# Area of Portion of Sphere from a inside Cube

The late painter Maqbool Fida Husain once coloured the surface of a huge hollow steel sphere, of radius $$1$$ metre, using just two colours, Red and Blue. As was his style however, both the red and blue areas were a bunch of highly irregular disconnected regions. The late sculptor Ramkinkar Baij then tried to fit in a cube inside the sphere, the eight vertices of the cube touching only red coloured parts of the surface of the sphere. Assume $$\pi=3.14$$ for solving this problem. Which of the following is true?

A)Baij is bound to succeed if the area of the red part is $$10 sq. metres$$;

B)Baij is bound to fail if the area of the red part is $$10 sq. metres$$;

C)Baij is bound to fail if the area of the red part is $$11 sq. metres$$;

D)Baij is bound to succeed if the area of the red part is $$11 sq. metres$$;

E)None of the above.

My solution is: Here given , radius of sphere  $$r=1m.$$

So, $$d=2m.$$

Now, say side of a cube is $$a$$ $$m.$$

So, diagonal of a square $$a\sqrt{2}m.$$

Now, see using Pythagoras Theorem

$$a^{2}+\left ( a\sqrt{2} \right )^{2}=2^{2}$$

$$\Rightarrow a=\frac{2}{\sqrt{3}}$$

Now, we have to divide the sphere such a way $$a=\frac{2}{\sqrt{3}}$$ is maximum cord.

Check the diagram, where red area inscribe the cube. and blue area outside the cube.

Now see the sphere divided inside hollow region of the cube in $$6$$ equal parts (1,2,3,4, shown as in diagram and 5th one in front and 6th one is back. )

So, we can say , it takes the (area of cube+$$\frac{5}{6}$$ . area of hollow sphere) [As only upper side of cube is blue region.]

So, area of cube$$=6\times \left ( \frac{2}{\sqrt{3}} \right )^{2}=8m.$$

Now $$\frac{5}{6}$$ of hollow region of sphere $$\left ( 4\pi \left ( 1 \right )^{2}-6\left ( \frac{2}{\sqrt{3}} \right )^{2} \right )\times \frac{5}{6}=3.809$$

So, total area will be $$=8+3.809=11.809m.$$

So, nearest option will be $$D)$$ or exactly $$E)$$

Is this correct approach?

• gateoverflow.in/25387/tifr2013-a-5 – Will Jagy Jun 17 at 17:46
• @WillJagy haha, that is me only :). I want to know, is my approach correct? – Srestha Jun 17 at 17:58
• @WillJagy Is the ans correct? – Srestha Jun 18 at 2:10