Geometry & Mensuration
Aptitude

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Q.

A chord $AB$ of a circle of radius 5.25 cm makes an angle of $60^\circ$ at the centre of the circle. Find the area of the major segment.

 A.

168 cm$^{2}$

 B.

100 cm$^{2}$

 C.

84 cm$^{2}$

 D.

70 cm$^{2}$

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Solution:
Option(C) is correct

The given circle is an equivlateral triangle 

Area of the minor sector

\(=\dfrac{60}{360}\times \pi\times 5.25^2\)

\(=14.4375\text{ cm}^2\)

Area of the triangle 

\(=\dfrac{\sqrt{3}}{4}\times 5.25^2\)

\(=11.93\text{ cm}^2\)

Area of the minor segment = Area of the minor sector - Area of the triangle 

\(=2.5\text{ cm}^2\)

Area of the major segment = Area of the circle - Area of the minor segment

\(=86.54\text{ cm}^2-2.5\text{ cm}^2\) 

\(84\text{ cm}^2\)


(1) Comment(s)


Prabhat
 ()

area of major sigment =area of triangle+ area of major sector

=11.93 cm2 + 300/360[π×(5.25)2]

= 11.93+72.1

=84 cm2