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

Answer 1

Answer: (A)

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