{"id":8940,"date":"2021-12-09T22:31:40","date_gmt":"2021-12-09T17:01:40","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8940"},"modified":"2022-01-16T17:12:12","modified_gmt":"2022-01-16T11:42:12","slug":"segment-of-a-circle-area","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/segment-of-a-circle-area\/","title":{"rendered":"Segment of a Circle Area – Formula and Examples"},"content":{"rendered":"
Here you will learn what is the formula for major and minor segment of a circle to find its area with examples.<\/p>\n
Let’s begin –<\/p>\n
A segment of a circle is defined as the part of circle bounded by a chord and the arc.<\/p>\n
In the figure, the part APB is a segment of circle.<\/p>\n
Let AB be a chord of circle with radius r. Let \\(\\angle\\) AOB = \\(\\theta\\) and 0 < \\(\\theta\\) 180.<\/p>\n
The minor segment corresponding to chord AB is shown in figure.<\/p>\n
Area of Minor Segment = Area of sector OAB – Area of triangle OAB<\/p>\n
Since area of sector = \\(\\theta\\over 360\\) \\(\\pi r^2\\) and <\/p>\n
area of triangle = \\(1\\over 2\\) \\(r^2 sin\\theta\\)<\/p>\n
\nHence, Area of Minor Segment = \\(\\theta\\over 360\\) \\(\\pi r^2\\) – \\(1\\over 2\\) \\(r^2 sin\\theta\\)<\/p>\n<\/blockquote>\n
Area of Major Segment = Area of Circle – Area of Minor Segment<\/p>\n
\nHence, Area of Major Segment = \\(\\pi r^2\\) – ( \\(\\theta\\over 360\\) \\(\\pi r^2\\) – \\(1\\over 2\\) \\(r^2 sin\\theta\\))<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : A chord 10 cm long is drawn in a circle whose radius is \\(\\sqrt{50}\\) cm. Find the area of segments.<\/p>\n
Solution<\/span><\/strong> : Radius of the circle = \\(\\sqrt{50}\\) cm<\/p>\n
\\(\\therefore\\) Area of circle = \\(22\\over 7\\) \\(\\times\\) \\((\\sqrt{50})^2\\) = \\(1100\\over 7\\) = 157.14 \\(cm^2\\)<\/p>\n
Since, OA = OB = \\(\\sqrt{50}\\) cm<\/p>\n
\\((OA)^2\\) + \\((OB)^2\\) = 50 + 50 = 100 cm<\/p>\n
\\((AB)^2\\) = 100<\/p>\n
\\(\\therefore\\) \\((OA)^2\\) + \\((OB)^2\\) = \\((AB)^2\\) \\(\\implies\\) \\(\\angle\\) AOB = 90<\/p>\n
Area of sector OAB = \\(90\\over 360\\) \\(\\times\\) \\(22\\over 7\\) \\(\\times\\) \\((\\sqrt{50})^2\\) = 39.29 \\(cm^2\\)<\/p>\n
Area of triangle OAB = \\(1\\over 2\\) \\(r^2 sin\\theta\\) = \\(1\\over 2\\) \\(\\times\\) (50 sin 90) = 25 \\(cm^2\\)<\/p>\n
\\(\\therefore\\) Area of Minor Segment = Area of sector OAB – Area of triangle OAB<\/p>\n
= 39.29 – 25 = 14.29 \\(cm^2\\)<\/p>\n
Area of Major Segment = Area of Circle – Area of Minor Segment<\/p>\n
= 157.14 – 14.29 = 142.85 \\(cm^2\\)<\/p>\n\n\n