CBSE Class 10 Mathematics Areas Related to Circles Notes Set 05

Area of Sector and Segment of a Circle

Sector of a Circle

The portion (or part) of the circular region enclosed between two radii and the corresponding arc is called a sector of the circle.

In the figure, the shaded region OAPB is a sector of the circle with centre O. \( \angle AOB \) is called the angle of the sector. OAPB is called the minor sector and OAQB is called the major sector. The angle of the major sector is \( 360^\circ - \angle AOB \).

Area of a Sector of Circle

Area of the sector \( = \frac{\theta}{360^\circ} \times \pi r^2 \), where r is the radius of the circle and \( \theta \) is the angle of the sector in degrees.
Length of the arc APB corresponding to the sector OAPB \( = \frac{\theta}{360^\circ} \times 2\pi r = \frac{\theta}{180^\circ} \times \pi r \).

The Segment of a Circle

The portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.

In the figure, two points A and B on the circumference of a circle are joined to form the line segment AB. This line segment AB is called the chord of the circle. Chord AB divides the circle into two parts, i.e., APB (minor segment) and AQB (major segment).

Area of the segment of a circle
Let O be the centre and r be the radius of the circle. Area of the segment APB \( = \) Area of the sector OAPB \( - \) Area of \( \triangle OAB \). Let \( \angle AOB = \theta \). Then area of the sector \( OAPB = \frac{\theta}{360^\circ} \times \pi r^2 \).

In \( \triangle OAB \), draw \( OM \perp AB \). Then \( \cos\frac{\theta}{2} = \frac{OM}{OA} \) and \( \sin\frac{\theta}{2} = \frac{AM}{OA} \).
So \( OM = r\cos\frac{\theta}{2} \) and \( AM = r\sin\frac{\theta}{2} \). Therefore \( AB = 2AM = 2r\sin\frac{\theta}{2} \).
\( \therefore \) Area of \( \triangle OAB = \frac{1}{2}(AB \times OM) = \frac{1}{2}\left(2r\sin\frac{\theta}{2} \times r\cos\frac{\theta}{2}\right) = r^2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \).

Therefore, area of the segment \( APB = \frac{\theta}{360^\circ} \times \pi r^2 - r^2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = \left[\frac{\pi}{360^\circ} \times \theta - \sin\frac{\theta}{2}\cos\frac{\theta}{2}\right]r^2 \)\[ \text{or, area of segment } APB = \frac{\theta}{360}\pi r^2 - \frac{1}{2}r^2\sin\theta \quad \left[\because \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right] \]Note: Area of the major segment \( AQB = \pi r^2 - \) Area of the minor segment APB.

Knowledge Hub

1. Angle subtended by the hour hand of a clock in 12 hours is \( 360^\circ \). Angle subtended by the hour hand in 1 hour \( = \frac{360^\circ}{12} = 30^\circ \). Angle subtended by the hour hand in 60 minutes \( = 30^\circ \). Angle subtended by the hour hand in 1 minute \( = \frac{30^\circ}{60} = \left(\frac{1}{2}\right)^\circ \).
2. Angle subtended by the minute hand of a clock in 60 minutes is \( 360^\circ \). Therefore, the angle subtended by the minute hand in one minute \( = \left(\frac{360}{60}\right)^\circ = 6^\circ \).

Example 1. If the angle of the sector is \( 120^\circ \) and the radius of the circle is 21 cm, then find the area of the segment of the circle. (Take \( \pi = \frac{22}{7} \))
Answer:
Radius of the circle \( = 21 \) cm and \( \theta = 120^\circ \).
Area of the segment \( = \left\{\frac{\pi}{360^\circ} \times \theta - \sin\frac{\theta}{2}\cos\frac{\theta}{2}\right\}r^2 = \left\{\frac{22}{7} \times \frac{120^\circ}{360^\circ} - \sin 60^\circ \cos 60^\circ\right\}(21)^2 \) cm².
\( = \left\{\frac{22}{7} \times \frac{1}{3} - \frac{\sqrt{3}}{2} \times \frac{1}{2}\right\}(21)^2 = \left\{\frac{22}{21} - \frac{\sqrt{3}}{4}\right\}(21)^2 = \left(\frac{88 - 21\sqrt{3}}{21 \times 4}\right)(21)^2 \) cm².
\( = \frac{21}{4}\left(88 - 21\sqrt{3}\right) \) cm².
Hence, the area of the segment is \( \frac{21}{4}(88 - 21\sqrt{3}) \) cm².

