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19 Nov 2021, 03:15
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Difficulty:
85%
(hard)
Question Stats:
36% (02:13) correct
64%
(02:18)
wrong
based on 25
sessions
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A square, whose side is 2 meters, has its corners cut away so as to form an octagon with all sides equal. Then the length of each side of the octagon in meters is.
A. \(\frac{\sqrt{2}}{\sqrt{2}+1}\)
B. \(\frac{2}{\sqrt{2}+1}\)
C. \(\frac{\sqrt{2}}{\sqrt{2}-1}\)
D. \(\frac{2}{\sqrt{2}-1}\)
E. None of these
A. \(\frac{\sqrt{2}}{\sqrt{2}+1}\)
B. \(\frac{2}{\sqrt{2}+1}\)
C. \(\frac{\sqrt{2}}{\sqrt{2}-1}\)
D. \(\frac{2}{\sqrt{2}-1}\)
E. None of these
18 Dec 2021, 14:02
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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, \(\angle QRB\) = \(\angle BRQ\) = 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, SC = RB = \(\frac{x }{ \sqrt[]2}\) {due to 45-45-90 triangle}
Hence,
x + 2 (\(\frac{x }{ \sqrt[]2}\)) = 2
x = \(\frac{ 2 }{ (\sqrt[]2 + 1)}\)
IMO - B
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, \(\angle QRB\) = \(\angle BRQ\) = 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, SC = RB = \(\frac{x }{ \sqrt[]2}\) {due to 45-45-90 triangle}
Hence,
x + 2 (\(\frac{x }{ \sqrt[]2}\)) = 2
x = \(\frac{ 2 }{ (\sqrt[]2 + 1)}\)
IMO - B
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