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?
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?
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Let the side of the square PQRS be s. Find the radius of the outer circle (half the diagonal) and the side of ABCD (it joins the midpoints of the sides of PQRS).
Updated On: Nov 4, 2026
- \(\frac{\pi}{4}\)
- \(\frac{3\pi}{2}\)
- \(\frac{\pi}{2}\)
- \(\pi\)
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The Correct Option is C
Solution and Explanation
Step 1: Outer circle:
Let the side of square PQRS be \(s\). The outer circle passes through P, Q, R, S, so its diameter is the diagonal \(s\sqrt{2}\). Its perimeter is \(\pi \cdot s\sqrt{2}\).
Step 2: Polygon ABCD:
The inner circle touches the sides of PQRS at their midpoints A, B, C, D. Joining adjacent midpoints of a square gives a square whose side is the hypotenuse of a right triangle with legs \(\frac{s}{2}\) and \(\frac{s}{2}\), so \(AB = \frac{s}{\sqrt{2}}\). Perimeter of ABCD \(= 4 \cdot \frac{s}{\sqrt{2}} = 2\sqrt{2}\,s\).
Step 3: Ratio:
\[\frac{\pi s\sqrt{2}}{2\sqrt{2}\,s} = \frac{\pi}{2}\]
Step 4: Check options:
The ratio is \(\frac{\pi}{2} \approx 1.57\), which matches option (3).
Final Answer:
The correct option is (3): \(\frac{\pi}{2}\)
Let the side of square PQRS be \(s\). The outer circle passes through P, Q, R, S, so its diameter is the diagonal \(s\sqrt{2}\). Its perimeter is \(\pi \cdot s\sqrt{2}\).
Step 2: Polygon ABCD:
The inner circle touches the sides of PQRS at their midpoints A, B, C, D. Joining adjacent midpoints of a square gives a square whose side is the hypotenuse of a right triangle with legs \(\frac{s}{2}\) and \(\frac{s}{2}\), so \(AB = \frac{s}{\sqrt{2}}\). Perimeter of ABCD \(= 4 \cdot \frac{s}{\sqrt{2}} = 2\sqrt{2}\,s\).
Step 3: Ratio:
\[\frac{\pi s\sqrt{2}}{2\sqrt{2}\,s} = \frac{\pi}{2}\]
Step 4: Check options:
The ratio is \(\frac{\pi}{2} \approx 1.57\), which matches option (3).
Final Answer:
The correct option is (3): \(\frac{\pi}{2}\)
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