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7.  A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
(Use \pi = 3.14 and \sqrt3 = 1.73)

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For the area of the segment, we need the area of sector and area of the associated triangle.

So, the area of the sector is :

                                           =\ \frac{120^{\circ}}{360^{\circ}}\times \pi \times 12^2

or                                        =\ 150.72\ cm^2

Now, consider the triangle:-  

          Draw a perpendicular from the centre of the circle on the base of the triangle (let it be h).

Using geometry we can write,

                                           \frac{h}{r}\ =\ \cos 60^{\circ}

or                                      h\ =\ 6\ cm

Similarly,                            \frac{\frac{b}{2}}{r}\ =\ \sin 60^{\circ}

or                                      b\ =\ 12\sqrt{3}\ cm

Thus the area of the triangle is :   

                                             =\ \frac{1}{2}\times 12\sqrt{3}\times 6

or                                         =\ 62.28\ cm^2

Hence the area of segment is:=\ 150.72\ -\ 62.28\ =\ 88.44\ cm^2.    

                                                                        

Posted by

Devendra Khairwa

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