Question

In the given figure, when each of the outer circles have radii $r$, then the radius of the inner circle will be:

• A. $\sqrt{2}r$
• B. $(\sqrt{2}-1)r$
• C. $\frac{1}{\sqrt{2}r}$
• D. $\frac{2}{(\sqrt{2}+1)r}$

Answer 1

The correct option is B $(\sqrt{2}-1)r$

Let the radius of the four circle ${}^{\prime }{r}^{\prime }$. Each side of the square is $2r$.

$\therefore$ Distance from top right circle to bottom left circle $=\sqrt{{\left(2r\right)}^2+{\left(2r\right)}^2}$

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

$=2\sqrt{2}r$

Now, take away the parts of that distance that overlap with the two outer circles $=2\sqrt{2}r-2r$, since the part taken out of that distance from the two circles was their common radius twice.

This distance is the diameter of inner circle.

So, radius of inner circle $=\frac{2r(\sqrt{2}-1)}{2}=r(\sqrt{2}-1)$