Circle Segments

Regions Bounded by a Chord and an Arc

A line that intersects a circle at two points forms a chord. The chord and either arc between its endpoints bound a circle segment. The minor arc bounds the minor segment, while the remaining major arc bounds the major segment. A diameter is the limiting case in which both segments are semicircles.

Difference Between Arc and Segment

An arc and a segment name different parts of a circle:

Summary of Differences:

• Arc: Curved line on the circumference of a circle (only has length, no area)
• Segment: Region surrounded by arc and chord (has area, can be calculated)

Types of Segment Based on Size

Based on their size, segments can be divided into two types with different characteristics:

For a fixed radius, the central angle distinguishes the two segments:

• Minor Segment: central angle $< 180^\circ$, segment area $< \frac{1}{2}\pi r^2$
• Major Segment: central angle $> 180^\circ$, segment area $> \frac{1}{2}\pi r^2$

Segment Area Formula

To calculate the area of a minor segment, subtract the triangle formed by the two radii and the chord from its sector.

Compare arc length and segment area before using their formulas.

Substituting the sector-area and triangle-area formulas gives:

Here $0^\circ < \alpha < 180^\circ$ is the minor central angle in degrees and $r$ is the circle radius.

For a major segment, first use the complementary minor angle $\beta=360^\circ-\alpha$, then subtract the corresponding minor segment from the whole circle:

Sector Triangle and Segment

The diagram labels the sector, the triangle formed by the two radii and chord, and the segment left after subtracting the triangle.

Segment Area Calculation

To understand how to calculate the area of a segment, use an example with a central angle of $\alpha = 90^\circ$ and radius $r = 10 \text{ cm}$, according to the visualization of segment formation components above.

Step 1: Calculate Sector Area

Step 2: Calculate Triangle Area

For a $90^\circ$ angle, the triangle formed is a right triangle with both perpendicular sides being radii:

Step 3: Calculate Segment Area

Exercises

Each problem asks for the area of a segment, so subtract the triangle from the sector that share the same central angle.

1. A circle has a radius of $7 \text{ cm}$. If there is a segment with a central angle of $60^\circ$, determine the area of the segment.

2. A minor segment has an area of approximately $15.3546 \text{ cm}^2$ in a circle of radius $5 \text{ cm}$. Determine its minor central angle.

3. A horizontal cylindrical tank of radius $2 \text{ m}$ contains water to a depth of $1.5 \text{ m}$ measured from the bottom. Determine the cross-sectional area occupied by the water.

Worked Solutions

1. Solution Steps:

   Given: $r = 7 \text{ cm}$, $\alpha = 60^\circ$

   Calculate sector area:

   $$\text{Sector Area} = \frac{60^\circ}{360^\circ} \times \pi \times 7^2$$

   $$= \frac{1}{6} \times \pi \times 49$$

   $$= \frac{49\pi}{6} \text{ cm}^2$$

   Calculate triangle area:

   $$\text{Triangle Area} = \frac{1}{2} r^2 \sin 60^\circ$$

   $$= \frac{1}{2} \times 49 \times \frac{\sqrt{3}}{2}$$

   $$= \frac{49\sqrt{3}}{4} \text{ cm}^2$$

   Calculate segment area:

   $$\text{Segment Area} = \frac{49\pi}{6} - \frac{49\sqrt{3}}{4}$$

   $$\approx 25.6563 - 21.2176 = 4.4387 \text{ cm}^2$$

2. Solution Steps:

   Given: segment area $\approx 15.3546 \text{ cm}^2$, $r = 5 \text{ cm}$

   Using the formula: $\text{Segment Area} = \frac{\alpha}{360^\circ} \times \pi r^2 - \frac{1}{2} r^2 \sin \alpha$

   Solve this equation numerically on the minor-angle interval $0^\circ<\alpha<180^\circ$. This gives:

   $\alpha \approx 120^\circ$

   Verification:

   $$\begin{aligned} \text{Sector Area} &= \frac{120^\circ}{360^\circ} \times \pi \times 25 \\ &= \frac{25\pi}{3} \end{aligned}$$

   $$\begin{aligned} &\text{Triangle Area} \\ &= \frac{1}{2} \times 25 \times \sin 120^\circ \\ &= \frac{25\sqrt{3}}{4} \end{aligned}$$

   $$\begin{aligned} &\text{Segment Area} \\ &= \frac{25\pi}{3} - \frac{25\sqrt{3}}{4} \\ &\approx 26.1799 - 10.8253 \\ &= 15.3546 \text{ cm}^2 \end{aligned}$$

   This matches the stated area after rounding, so the central angle is approximately $120^\circ$.

3. Solution Steps:

   Given: tank radius = $2 \text{ m}$, water depth = $1.5 \text{ m}$

   Since the water depth ($1.5 \text{ m}$) is less than the radius ($2 \text{ m}$), the filled part of the circular cross-section is a minor segment.

   For horizontal cylindrical tanks, we use a special segment formula with a specific height.

   Determine segment height from tank bottom:

   The water height measured from the bottom is $h=1.5\text{ m}$.

   Using the segment area formula for water in a horizontal tank:

   Write $A$ for the segment area and define the height ratio $u$:

   $$u=\frac{r-h}{r}$$

   $$A=r^2\left(\arccos u-u\sqrt{1-u^2}\right)$$

   Here the angle returned by $\arccos$ is measured in radians, as required by the area formula.

   Substitute values:

   $$\frac{r-h}{r} = \frac{2-1.5}{2} = 0.25$$

   $$\arccos(0.25) \approx 1.3181 \text{ radians}$$

   $$\sqrt{1-0.25^2} = \sqrt{0.9375} \approx 0.9682$$

   Calculate the water-filled cross-sectional area:

   $$\theta=\arccos(0.25)$$

   $$v=0.25\sqrt{0.9375}$$

   $$A=4(\theta-v)$$

   $$\approx 4 \times 1.0760546$$

   $$\approx 4.3042 \text{ m}^2$$