Here you will learn what is the formula for major and minor segment of a circle to find its area with examples.
Letโs begin โ
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 โ Area of triangle OAB
Since area of sector = \(\theta\over 360\) \(\pi r^2\) andย
area of triangle = \(1\over 2\) \(r^2 sin\theta\)
Hence, Area of Minor Segment = \(\theta\over 360\) \(\pi r^2\) โ \(1\over 2\) \(r^2 sin\theta\)
Area of Major Segment = Area of Circle โ Area of Minor Segment
Hence, Area of Major Segment = \(\pi r^2\) โ ( \(\theta\over 360\) \(\pi r^2\) โ \(1\over 2\) \(r^2 sin\theta\))
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
\(\therefore\)ย ย 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
\(\therefore\)ย \((OA)^2\) + \((OB)^2\) = \((AB)^2\)ย \(\implies\)ย \(\angle\) AOB = 90
Area of sector OAB = \(90\over 360\) \(\times\) \(22\over 7\) \(\times\)ย \((\sqrt{50})^2\) = 39.29 \(cm^2\)
Area of triangle OAB = \(1\over 2\) \(r^2 sin\theta\) = \(1\over 2\) \(\times\) (50 sin 90) = 25 \(cm^2\)
\(\therefore\)ย Area of Minor Segment = Area of sector OAB โ Area of triangle OAB
= 39.29 โ 25 = 14.29ย \(cm^2\)
Area of Major Segment = Area of Circle โ Area of Minor Segment
= 157.14 โ 14.29 = 142.85 \(cm^2\)