question_answer

The circumference of a circle is 11 cm and the angle of a sector of the circle is \[60{}^\circ .\] The area of the sector is \[(use\,\,\pi =\frac{22}{7})\]

A)  \[1\frac{29}{48}\,\,\text{c}{{\text{m}}^{\text{2}}}\]   

B)  \[2\frac{29}{48}\,\,\text{c}{{\text{m}}^{\text{2}}}\]

C)  \[1\frac{27}{48}\,\,\text{c}{{\text{m}}^{\text{2}}}\]               

D)  \[2\frac{27}{48}\,\,\text{c}{{\text{m}}^{\text{2}}}\]

Correct Answer: A

Solution :

Let the radius of the circle be real. According to the question,             \[2\pi r=11\] \[\Rightarrow \]   \[2\times \frac{22}{7}r=11\] \[\Rightarrow \]   \[r=\frac{11\times 7}{2\times 22}=\frac{7}{4}\,cm\] \[\therefore \] Area of the sector AOB \[\therefore \]      \[=\frac{\theta }{360{}^\circ }\times \pi {{r}^{2}}\]             \[=\frac{60{}^\circ }{360{}^\circ }\times \frac{22}{7}\times \frac{7}{4}\times \frac{7}{4}\,sq\,cm\]             \[=\frac{77}{48}=1\frac{29}{48}\,\,\text{sq}\,\,\text{cm}\]