SOLUTION: Find the side of a regular octagon inscribed in a circle of radius 19 cm.

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Question 1011941: Find the side of a regular octagon inscribed in a circle of radius 19 cm.
Found 2 solutions by It is costly, KMST:
Answer by It is costly(175) About Me  (Show Source):
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Think TRIANGLES. Triangles are the key to solving many problems.
They are at the heart of geometry, trigonometry and engineering.
Or think TRIGONOMETRY, if you wish.
Use LAW OF COSINES, if that is what you were taught recently.

Here is the circle, with its center, the inscribed regular octagon,
and some line segments that are at the same time
diameters of the circle and diagonals of the octagon.



The diagonals connecting opposite octagon vertices go through the center of the circle,
and split the octagon into 8 isosceles triangles.
Each of those triangles has two sides measuring 19cm
(the radius of the circle), and an angle measuring
360%5Eo%2F8=45%5Eo (or 2pi%2F8=pi%2F4 if you prefer radians)
at the center of the circle.

USING TRIANGLES:
Let's look at one of those triangles, and use the Pythagorean theorem
x%5E2%2Bx%5E2=19%5E2--->2x%5E2=19%5E2--->x%5E2=19%5E2%2F2--->x=19%2Fsqrt%282%29 for the larger right triangle.
For the smaller right triangle, x%5E2%2B%2819-x%29%5E2=side%5E2 .
We can substitute 19%2Fsqrt%282%29 for x (and 19%5E2%2F2 for x%5E2 if you wish)
sooner or later, as you wish.
x%5E2%2B%2819-x%29%5E2=side%5E2<-->x%5E2%2B19%5E2-2%2A19x%2Bx%5E2=side%5E2<-->2x%5E2%2B19%5E2-2%2A19x=side%5E2<-->2%28x%5E2-19x%29%2B19%5E2=side%5E2
2%2819%5E2%2F2-19%2A19%2Fsqrt%282%29%29%2B19%5E2=side%5E2
19%5E2-19%5E2sqrt%282%29%2B19%5E2=side%5E2
2%2A19%5E2-19%5E2sqrt%282%29=side%5E2
19%5E2%282-sqrt%282%29%29=side%5E2--->side=sqrt%2819%5E2%282-sqrt%282%29%29%29--->highlight%28side=19sqrt%282-sqrt%282%29%29%29 or for an approximate measure highlight%28side=about14.54cm%29(rounded)

USING TRIGONOMETRY:
We can put one of those triangles on a set of coordinates to find the distance between the vertices of the octagon.
Here are two similar triangles:
B%28cos%2845%5Eo%29%2Csin%2845%5Eo%29%29=B%28sqrt%282%29%2F2%2Csqrt%282%29%2F2%29 , and A%281%2C0%29 ,
so

CD=highlight%2819sqrt%282-sqrt%282%29%29%29 is the length of the side of the octagon, in cm.

USING LAW OF COSINES:
Law of cosine says that a%5E2=b%5E2%2Bc%5E2-2bc%2Acos%28A%29 , so with length measured in cm,
a%5E2=19%5E2%2B19%5E2-2%2A19%2A19%2Acos%2845%5Eo%29
a%5E2=2%2A19%5E2-2%2A19%5E2%2A%28sqrt%282%29%2F2%29
a%5E2=2%2A19%5E2-19%5E2%2Asqrt%282%29
a%5E2=19%5E2%282-sqrt%282%29%29
a=sqrt%2819%5E2%282-sqrt%282%29%29%29 ,
and a=highlight%2819sqrt%282-sqrt%282%29%29%29 is the length of the side of the octagon, in cm.