Area of a Segment

The area of a segment of a circle is the area enclosed by a chord and the corresponding arc. It’s found by subtracting the area of the triangle formed by the chord and the center of the circle from the area of the sector that encompasses the segment. This concept builds upon the understanding of sectors and triangles within a circle.

Formulae

The core formula for the area of a segment is:

Area of Segment = Area of Sector – Area of Triangle

Where:

• Area of Sector = $\frac{\theta}{360} \times \pi r^2$ (where $\theta$ is the central angle in degrees and $r$ is the radius).
• Area of Triangle (formed by the chord and radii) = $\frac{1}{2} \times r^2 \times sin(\theta)$ (when the angle is in degrees). Note that for other scenarios the area might differ based on the triangle type (e.g., equilateral, isosceles, etc.).

Therefore, the complete formula for Area of Segment, when the angle is in degrees, is:

Area of Segment = $\frac{\theta}{360} \times \pi r^2 – \frac{1}{2} r^2 sin(\theta)$

Examples

Example-1:

Consider a circle with a radius of 10 cm. A chord subtends a central angle of 60 degrees. Calculate the area of the segment.

Solution:

• Area of Sector = $\frac{60}{360} \times \pi \times 10^2 = \frac{1}{6} \times 100\pi \approx 52.36 cm^2$
• Area of Triangle = $\frac{1}{2} \times 10^2 \times sin(60) = 50 \times \frac{\sqrt{3}}{2} \approx 43.30 cm^2$
• Area of Segment = $52.36 – 43.30 \approx 9.06 cm^2$

Example-2:

A circle has a radius of 5 cm. A chord of length 5 cm is drawn. Find the area of the minor segment.

Solution:

Since the chord’s length equals the radius, the triangle formed is an equilateral triangle. Therefore, the central angle is 60 degrees.

• Area of Sector = $\frac{60}{360} \times \pi \times 5^2 = \frac{1}{6} \times 25\pi \approx 13.09 cm^2$
• Area of Triangle = $\frac{\sqrt{3}}{4} \times 5^2 = \frac{25\sqrt{3}}{4} \approx 10.83 cm^2$ (using the area of an equilateral triangle formula)
• Area of Segment = $13.09 – 10.83 \approx 2.26 cm^2$

Common mistakes by students

• Incorrect Angle Units: Using radians when degrees are required, or vice versa. Always ensure the angle unit used in calculations is consistent with the formula.
• Confusing Sector and Segment: Mixing up the formulas for sector area and segment area. Remember, segment area involves subtracting the triangle area.
• Incorrect Triangle Area Calculation: Using the wrong formula for the area of the triangle, especially when the triangle isn’t a right-angled triangle. Make sure to use the formula that corresponds to the type of triangle present (e.g., equilateral, isosceles).
• Forgetting the Units: Omitting the correct units (e.g., $cm^2$, $m^2$) in the final answer.

Real Life Application

The area of a segment has real-world applications, such as:

• Architecture and Design: Calculating the area of curved glass in windows or doors.
• Engineering: Determining the surface area of curved sections in structures.
• Land Surveying: Calculating land areas bounded by curves.
• Manufacturing: Calculating the amount of material needed for a curved part.

Fun Fact

The area of a segment is a classic example of how geometry blends with trigonometry. The key is to realize how geometric shapes can be broken down into more basic forms like sectors and triangles, leading to elegant solutions.

Recommended YouTube Videos for Deeper Understanding

Q.1 The radius of a circle is 10 cm. A chord of the circle subtends an angle of $60^\circ$ at the center. Find the area of the minor segment.

