# Segment of a Circle Area – Formula and Examples

Here you will learn what is the formula for major and minor segment of a circle to find its area with examples.

Let’s begin –

## What is the Segment of a Circle ?

A segment of a circle is defined as the part of circle bounded by a chord and the arc.

In the figure, the part APB is a segment of circle.

## Area of Segments Formula

Let AB be a chord of circle with radius r. Let $$\angle$$ AOB = $$\theta$$ and 0 < $$\theta$$ 180.

The minor segment corresponding to chord AB is shown in figure.

Area of Minor Segment = Area of sector OAB – Area of triangle OAB

Since area of sector = $$\theta\over 360$$ $$\pi r^2$$ and

area of triangle = $$1\over 2$$ $$r^2 sin\theta$$

Hence, Area of Minor Segment = $$\theta\over 360$$ $$\pi r^2$$ – $$1\over 2$$ $$r^2 sin\theta$$

Area of Major Segment = Area of Circle – Area of Minor Segment

Hence, Area of Major Segment = $$\pi r^2$$ – ( $$\theta\over 360$$ $$\pi r^2$$ – $$1\over 2$$ $$r^2 sin\theta$$)

Example : A chord 10 cm long is drawn in a circle whose radius is $$\sqrt{50}$$ cm. Find the area of segments.

Solution : Radius of the circle = $$\sqrt{50}$$ cm

$$\therefore$$   Area of circle = $$22\over 7$$ $$\times$$ $$(\sqrt{50})^2$$ = $$1100\over 7$$ = 157.14 $$cm^2$$

Since, OA = OB = $$\sqrt{50}$$ cm

$$(OA)^2$$ + $$(OB)^2$$ = 50 + 50 = 100 cm

$$(AB)^2$$ = 100

$$\therefore$$  $$(OA)^2$$ + $$(OB)^2$$ = $$(AB)^2$$  $$\implies$$  $$\angle$$ AOB = 90

Area of sector OAB = $$90\over 360$$ $$\times$$ $$22\over 7$$ $$\times$$  $$(\sqrt{50})^2$$ = 39.29 $$cm^2$$

Area of triangle OAB = $$1\over 2$$ $$r^2 sin\theta$$ = $$1\over 2$$ $$\times$$ (50 sin 90) = 25 $$cm^2$$

$$\therefore$$  Area of Minor Segment = Area of sector OAB – Area of triangle OAB

= 39.29 – 25 = 14.29  $$cm^2$$

Area of Major Segment = Area of Circle – Area of Minor Segment

= 157.14 – 14.29 = 142.85 $$cm^2$$