question_answer

In the adjoining figure, if the radius of each of the four outer circles is r, what is the radius of the inner circle?

A) \[\frac{2r}{(\sqrt{2}+1)}\]

B) \[\frac{r}{\sqrt{2}}\]

C) \[(\sqrt{2}-1)r\]

D) \[\sqrt{2}r\]

Correct Answer: C

Solution :

[c] Each side of square\[ABCD=2r\] Diagonal, \[BD=\sqrt{2}(2r)=2\sqrt{2}r\] \[BO=\frac{1}{4}\times BD=\sqrt{2}r\] \[\Rightarrow \]   \[BP+PO=\sqrt{2}r\] \[\Rightarrow \]   \[r+PO=\sqrt{2}r\] \[\Rightarrow \]   \[PO=(\sqrt{2}-1)r\]