Question

A circle of radius $r$ is inscribed in a square. The mid points of sides of the square have been connected by line segment and a new square resulted. The sides of the resulting square were also connected by segments so that a new square was obtained and so on, then the radius of the circle inscribed in the $n ^{\text {th }}$ square is

• A. $\left[2^{\frac{1- n }{2}}\right] r$
• B. $\left[2^{\frac{3-3 n }{2}}\right] r$
• C. $\left[2^{-\frac{ n }{2}}\right] r$
• D. $\left[2^{-\frac{5-3 n }{2}}\right] r$

Answer 1

Answer: (A)

Side of square $S_1=2 r$
side of square $S_2=r \sqrt{2} \left(a^2+a^2=4 r^2 \Rightarrow a=r \sqrt{2}\right)$
$=\frac{2 r }{2}\left(\frac{1}{\sqrt{2}}\right)^{2-1}=2 r \left[\frac{1}{\sqrt{2}}\right]^{2-1}$
side of square $S_3=2 r\left(\frac{1}{\sqrt{2}}\right)^{3-1}=2 r\left(\frac{1}{\sqrt{2}}\right)^2$ and so on,
side of square $S _{ n }=2 r \left(\frac{1}{\sqrt{2}}\right)^{ n -1} ; $
$ radius $= r \left(2^{-\frac{1}{2}}\right)^{ n -1}= r \left(2^{\frac{1- n }{2}}\right)$ and so on,
side of square $\left.S_n=r\left(2^{-1 / 2}\right)^{n-1}=r\left(2^{\frac{1-n}{2}}\right)\right]$