Question

A circle of radius r is inscribed in a square . The mid point of slides 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 nth 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

The correct option is A $\left[2^{\frac{1-n}{2}}\right]r$

For $1$st circle radius=$r$

For $2$nd circle radius=$\frac{r}{\sqrt{2}}$

Common ratio=$\frac{1}{\sqrt{2}}$

$\therefore$ for $n$th circle

radius=$r\times (\frac{1}{\sqrt{2}}{)}^{n-1}$

=$(2^{-\frac{1}{2}}{)}^{n-1}r$

=$r\left(2^{\frac{1-n}{2}}\right)$