calculate (to 1 decimal accuracy) the perimeter of a regular octagon inscribed in a circle of radius 10cm
2007-09-08 10:24:36 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Each angle at centre = 45°
There are 8 triangles.
Each triangle has legs of 10 cm containing an angle of 45°
Remaining side = x say
x ² = 10 ² + 10 ² - (2) (10) (10) cos 45°
x ² = 200 - 200 / √2
x ² = 58.6
x = 7.65 cm
Perimeter = 8 x 7.65 cm
Perimeter = 61.2 cm
2007-09-12 07:23:30 · answer #1 · answered by Como 7 · 0⤊ 0⤋
octagon has eight sides so divide the circle up by 8 parts
360/8 = 45 the interior angle at the origin.
the other two angle will be the equal so 180-45 = 135
135/2 = 67.5
let L be one side of the octagon
sin(45)/L = sin(67.5)/10
L = 7.653668647
L * 8 = 61.22934918 or 61.2 cm
2007-09-08 17:34:41 · answer #2 · answered by z32486 3 · 0⤊ 0⤋
the perimeter
= 8[(2)(10)sin22.5], 8 times the side measure
= 61.23 cm
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Ideas: Cut the octagon into 16 congruent right triangles. 1/16 of the whole circle has a central angle of 360/16 = 22.5 degrees.
2007-09-08 17:31:59 · answer #3 · answered by sahsjing 7 · 0⤊ 0⤋
A= 8 · A_triangle = 8 · R^2 · sin(2·pi/8) /2 = 400· sin(pi/4) =
200·â2=282.842712.... ~ 282.4 cm^2
P=8·L= 8·R·sin45º / sin(67,5º)~ 61,2 cm
Saludos.
2007-09-08 17:32:54 · answer #4 · answered by lou h 7 · 0⤊ 0⤋
a=lenght of a octagon side
r=radius
360digree/8=pi/4
a^2=r^2+r^2-2rcospi/4
a^2=100+100-200(sqrt2/2)
a^2=200-100(sqrt2)
a=10sqrt[2-(sqrt2)]
perimeter of a regular octagon=
8a=80*sqr[2-(sqrt2)]=
80(2-1.4)=80sqrt(0.6)
8a=61.96
2007-09-08 18:02:47 · answer #5 · answered by koh_arian 2 · 0⤊ 0⤋
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2007-09-08 17:35:06 · answer #6 · answered by Anonymous · 0⤊ 0⤋