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.segment of circle

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 \(cm^2\)

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