The length of the largest possible rod that can be placed in a cubical room is 35√3 m. The surface area of the largest possible sphere that fit within the cubical room $$\left( {{\text{assuming }}\pi = \frac{{22}}{7}} \right)$$    (in sq. m) is

Solution (By Examveda Team)

$$\eqalign{ & {\text{Diagonal}} = 35\sqrt 3 \cr & \therefore {\text{The length of largest rod}} = {\text{Diagonal}} = {\text{side}}\sqrt 3 \cr & {\text{side}}\sqrt 3 = 35\sqrt 3 \cr & {\text{side}} = \frac{{35\sqrt 3 }}{{\sqrt 3 }} = 35 \cr & {\text{Side of cube}} = 35 \cr} $$

$$\eqalign{ & {\text{Diameter of the sphere }} = {\text{ side of the cube}} \cr & 2 \times {\text{radius}} = {\text{side}} \cr & {\text{radius}} = \frac{{35}}{2}{\text{ cm}} \cr & {\text{Surface area of the sphere}} = 4\pi {r^2} \cr & = 4 \times \frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \cr & = 3850{\text{ }}{{\text{m}}^2} \cr} $$