question_answer

The area of the greatest circle, which can be inscribed in a square whose perimeter is 120 cm, is

A)  \[\frac{22}{7}\times {{(15)}^{2}}c{{m}^{2}}\]

B)  \[\frac{22}{7}\times {{\left( \frac{7}{2} \right)}^{2}}c{{m}^{2}}\]

C)  \[\frac{22}{7}\times {{\left( \frac{15}{2} \right)}^{2}}c{{m}^{2}}\]    

D)  \[\frac{22}{7}\times {{\left( \frac{9}{2} \right)}^{2}}c{{m}^{2}}\]

Correct Answer: A

Solution :

Side of the square \[=\frac{120}{4}=30\,cm\] Clearly, diameter of the greatest circle = Side of the square \[=30\,cm\] \[\therefore \]      Radius \[=\frac{30}{2}=15\,cm\] Required area \[=\pi \times {{(radius)}^{2}}\]             \[=\frac{22}{7}\times {{(15)}^{2}}\,c{{m}^{2}}\]