Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
In the adjoining figure, if the radius of each of the four outer circles is $r$, what is the radius of the inner circle? $\frac{2}{\sqrt{2}+1} r$
(b)$\frac{1}{\sqrt{2}} r$
(c)$(\sqrt{2}-1) r$
(d)$\sqrt{2} r$
Answer
Answer (as printed): C
Explanation
Side of the square $=2 r$. Diagonal of the square $=2 \sqrt{2} r$. $$\begin{aligned} \therefore \quad & \text { Diameter of the inner circle }=(2 \sqrt{2} r-2 r)=2 r(\sqrt{2}-1) \\ & \text { Radius of the inner circle }=r(\sqrt{2}-1) . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.