question_answer

A chord of a circle of radius \[\text{28 cm}\] subtends an angle of \[60{}^\circ \]at the centre of the circle. The area of the minor segment is: (Take \[\sqrt{3}=1.732\])

A) \[60.256\,c{{m}^{2}}\]

B) \[339.47\,c{{m}^{2}}\]

C) \[410.67\,c{{m}^{2}}\]

D) \[71.20\,c{{m}^{2}}\]

Correct Answer: D

Solution :

    

[d] Here, \[r=28\,cm\] and \[\theta =60{}^\circ \]

Since   \[OA=OB\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\angle A=\angle B=60{}^\circ \]

\[\Rightarrow \,\,\,\,\Delta AOB\] is an equilateral triangle.

\[\therefore \]Area of \[\Delta AOB\]

\[=\frac{\sqrt{3}}{4}{{r}^{2}}=\frac{1732}{4}\times 28\times 28=339.472c{{m}^{2}}\]

Area of the minor segment \[=\frac{\theta }{360{}^\circ }\pi {{r}^{2}}-\] Area of \[\Delta AOB\]

\[=\frac{60{}^\circ }{360{}^\circ }\times \frac{22}{7}\times 28\times 27-339.4372\]

\[=(410.67-339.47)c{{m}^{2}}=71.20c{{m}^{2}}\]