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  • A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use pi = 3.14 and root 3 = 1.73)

6.  A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use \pi = 3.14 and \sqrt 3 = 1.73)

Answers (1)

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The area of the sector is : 

                                                 \\=\ \frac{\Theta }{360^{\circ}}\times \pi r^2\\\\=\ \frac{60^{\circ} }{360^{\circ}}\times \pi \times 15^2\\\\=\ 117.85\ cm^2                                                                                         

Now consider the triangle, the angle of the sector is 600.

This implies it is an equilateral triangle. (As two sides are equal so will have the same angle. This possible only when all angles are equal i.e., 60o.) 

Thus, the area of the triangle is:- 

                                               =\ \frac{\sqrt{3}}{4}\times 15^2

or                                           =\ 56.25 \sqrt{3}\ =\ 97.31\ cm^2

Hence area of the minor segment :   =\117.85\ -\ 97.31\ =\ 20.53\ cm^2

And the area of the major segment is :

                                                          =\ \pi r^2\ -\ 20.53

or                                                      =\ \pi\times 15^2\ -\ 20.53

or                                                       =\ 707.14\ -\ 20.53

or                                                       =\ 686.6\ cm^2

Posted by

Devendra Khairwa

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