The perimeters of a regular hexagon and a regular octagon are equal. Find the ratio of the area of the octagon to that of the hexagon. [Use $\tan 67.5° = (\sqrt{2} + 1)$]

Perimeter of a regular hexagon = Perimeter of a regular octagon.

Value to use: $\tan 67.5^\circ = (\sqrt{2} + 1)$.

Let the side length of the regular hexagon be $a$ and the side length of the regular octagon be $b$.

Since their perimeters are equal:

$6a = 8b$

$a = \frac{8b}{6}$

$a = \frac{4b}{3}$

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The area of a regular hexagon with side $a$ is given by:

$A_H = \frac{3\sqrt{3}}{2} a^2$

Substituting $a = \frac{4b}{3}$:

$A_H = \frac{3\sqrt{3}}{2} \left( \frac{4b}{3} \right)^2$

$A_H = \frac{3\sqrt{3}}{2} \times \frac{16b^2}{9}$

$A_H = \frac{8\sqrt{3}}{3} b^2$

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A regular octagon can be divided into 8 congruent isosceles triangles with the center of the octagon as the common vertex.

Base of each triangle = side of octagon = $b$

Central angle = $\frac{360^\circ}{8} = 45^\circ$

Base angles of the triangle = $\frac{180^\circ - 45^\circ}{2} = 67.5^\circ$

Let $h$ be the height (apothem) of one such triangle.

Using trigonometry in the right-angled half of the triangle:

$\tan 67.5^\circ = \frac{h}{(b/2)}$

$h = \frac{b}{2} \tan 67.5^\circ$

Area of one triangle:

$= \frac{1}{2} \times b \times \frac{b}{2} \tan 67.5^\circ$

$= \frac{b^2}{4} \tan 67.5^\circ$

Total Area of Octagon:

$A_O = 8 \times \frac{b^2}{4} \tan 67.5^\circ$

$A_O = 2b^2 \tan 67.5^\circ$

Using $\tan 67.5^\circ = (\sqrt{2} + 1)$:

$A_O = 2b^2 (\sqrt{2} + 1)$

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The ratio $\frac{A_O}{A_H}$:

$\text{Ratio} = \frac{2b^2(\sqrt{2} + 1)}{\frac{8\sqrt{3}}{3}b^2}$

$\text{Ratio} = \frac{2(\sqrt{2} + 1) \times 3}{8\sqrt{3}}$

$\text{Ratio} = \frac{6(\sqrt{2} + 1)}{8\sqrt{3}}$

$\text{Ratio} = \frac{3(\sqrt{2} + 1)}{4\sqrt{3}}$

Rationalizing the denominator:

$\text{Ratio} = \frac{3\sqrt{3}(\sqrt{2} + 1)}{4 \times 3}$

$\text{Ratio} = \frac{\sqrt{3}(\sqrt{2} + 1)}{4}$

$\text{Ratio} = \frac{\sqrt{6} + \sqrt{3}}{4}$