Question 16.

Given that, The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm

So, $2(lb+bh+hl)=846$. 

And $4(l+b+h)=144$

$(l+b+h)=36$

$\left(l+b+h\right)^2=l^2+b^2+h^2+2\left(lb+bh+hl\right)$

$1296=\left(l^2+b^2+h^2\right)+846$

$450=l^2+b^2+h^2$

We are told that this cuboid is inscribed in a sphere, the body diagonal of the cuboid equals the diameter of the sphere, this can be visualised as:

This is nothing but, $\sqrt{l^2+b^2+h^2}=2R$

$l^2+b^2+h^2=4R^2$

$450=4R^2$

$R^2=\frac{225}{2}$

$R=\frac{15}{\sqrt{2}}$

Volume of sphere will be $\dfrac{4}{3}\times\ \pi\ \times\ \left(\dfrac{15}{\sqrt{2}}\right)^3$

$\dfrac{4}{3}\pi\ \left(\dfrac{3375}{2\sqrt{\ 2}}\right)$

$\pi\ \times\ 1125\sqrt{\ 2}$