of \( 7 \) cm has an area of \( 42 \) cm². Find the angle of the sector.

Using the area formula:
\[
\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2
\]
Rearrange to solve for \( \theta \):
\[
\theta = \frac{\text{Area} \times 360^\circ}{\pi r^2}
\]
Substitute \( \text{Area} = 42 \) and \( r = 7 \):
\[
\theta = \frac{42 \times 360^\circ}{\pi \times 7^2}
\]
Simplify:
\[
\theta = \frac{15120^\circ}{49\pi}
\]
Approximate \( \pi \approx 3.14159 \):
\[
\theta \approx \frac{15120}{49 \times 3.14159} \approx \frac{15120}{153.94} \approx 98.2
\]
Round to 1 significant figure:
\[
\theta \approx 100^\circ
\]

---

Problem 7:

A \( 125^\circ \) sector has an area of \( 58 \) cm². Find the diameter of the sector.

Using the area formula:
\[
\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2
\]
Rearrange to solve for \( r \):
\[
r^2 = \frac{\text{Area} \times 360^\circ}{\pi \theta}
\]
Substitute \( \text{Area} = 58 \) and \( \theta = 125^\circ \):
\[
r^2 = \frac{58 \times 360^\circ}{\pi \times 125}
\]
Simplify:
\[
r^2 = \frac{20880}{125\pi}
\]
Approximate \( \pi \approx 3.14159 \):
\[
r^2 \approx \frac{20880}{125 \times 3.14159} \approx \frac{20880}{392.7} \approx 53.17
\]
Take the square root:
\[
r \approx \sqrt{53.17} \approx 7.3
\]
Now, find the diameter:
\[
\text{Diameter} = 2r \approx 2 \times 7.3 = 14.6
\]
Round to 2 significant figures:
\[
\text{Diameter} \approx 15 \, \text{cm}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \, \text{a. } 25.1 \, \text{cm}, \, \text{b. } 43.1 \, \text{cm}, \, \text{c. } 113.1 \, \text{cm}^2 \\
2. & \, 137.6^\circ \\
3. & \, 93^\circ \\
4. & \, 10.7 \, \text{cm} \\
5. & \, 5.1 \, \text{cm} \\
6. & \, 100^\circ \\
7. & \, 15 \, \text{cm}
\end{aligned}
}
\]