Question

Formula for length of an arc,area and perimeter of sector,area and perimeter of segment and Examples

Answer 1

Sure! Here are the formulas and explanations for the length of an arc, the area and perimeter of a sector, and the area and perimeter of a segment of a circle.

### 1. Length of an Arc

The length of an arc (L) of a circle depends on the radius (r) and the angle (θ) in radians subtended by the arc at the center of the circle.

**Formula:**
\[
L = r \cdot \theta
\]

**Example**:
If a circle has a radius of \( r = 5 \) cm and the angle θ = \( \frac{\pi}{3} \) radians, then the length of the arc is:
\[
L = 5 \cdot \frac{\pi}{3} = \frac{5\pi}{3} \text{ cm} \approx 5.24 \text{ cm}
\]

### 2. Area of a Sector

The area of a sector (A) of a circle is determined by the radius and the angle in radians.

**Formula:**
\[
A = \frac{1}{2} r^2 \theta
\]

**Example**:
Using the same example with \( r = 5 \) cm and \( \theta = \frac{\pi}{3} \):
\[
A = \frac{1}{2} \cdot 5^2 \cdot \frac{\pi}{3} = \frac{25\pi}{6} \text{ cm}^2 \approx 13.09 \text{ cm}^2
\]

### 3. Perimeter of a Sector

The perimeter (P) of a sector includes the length of the arc plus the two radii.

**Formula:**
\[
P = L + 2r = r \cdot \theta + 2r
\]
Or explicitly:
\[
P = r(\theta + 2)
\]

**Example**:
Using \( r = 5 \) cm and \( \theta = \frac{\pi}{3} \):
\[
L = 5 \cdot \frac{\pi}{3} = \frac{5\pi}{3} \text{ cm}
\]
So,
\[
P = \frac{5\pi}{3} + 2 \cdot 5 = \frac{5\pi}{3} + 10 \approx 5.24 + 10 = 15.24 \text{ cm}
\]

### 4. Area of a Segment

The area of a segment of a circle is the area of the sector minus the area of the triangular portion.

**Formula**:
\[
\text{Area of Segment} = A_{\text{sector}} - A_{\text{triangle}}
\]
Where:
- The area of the sector \( A_{\text{sector}} = \frac{1}{2} r^2 \theta \)
- The area of the triangle can be found using the formula:
\[
A_{\text{triangle}} = \frac{1}{2} r^2 \sin(\theta)
\]

**Example**:
Using \( r = 5 \) cm and \( \theta = \frac{\pi}{3} \):
\[
A_{\text{sector}} = \frac{25\pi}{6} \text{ cm}^2
\]
\[
A_{\text{triangle}} = \frac{1}{2} \cdot 5^2 \cdot \sin\left(\frac{\pi}{3}\right) = \frac{25}{2} \cdot \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{4} \text{ cm}^2 \approx 10.83 \text{ cm}^2
\]
Thus,
\[
\text{Area of Segment} = \frac{25\pi}{6} - \frac{25\sqrt{3}}{4} \approx 13.09 - 10.83 \approx 2.26 \text{ cm}^2
\]

### 5. Perimeter of a Segment

The perimeter of a segment consists of the length of the arc plus the chord.

**Formula**:
\[
\text{Perimeter of Segment} = L + 2a
\]
where \( L \) is the length of the arc, and \( a \) is the length of the chord.
The length of the chord can be calculated using:
\[
a = 2r \sin\left(\frac{\theta}{2}\right)
\]

**Example**:
For \( r = 5 \) cm and \( \theta = \frac{\pi}{3} \):
1. Calculate \( L \) (as done above): \( L = \frac{5\pi}{3} \text{ cm} \).
2. Calculate the chord length:
\[
a = 2 \cdot 5 \cdot \sin\left(\frac{\pi}{6}\right) = 10 \cdot \frac{1}{2} = 5 \text{ cm}
\]
Thus,
\[
\text{Perimeter of Segment} = \frac{5\pi}{3} + 5 \approx 5.24 + 5 = 10.24 \text{ cm}
\]

### Recap

- **Arc Length**: \( L = r \cdot \theta \)
- **Area of Sector**: \( A = \frac{1}{2} r^2 \theta \)
- **Perimeter of Sector**: \( P = r(\theta + 2) \)
- **Area of Segment**: \( \text{Area of Segment} = A_{\text{sector}} - A_{\text{triangle}} \)
- **Perimeter of Segment**: \( \text{Perimeter of Segment} = L + a \)

Feel free to ask if you need more examples or clarifications!