Question

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

• A. \({36(1 + \sqrt{2})}\)
• B. \({72(2 + \sqrt{2})}\)
• C. \({72(1 + \sqrt{2})}\)
• D. \({36(2 + \sqrt{2})}\)

Answer 1

The Correct Option is D

A regular octagon can be divided into a central square and four pairs of isosceles right triangles, where one triangle from each pair is adjacent to one side of the square. To find the area of square ACEG, we begin by calculating the side length of the square formed by the alternate vertices of the octagon, specifically vertices A, C, E, and G.

The octagon can be divided into a larger square by joining opposite corners, forming an inner square ACEG.

Since the octagon is regular and each side has a length of 6 cm, we use the formula for the side of the inner square in a regular octagon:

\[Side\ of\ the\ square = \frac{a}{\sqrt{2} - 1}\]

Here, \(a\) is the side length of the octagon. Substituting \(a=6\), we have:

\[Side\ of\ the\ square = \frac{6}{\sqrt{2} - 1}\]

To rationalize the denominator, multiply the numerator and the denominator by the conjugate:

\[\frac{6(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{6(\sqrt{2} + 1)}{2-1} = 6(\sqrt{2} + 1)\]

The area of the square ACEG is:

\[Area = (6(\sqrt{2} + 1))^2\]

Expand the expression:

\[= 36(\sqrt{2} + 1)^2 = 36(2 + 2\sqrt{2} + 1)\]

\[= 36(3 + 2\sqrt{2})\]

\[= 108 + 72\sqrt{2}\]

Thus, the area of the square ACEG, expressed with the options given, is:

\[= 36(2 + \sqrt{2})\]

This matches the correct answer choice \({36(2 + \sqrt{2})}\).