Asked by ashley
find the ara of a regular octagon inscribed in a circle with a radius of 1 cm.
Answers
Answered by
drwls
The octagon area can be broken into eight congruent isosceles triangles or sixteen congruent right triangles. The hypotenuse of each of the right trangles is r = 1 cm and the smallest angle is 360/16 = 22.5 degrees.
The area of each of the sixteen triangles is
A = (1/2)*r*cos 22.5*r sin 22.5
= (r^2/4)sin 45
Multiply this by 16 for the full area:
4 r^2*(sqrt 2)/2 = 2*sqrt 2* r^2 = 2.828 r^2
Note that this is less than pi*r^2, the circumscribed circle area, as it must be
The area of each of the sixteen triangles is
A = (1/2)*r*cos 22.5*r sin 22.5
= (r^2/4)sin 45
Multiply this by 16 for the full area:
4 r^2*(sqrt 2)/2 = 2*sqrt 2* r^2 = 2.828 r^2
Note that this is less than pi*r^2, the circumscribed circle area, as it must be
Answered by
Reiny
If you know two sides of a triangle and the angle contained between those sides, then the area is
(1/2)(side1)(side2)sin(angle)
This is the case for one of the eight isosceles triangles
Each of those is (1/2)(1)(1)sin45º or .353553 or √2/4
so the octogon is 8*√2/4 or 2.828
(1/2)(side1)(side2)sin(angle)
This is the case for one of the eight isosceles triangles
Each of those is (1/2)(1)(1)sin45º or .353553 or √2/4
so the octogon is 8*√2/4 or 2.828
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