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?