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4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding :

  (i) minor segment

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The angle in the minor sector is 90o.

Thus the area of the sector is given by :-  

                                                   \\Area\ =\ \frac{\Theta}{360^{\circ}}\times \pi r^2\\\\Area=\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 10^2\\\\Area=\ \frac{1100}{14}\ cm^2\ =\ 78.5\ cm^2                                                    

Now area of triangle is  :-          

                                                  Area\ =\ \frac{1}{2}bh\ =\ \frac{1}{2}\times 10\times 10\ =\ 50\ cm^2

Thus area of minor segment   =   Area of sector    -   Area of triangle

or                                            =\ 78.5\ -\ 50\ =\ 28.5\ cm^2

Posted by

Devendra Khairwa

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