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The area of the greatest circle which can be inscribed in a square whose perimeter is 120 cm, is :
Options:
A .  $$\frac{{22}}{7} \times {\left( {\frac{7}{2}} \right)^2}c{m^2}$$
B .  $$\frac{{22}}{7} \times {\left( {\frac{9}{2}} \right)^2}c{m^2}$$
C .  $$\frac{{22}}{7} \times {\left( {\frac{15}{2}} \right)^2}c{m^2}$$
D .  $$\frac{{22}}{7} \times {\left( {15} \right)^2}c{m^2}$$
Answer: Option D
Side of the square :$$\eqalign{
& = \frac{{120}}{4}cm \cr
& = 30\,cm \cr} $$
Radius of the required circle :$$\eqalign{
& = \left( {\frac{1}{2} \times 30} \right)cm \cr
& = 15\,cm \cr
& = \pi \times {r^2} \cr
& = \left[ {\frac{{22}}{7} \times {{\left( {15} \right)}^2}} \right]c{m^2} \cr} $$

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