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The figure below shows two concentric circle with centre O. PQRS 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?

Image for Geometry & Mensuration, Aptitude:2199-1


$\pi / 4$


$3 \pi / 2$


$\pi / 2$



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Option(C) is correct

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

Let PQ be $r$. Then, the radius of the outer circle,

\(= \dfrac{r\sqrt{2}}{2}\\ = \dfrac{r}{\sqrt{2}}\)
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 inner square is $r/\sqrt{2}$. \(\therefore\) Ratio of perimeter of outer circle to that of
Polygon ABCD,

\(=\left(\dfrac{2\pi \dfrac{r}{\sqrt2}}{4\dfrac{r}{\sqrt2}}\right)\\ =\dfrac{\pi}{2}\)

Thus option (C) is the right choice.

Edit: As pointed by KARTIK correct typo in the solution. (Changed PO $=r$ to PQ $=r$)

(4) Comment(s)


There is an awesome shortcut

Since we know 2πr is the perimeter of the circle

while 2*s (s being a side of the square) is perimeter for that inner square

since the center is common for all shapes here, it can be inferred that the radius when doubled will equate s; thus by comparing the both required perimeters

2πr = 2(2r)

2πr = 4r

2πr/4r => π/2 Answer :)


typo error let PO is $r$, should be PQ is $r$. :)


Thank you for letting me know the anomaly. Corrected it.


No problem ;)

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