The sum of perimeters of the six faces of a cuboid is 72 cm and the total surface area of the cuboid is 16 cm². Find the longest possible length that can be kept inside the cuboid :
a) 5.2 cm
b) 7.8 cm
c) 8.05 cm
d) 8.36 cm

Answer: c
Explanation: Sum of perimeters of the six faces :
$$\eqalign{ & = 2\left[ {2\left( {l + b} \right) + 2\left( {b + h} \right) + 2\left( {l + h} \right)} \right] \cr & = 4\left( {2l + 2b + 2h} \right) \cr & = 8\left( {l + b + h} \right) \cr} $$
Total surface area $$ = 2\left( {lb + bh + lh} \right)$$
$$\eqalign{ & 8\left( {l + b + h} \right) = 72 \cr & \Rightarrow l + b + h = 9 \cr & 2\left( {lb + bh + lh} \right) = 16 \cr & \Rightarrow lb + bh + lh = 8 \cr} $$
Now,
$${\left( {l + b + h} \right)^2} = {l^2} + {b^2} + {h^2} + 2$$     $$\left( {lb + bh + lh} \right)$$
$$\eqalign{ & \Rightarrow {\left( 9 \right)^2} = {l^2} + {b^2} + {h^2} + 16 \cr & {l^2} + {b^2} + {h^2} = 81 – 16 \cr & {l^2} + {b^2} + {h^2} = 65 \cr} $$
Required length :
$$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {65} \cr & = 8.05\,cm \cr} $$