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  3. 기하학 Geometry

Area of a ​​regular octagon in contact with a square

작성자 Uploader : rainman 작성일 Upload Date: 2019-03-04변경일 Update Date: 2019-03-05조회수 View : 270

The area of ​​an octagon in which a side having a length of a is in contact with a square, as shown in the digure, can be obtained as follows.

If the length of one side of the octagon is b and the length of one side of the triangle in which the square and the octagon do not overlap is c,

then a = b + 2c

The hypotenuse of the triangle has a length of 2^(1/2)*c = b, then inserting this into the above equation,

a = (2 + (2)^(1/2))*c

From this,

c = a / (2 + (2)^(1/2))

Then, the area to be obtained is:

Area = a^2 - 2*c^2

That is,

Area = a^2 - 2*c^2
     = a^2 - 2*a^2/(2+(2)^(1/2))^2
     = a^2*(1- 2/(2+(2)^(1/2))^2)
     = a^2*(1- 1/(3+2*(2)^(1/2)))

*** 참고문헌[References] ***
Area = a^(2)*(1- 1/(3+2*(2)^(1/2)))
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변수명 Variable 변수값 Value 변 수 설 명 Description of the variable


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