Question

The figure below shows two concentric circles with centre 'O'. $PQRS$ is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at points $B,C,D$ and $A$. What is the ratio of the perimeter of the outer circle to that of polygon $ABCD$?

• A. $\frac{\pi }{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $\frac{3\pi }{2}$

Answer 1

Answer: (A)

$\frac{\pi }{2}$

$A,B,C$ and $D$ must be the mid-points of $PS,PQ,QR$ and $RS$ and $ABCD$ be a square.

Let $PQ$ be $r.$

$\Rightarrow$ Then, the radius of the outer circle $=\frac{r\sqrt{2}}{2}=\frac{r}{\sqrt{2}}$

The diameter of the inner circle is equal to the side of the outer square, that is $r.$

The diameter of the $7$ inner circle is equal to the diagonal of the inner square. So, the diagonal of the inner square is $r.$

Hence, the side of the inne square is $\frac{r}{\sqrt{2}}$

$\therefore$ Ratio of perimeter of outer circle to that of polygon $ABCD$ $=\frac{2\pi \frac{r}{\sqrt{2}}}{4\frac{r}{\sqrt{2}}}=\frac{\pi }{2}$