Given the side of the square is 2 meters. Because the corners are cut to form an octagon with equal sides, the octagon formed is a regular octagon.
The cut out corners form a 45-45-90 triangle and this will help us find the side of the octagon.
In the triangle QBR,\( \angleQRB\) =\( \angleBRQ\) = exterior angle of the octagon.
Because the figure is a regular octagon, each exterior angle =\(\frac{ 360 }{8}\) = 45
From the figure we know that SC + SR + RB = 2
Let each side of the octagon be x
ST = RQ = x
Consequently,
...
Attachments![Capture.JPG]( "Capture.JPG")
Capture.JPG [ 24.11 KiB | Viewed 13 times ]
The cut out corners form a 45-45-90 triangle and this will help us find the side of the octagon.
In the triangle QBR,\( \angleQRB\) =\( \angleBRQ\) = exterior angle of the octagon.
Because the figure is a regular octagon, each exterior angle =\(\frac{ 360 }{8}\) = 45
From the figure we know that SC + SR + RB = 2
Let each side of the octagon be x
ST = RQ = x
Consequently,
...
Attachments
Capture.JPG [ 24.11 KiB | Viewed 13 times ]
