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 ?

Updated On: Jul 21, 2025
  • \(1125\pi\sqrt{2}\)
  • \(750\pi\)
  • \(750\pi\sqrt{2}\)
  • \(1125\pi\)
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The Correct Option is A

Approach Solution - 1

To find the volume of the sphere in which the rectangular box is inscribed, we begin with the given conditions: the surface area (SA) of the rectangular box is 846 sq cm, and the sum of the lengths of all its edges (P) is 144 cm.
Let's denote the dimensions of the box as \(a\), \(b\), and \(c\). The following equations arise from the problem statement:
  • SA Equation: \[2(ab+bc+ca)=846 \rightarrow ab+bc+ca=423\]
  • Perimeter Equation: \[4(a+b+c)=144 \rightarrow a+b+c=36\]
The box is inscribed in a sphere. Thus, the diagonal of the box equals the diameter of the sphere. Using the Pythagorean theorem in three dimensions, the diagonal \(d\) is:
\[d=\sqrt{a^2+b^2+c^2}\]
Now, we use the identity:
\[(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\]
Substituting the known values:
\[36^2=a^2+b^2+c^2+2 \times 423\]
\[1296=a^2+b^2+c^2+846\]
Therefore:
\[a^2+b^2+c^2=450\]
Thus, the diagonal (the diameter of the sphere) is:
\[d=\sqrt{450}=15\sqrt{2}\]
The radius \(r\) of the sphere is \(\frac{d}{2}=\frac{15\sqrt{2}}{2}\).
The volume \(V\) of the sphere is given by the formula:
\[V=\frac{4}{3}\pi r^3\]
Substituting \(r=\frac{15\sqrt{2}}{2}\), we get:
\[V=\frac{4}{3}\pi \left(\frac{15\sqrt{2}}{2}\right)^3\]
\[=\frac{4}{3}\pi \times \left(\frac{3375 \times 2\sqrt{2}}{8}\right)\]
\[=\frac{4}{3}\pi \times \frac{6750\sqrt{2}}{8}\]
\[=\pi \times 1125\sqrt{2}\]
Thus, the volume of the sphere is \(\boxed{1125\pi\sqrt{2}}\) cubic cm.
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Approach Solution -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.

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