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 \[=\pi \times {{(radius)}^{2}}\] \[=\frac{22}{7}\times {{(15)}^{2}}\,c{{m}^{2}}\]You need to login to perform this action.
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