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

Question

Accepted Answer

The problem involves calculating the volume of a sphere that inscribes a rectangular box. Here's the step-by-step solution:

**Given Data:**

- Surface area of the rectangular box:
     $$
     2(lb + lh + bh) = 846 \, 	ext{sq cm}.
     $$
   - Sum of all edges:
     $$
     4(l + b + h) = 144 \, 	ext{cm}.
     $$

**Calculate $$l + b + h$$:**

- From the sum of edges:
     $$
     l + b + h = \frac{144}{4} = 36 \, 	ext{cm}.
     $$

**Relation Between Dimensions:**

- Use the equation:
     $$
     l^2 + b^2 + h^2 + 2(lb + lh + bh) = (l + b + h)^2.
     $$
   - Substitute values:
     $$
     l^2 + b^2 + h^2 + 846 = 36^2.
     $$
   - Simplify:
     $$
     l^2 + b^2 + h^2 = 1296 - 846 = 450.
     $$

**Diameter of Sphere:**

- The sphere's diameter is given by:
     $$
     	ext{Diameter} = \sqrt{l^2 + b^2 + h^2} = \sqrt{450}.
     $$
   - Simplify:
     $$
     	ext{Diameter} = 15\sqrt{2}.
     $$

**Radius of Sphere:**

- Radius is half the diameter:
     $$
     r = \frac{	ext{Diameter}}{2} = \frac{15\sqrt{2}}{2}.
     $$

**Volume of Sphere:**

- Use the formula for the volume of a sphere:
     $$
     V = \frac{4}{3} \pi r^3.
     $$
   - Substitute $$r = \frac{15\sqrt{2}}{2}$$:
     $$
     V = \frac{4}{3} \pi \left(\frac{15\sqrt{2}}{2}ight)^3.
     $$
   - Simplify:
     $$
     V = \frac{4}{3} \pi \cdot \frac{3375\sqrt{2}}{8}.
     $$
   - Final value:
     $$
     V = 1125\pi\sqrt{2}.
     $$

### Final Answer:

The volume of the sphere is **$$1125\pi\sqrt{2}$$ cubic cm**.  
Correct option: **B**

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