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02 Apr 2020, 07:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
66% (03:00) correct
34%
(03:11)
wrong
based on 68
sessions
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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. \(\frac{\pi}{4}\)
B. \(\frac{\pi}{2}\)
C. \(\pi\)
D. \(\frac{3\pi}{2}\)
E. \(2\pi\)
Attachment:
download.png [ 7.01 KiB | Viewed 7951 times ]
02 Apr 2020, 08:24
Kudos
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Bunuel
Let the radius of inner & outer circle be 'r' & 'R' respectively
and the side of inner & outer squares be 'a' & 'b' respectively
To Find: \(\frac{2πR}{4a} = \frac{πR}{2a}\)
Diagonal of ABCD = diamter of inner circle
--> \(\sqrt{2}a = 2r\)
--> \(a = \sqrt{2}r\)
Also, diagonal of PQRS = diameter of outer circle
--> \(\sqrt{2}b = 2R\)
--> \(b = \sqrt{2}R\) ....... (1)
Triangle APB is a right angled isosceles triangle
--> \(AB^2 = AP^2 + BP^2 = (b/2)^2 + (b/2)^2 = b^2/2\)
--> \(a^2 = b^2/2\)
--> \(b = \sqrt{2}a\)
From (1),
\(b = \sqrt{2}R\)
--> \(\sqrt{2}a = \sqrt{2}R\)
--> \(a = R\)
\(\frac{πR}{2a}\) = \(\frac{π}{2}\)
Option B
02 Apr 2020, 09:06
Kudos
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Bunuel
Let PQ = 2a
Radius of the outer circle = \(a\sqrt{2}\)
Perimeter of the outer circle = \(2\pi a\sqrt{2}\)
Side of square ABCD = AB = \(a\sqrt{2}\)
Perimeter of square ABCD = \(4a\sqrt{2}\)
The ratio of the perimeter of the outer circle to that of polygon ABCD = \(2\pi a\sqrt{2}/4a\sqrt{2} = \pi/2\)
IMO B
