A spherical disc of radius 5 cm is placed on a horizontal surface against a vertical wall as given.
Another small disc is placed in the gap between the right angle and the disc such that the horizontal surface, vertical wall, the larger disc, and the smaller disc touch each other. What is the radius (in cm) of the smaller disc ?
Solution
The radius of the bigger circle is 5 cm while of the smaller circle is r cm.
Since the two sides of the wall are tangential to the circles, we get a square formed by the radius of the bigger circle and the two sides wall, each with the length of 5 cm.
Now, the diagonal of this square (s1) shall be $$5\sqrt{\ 2}$$.
Focusing on the smaller circle, on which the wall is tangential again. We get a square here as well, the diagonal of which is $$r\sqrt{\ 2}$$.
Hence, the diagonal of s1: $$5\sqrt{\ 2}$$ = r + $$r\sqrt{\ 2}$$ + 5
On solving this, we get r = $$15 - 10sqrt{\ 2}$$
