Question

A circle is inscribed in a regular octagon. The same circle circumscribes a regular hexagon. Find the ratio of the areas of the circle, the hexagon and the octagon.

• A. \(2\pi : 3\sqrt{3} : 16(\sqrt{2}-1)\)
• B. \(\pi : 3\sqrt{3} : 4(\sqrt{2}-1)\)
• C. \(\dfrac{2\pi}{3} : 2\sqrt{3} : 4(\sqrt{2}-1)\)
• D. None of these

Hint:

Write each polygon's area using the circle's radius, as the octagon's apothem and the hexagon's circumradius, then simplify tan(22.5 degrees) to root 2 minus 1.

Answer 1

The Correct Option is A

Concept:
Let the circle have radius \(r\). Since the circle is inscribed in the octagon, \(r\) equals the octagon's apothem (the perpendicular distance from centre to a side). Since the same circle circumscribes the hexagon, \(r\) equals the hexagon's circumradius (the distance from centre to a vertex).

Step 1: Write the area of a regular polygon in terms of its apothem.
For a regular polygon with \(n\) sides and apothem \(a\), the side length is \(s = 2a\tan(\pi/n)\), and the area is \(\text{Area} = \frac{1}{2} \times \text{perimeter} \times \text{apothem} = \frac{1}{2}(ns)a = n a^2 \tan(\pi/n)\).
For the octagon, \(n = 8\) and \(a = r\), so \(\text{Area}_{\text{octagon}} = 8r^2\tan(\pi/8)\).

Step 2: Write the area of a regular polygon in terms of its circumradius.
A regular \(n\)-gon with circumradius \(R\) splits into \(n\) isosceles triangles, each with a vertex angle of \(2\pi/n\) at the centre, so \(\text{Area} = \frac{n}{2}R^2\sin(2\pi/n)\).
For the hexagon, \(n = 6\) and \(R = r\), so \(\text{Area}_{\text{hexagon}} = \frac{6}{2}r^2\sin(60^{\circ}) = 3r^2 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}r^2\).

Step 3: Write the area of the circle and form the ratio.
\(\text{Area}_{\text{circle}} = \pi r^2\). So the ratio of areas is:
\[ \pi r^2 : \frac{3\sqrt{3}}{2}r^2 : 8\tan(\pi/8)\,r^2 \]
Cancel \(r^2\) and multiply every term by 2 to clear the fraction:
\[ 2\pi : 3\sqrt{3} : 16\tan(\pi/8) \]

Step 4: Simplify \(\tan(\pi/8)\).
\(\pi/8\) radians is \(22.5^{\circ}\), and the half-angle formula gives \(\tan(22.5^{\circ}) = \sqrt{2} - 1\). Substituting this in:
\[ 2\pi : 3\sqrt{3} : 16(\sqrt{2}-1) \]

Final Answer:
The ratio of the areas of the circle, hexagon and octagon is \(2\pi : 3\sqrt{3} : 16(\sqrt{2}-1)\), which is option (1). Option (2) drops the factor of 2 needed to clear the apothem's fraction, and option (3) mismatches the hexagon and octagon terms, so both are incorrect. \[ \boxed{2\pi : 3\sqrt{3} : 16(\sqrt{2}-1)} \]