Question

Given a square of length 2m. its corners are cut such that to represent a regular octagon. find the length of side of octagon

Answer 1

Given: A square of side 2 m. Its corners are cut such that to represent a regular octagon.

Solution:

Refer to the attached image.

Let the side of regular octagon be 'x' meters.

and let 'y' be the side of the corner of the square which are cutted.

Since, the side of the square = y+x+y

And side of square = 2 m.

Therefore, y+x+y = 2

2y+x = 2 (Equation 1)

Now, consider triangle ABC,

By Pythagoras theorem,

$(AC)^2 = (AB)^2+(BC)^2$

$(x)^2 = (y)^2+(y)^2$

$(x)^2 = 2(y)^2$

$y^2 = \frac{x^2}{2}$

$y = \frac{ x} {\sqrt 2}$

Substituting the value of 'y' in equation 1.

$2y+x =2$

$(2 \times \frac{ x}{\sqrt 2})+x = 2$

$(\sqrt 2 \times x)+x = 2$

$x(\sqrt 2 +1) = 2$

$x=\frac{2}{(\sqrt 2 +1)}$

$x=\frac{2}{(\sqrt 2 +1)} \times \frac{\sqrt2 -1}{\sqrt2 -1}$

$x={2(\sqrt 2 -1)}$

Therefore, the side of regular octagon is $x={2(\sqrt 2 -1)}$ meters.