Question

The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is ?

• A. $1125\pi\sqrt{2}$
• B. $750\pi$
• C. $750\pi\sqrt{2}$
• D. $1125\pi$

Answer 1

Answer: (A)

$1125\pi\sqrt{2}$

Answer 2

Let the dimensions of the rectangular box be $a$, $b$, and $c$. The surface area $S$ and sum of the lengths of all edges $L$ are given by:

$S = 2(ab + bc + ca) = 846$,
$L = 4(a + b + c) = 144$.

From the second equation, we get:

$a + b + c = 36$.

Now, the box is inscribed in a sphere, so the diagonal of the box is the diameter of the sphere. The diagonal of the box is:

$\sqrt{a^2 + b^2 + c^2}$.

Let $D$ be the diameter of the sphere. Thus, the radius $r$ of the sphere is:

$r = \frac{D}{2} = \frac{\sqrt{a^2 + b^2 + c^2}}{2}$.

The volume $V$ of the sphere is:

$V = \frac{4}{3}\pi r^3$

15

Using the given information, we can solve for $a$, $b$, and $c$, and then find the volume of the sphere.