Check Solution

Ans: D

Area of sector = $\frac{\theta}{360} \pi r^2 = \frac{60}{360} \pi (10^2) = \frac{100\pi}{6} \text{ cm}^2$. Area of triangle = $\frac{1}{2}r^2 \sin\theta = \frac{1}{2} (10^2) \sin 60^\circ = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \text{ cm}^2$. Area of segment = Area of sector – Area of triangle = $\frac{100\pi}{6} – 25\sqrt{3} \approx 52.36 – 43.3 = 9.06 \text{ cm}^2$

Q.2 A sector of a circle of radius 14 cm has an angle of $45^\circ$. What is the area of the corresponding minor segment?

Check Solution

Ans: B

Area of sector = $\frac{45}{360} \pi (14^2) = \frac{1}{8} \pi (196) = 24.5 \pi \text{ cm}^2$. Area of triangle = $\frac{1}{2}(14)(14)\sin 45^\circ = 98 \times \frac{1}{\sqrt{2}} = 49\sqrt{2} \text{ cm}^2$. Area of segment = $24.5\pi – 49\sqrt{2} \approx 76.97 – 69.28 \approx 7.69 \text{ cm}^2$

Q.3 A circle has a radius of 6 cm. A chord of the circle is 6 cm long. Find the area of the minor segment.

Check Solution

Ans: A

The triangle formed by the chord and the radii is an equilateral triangle. Angle at the center is $60^\circ$. Area of sector = $\frac{60}{360} \pi (6^2) = 6\pi \text{ cm}^2$. Area of triangle = $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (6^2) = 9\sqrt{3} \text{ cm}^2$. Area of segment = $6\pi – 9\sqrt{3} \approx 18.85 – 15.59 = 3.26 \text{ cm}^2$

Q.4 The area of a segment of a circle is $25 \text{ cm}^2$. The radius of the circle is 10 cm and the central angle is $120^\circ$. Find the area of the corresponding sector.

Check Solution

Ans: A

Area of segment = Area of sector – Area of triangle. Let Area of sector = A. The area of the triangle = $\frac{1}{2}r^2 \sin\theta = \frac{1}{2}(100) \sin 120^\circ = 50 \frac{\sqrt{3}}{2} = 25\sqrt{3}$. A – $25\sqrt{3} = 25$. Area of sector = Area of segment + Area of triangle. Thus, Area of sector = $25+25\sqrt{3} \approx 25+43.3 \approx 68.3 \text{ cm}^2$. Alternatively, $A = 25 + 25\sqrt{3} \approx 68.3 \text{ cm}^2$ Area of sector = $\frac{120}{360} \pi r^2 = \frac{1}{3}\pi (100) = \frac{100\pi}{3} \approx 104.7 \text{ cm}^2$.

Q.5 A circle has radius 5 cm. If the area of the minor sector is $20 \text{ cm}^2$, and the area of the minor segment is $5.36 \text{ cm}^2$, then what is the angle of the sector?

Check Solution

Ans: C

Area of the segment = Area of sector – Area of triangle. Area of the triangle = Area of sector – Area of the segment. Hence, Area of triangle = $20 – 5.36 = 14.64$. Thus, $1/2 r^2 \sin\theta = 14.64$. Then $\sin\theta = \frac{2(14.64)}{5^2} = \frac{29.28}{25} \approx 1.17$, which is impossible since $\sin\theta$ cannot exceed 1. Area of segment = Area of sector – Area of triangle. We have area of sector $20 \text{ cm}^2$. $\frac{\theta}{360} \pi (5^2) = 20$. Then $\theta = \frac{20*360}{25\pi} \approx 91.67^\circ$. Also, the area of the triangle is $20-5.36 = 14.64$. $\frac{1}{2} r^2 \sin\theta = 14.64$. Then $\sin\theta = \frac{2 \cdot 14.64}{25} = 1.17$. Instead, we can calculate the angle directly. $20 = \frac{\theta}{360} \times \pi \times 5^2$, where $\theta$ is in degrees. $\frac{\theta}{360} = \frac{20}{25\pi} = \frac{4}{5\pi}$. Then $\theta = \frac{4 \times 360}{5\pi} \approx 91.67^\circ$.

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