**How To Find The Area Of A Segment Of A Circle**

Area of the sector OPRQ = Area of the segment PRQ + Area of ∆OPQ

⇒ Area of segment PRQ = \(\left\{ \frac{\pi }{360}\times \theta -\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right\}{{r}^{2}} \)

**Read More:**

- Parts of a Circle
- Perimeter of A Circle
- Common Chord of Two Intersecting Circles
- Construction of a Circle
- The Area of A Circle
- Properties of Circles
- Sector of A Circle
- The Area of A Sector of A Circle

**Area Of A Segment Of A Circle With Examples**

**Example 1: ** Find the area of the segment of a circle,given that the angle of the sector is 120º and the radius of the circle is 21 cm. (Take π = 22/7)

**Sol. ** Here, r = 21 cm and π = 120

∴ Area of the segment

= \(\left\{ \frac{\pi }{360}\times \theta -\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right\}{{r}^{2}} \)

\( =\left\{ \frac{22}{7}\times \frac{120}{360}-\sin 60{}^\text{o}\cos 60{}^\text{o} \right\}{{\left( 21 \right)}^{2}}c{{m}^{2}} \)

\( =\left\{ \frac{22}{21}-\frac{1}{2}\times \frac{\sqrt{3}}{2} \right\}{{\left( 21 \right)}^{2}}c{{m}^{2}} \)

\( =\left\{ \frac{22}{21}\times {{(21)}^{2}}-{{(21)}^{2}}\times \frac{\sqrt{3}}{4} \right\}c{{m}^{2}} \)

\( =\left( 462-\frac{441}{4}\sqrt{3} \right)=\frac{21}{4}(88-21\sqrt{3})\text{ c}{{\text{m}}^{\text{2}}} \)

**Example 2: ** A chord AB of a circle of radius 10 cm makes a right angle at the centre of the circle. Find the area of major and minor segments (Take π = 3.14)

**Sol.** We know that the area of a minor segment of angle θº in a circle of radius r is given by

\(A=\left\{ \frac{\pi }{360}\times \theta -\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right\}{{r}^{2}} \)

Here, r = 10 and θ = 90º

\( \therefore A=\left\{ \frac{3.14\times 90}{4}-\sin 45{}^\text{o}\,\,\cos 45{}^\text{o} \right\}{{\left( 10 \right)}^{2}} \)

\( \Rightarrow A=\left\{ \frac{3.14}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}} \right\}{{\left( 10 \right)}^{2}} \)

A = {3.14 × 25 – 50} cm^{2} = (78.5 – 50) cm^{2}

= 28.5 cm^{2}

Area of the major segment = Area of the circle – Area of the minor segment

= {3.14 × 10^{2} – 28.5} cm^{2}

= (314 – 28.5) cm^{2} = 285.5 cm^{2}

**Example 3: ** The diagram shows two arcs, A and B. Arc A is part of the circle with centre O and radius of PQ. Arc B is part of the circle with centre M and radius PM, where M is the mid-point of PQ. Show that the area enclosed by the two arcs is equal to \( 25\left\{ \sqrt{3}-\frac{\pi }{6} \right\}\text{ c}{{\text{m}}^{\text{2}}} \)

**Sol. ** We have,

Area enclosed by arc B and chord PQ = Area of semi-circle of radius 5 cm

\( =\frac{1}{2}\times \pi \times {{5}^{2}}=\frac{25\pi }{2}c{{m}^{2}} \)

Let ∠MOQ = ∠MOP = θ

In ∆OMP, we have

\( \sin \theta =\frac{PM}{OP}=\frac{5}{10}=\frac{1}{2} \)

⇒ θ = 30º ⇒ ∠POQ = 2θ = 60º

∴ Area enclosed by arc A and chord PQ.

= Area of segment of circle of radius 10 cm and sector containing angle 60º

\( =\left\{ \frac{\pi \times 60}{360}-\sin 30{}^\text{o}\times \cos 30{}^\text{o} \right\}\times \text{ }{{10}^{2}} \)

\( \left[ \because A=\left\{ \frac{\pi \theta }{360}-\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right\}{{r}^{2}} \right] \)

\( =\left\{ \frac{50\pi }{3}-25\sqrt{3} \right\}c{{m}^{2}} \)

Hence, Required area

\( =\left\{ \frac{25\pi }{2}-\left( \frac{50\pi }{3}-25\sqrt{3} \right) \right\} \)

\( =\left\{ 25\sqrt{3}-\frac{25\pi }{6} \right\} \)

\( =25\left\{ \sqrt{3}-\frac{\pi }{6} \right\}\text{ c}{{\text{m}}^{\text{2}}} \)