A regular octagon ABCDEFGH has sides on length 6 cm each. Then the area, in sq. cm, of the square ACEG is

We have a regular octagon and need to find the area of square ACEG, which is formed by connecting every other vertex of the octagon.

The side of our square is the distance AC. Since we're dealing with a regular octagon, we can use the interior angle and apply the Law of Cosines to find this distance.

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First, let's find the interior angle of a regular octagon:

Interior angle = $\frac{n-2}{n} \times 180°$

For an octagon (n = 8):

Interior angle = $\frac{8-2}{8} \times 180° = \frac{6}{8} \times 180° = 135°$

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Now we'll consider triangle BAC, where:

• B is the center of the octagon

• A and C are vertices of the octagon separated by one vertex

• BA = BC = 6 (radius from center to vertex, equal to the side length for this construction)

• Angle ABC = 135°

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Using the Law of Cosines to find AC:

$c^2 = a^2 + b^2 - 2ab \cos C$

In our case:

$AC^2 = BA^2 + BC^2 - 2(BA)(BC) \cos(135°)$

$AC^2 = 6^2 + 6^2 - 2(6)(6) \cos(135°)$

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Since $\cos(135°) = -\frac{1}{\sqrt{2}}$:

$AC^2 = 36 + 36 - 72 \times \left(-\frac{1}{\sqrt{2}}\right)$

$AC^2 = 72 - \left(72 \times \frac{-1}{\sqrt{2}}\right)$

$AC^2 = 72 + \frac{72}{\sqrt{2}}$

$AC^2 = 72\left(1 + \frac{1}{\sqrt{2}}\right)$

$AC^2 = 36\left(2 + \sqrt{2}\right)$

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Since ACEG is a square with side length AC:

Area of square ACEG = $(AC)^2 = 36(2 + \sqrt{2})$