Example 2. A sector has been cut from a circle of radius 21 cm. If the angle of the sector is \( 150^\circ \), then find the length of its arc and area.
Answer:
Let r be the radius and \( \theta \) the angle of the sector. The arc length l and area A of the sector are \( l = \frac{\theta}{360^\circ} \times 2\pi r \) and \( A = \frac{\theta}{360^\circ} \times \pi r^2 \) respectively.
Given \( r = 21 \) cm and \( \theta = 150^\circ \).
\( \therefore l = \left\{\frac{150}{360} \times 2 \times \frac{22}{7} \times 21\right\} \) cm \( = 55 \) cm.
\( A = \frac{150}{360} \times \frac{22}{7} \times (21)^2 \) cm² \( = \frac{1155}{2} \) cm² \( = 577.5 \) cm².
Hence, the length of the arc is 55 cm and the area is 577.5 cm².

Example 3. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the area of the sector.
Answer:
Radius of the circle \( = 5.2 \) cm and perimeter of the sector \( = 16.4 \) cm.
Perimeter of the sector \( = \) length of the arc PRQ \( + 2r \). \( \therefore \) Length of the arc \( PRQ = 16.4 - (2 \times 5.2) = 16.4 - 10.4 = 6 \) cm ...(i)
Area of the sector \( OPRQ = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{\theta}{360^\circ} \times 2\pi r \times \frac{r}{2} \) (multiplying numerator and denominator by 2).
\( = \) length of the arc \( PRQ \times \frac{r}{2} = 6 \times \frac{5.2}{2} = 15.6 \) cm² (from (i)).
Therefore, the area of the sector is 15.6 cm².

Example 4. The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 9 A.M. and 9:35 A.M.
Answer:
Angle described by the minute hand in one minute \( = \frac{360^\circ}{60} = 6^\circ \).
\( \therefore \) Angle described by the minute hand in 35 minutes \( = (6 \times 35)^\circ = 210^\circ \).
Area swept by the minute hand in 35 minutes \( = \) area of a sector of angle \( 210^\circ \) in a circle of radius 10 cm \( = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{210^\circ}{360^\circ} \times \frac{22}{7} \times (10)^2 \) cm² \( = 183.3 \) cm².
Hence, the required area is about 183.3 cm².

Example 5. The given diagram shows two arcs, P and Q. Arc P is part of the circle with centre O and radius OB. Arc Q is part of the circle with centre M and radius AM, where M is the mid-point of AB. Show that the area enclosed by the two arcs is equal to \( 25\left(\sqrt{3} - \frac{\pi}{6}\right) \) cm².
Answer:
From the figure, area enclosed by arc Q and chord AB \( = \) area of the semicircle of radius 5 cm \( = \frac{1}{2} \times \pi \times 5^2 = \frac{25\pi}{2} \) cm².
Let \( \angle MOB = \angle MOA = \theta \). In \( \triangle OMA \), \( \sin\theta = \frac{AM}{OA} = \frac{5}{10} = \frac{1}{2} \Rightarrow \theta = 30^\circ \Rightarrow \angle AOB = 2\theta = 60^\circ \).
Area enclosed by arc P and chord AB \( = \) area of the segment APB of the circle with centre O and \( \angle AOB = 60^\circ \) \( = \frac{60^\circ}{360^\circ} \times \pi \times 10^2 - \frac{1}{2} \times 10^2 \times \sin 60^\circ = \frac{50\pi}{3} - 25\sqrt{3} \) cm².
Required area \( = \) area enclosed by arc Q and chord AB \( - \) area enclosed by arc P and chord AB \( = \left\{\frac{25\pi}{2} - \frac{50\pi}{3} + 25\sqrt{3}\right\} = \left\{25\sqrt{3} - \frac{25\pi}{6}\right\} = 25\left(\sqrt{3} - \frac{\pi}{6}\right) \) cm².
Hence proved.