Lesson The area of a regular octagon

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The area of a regular octagon


Problem 1

We cut a regular octagon  ABCDEFGH  out of a piece of cardboard.
If  AB = 1 unit,  what is the area of the octagon?

Solution 

Let R be the radius of the circumscribed circle around our octagon.


Let's find its radius via the side length s.


The octagon consists of 8 congruent isosceles triangles, having 
the common vertex in the center.


Each such a triangle is an isosceles triangle with the lateral sides of the length R
and the angle at the vertex of 45°.  Write the cosine rule equation for such a triangle 

    s%5E2 = R%5E2 + R%5E2 - 2%2AR%2AR%2Acos%2845%5Eo%29,

    s%5E2 = 2R%5E2%2A%281-cos%2845%5Eo%29%29,

    R%5E2 = s%5E2%2F%282%2A%281-cos%2845%5Eo%29%29%29.    (1)


In our case with s = 1, the last formula takes the form

    R%5E2 = 1%2F%282%2A%281-cos%2845%5Eo%29%29%29 = 1%2F%282%2A%281-sqrt%282%29%2F2%29%29%29 = 2%2F%282%2A%282-sqrt%282%29%29%29 = 1%2F%282-sqrt%282%29%29 = %282%2Bsqrt%282%29%29%2F%284-2%29 = %282%2Bsqrt%282%29%29%2F2.    (2)


Now the area of one such a triangle is

    area%5Btriangle%5D = %281%2F2%29%2AR%2AR%2Asin%2845%5Eo%29 = %281%2F2%29%2AR%5E2%2Asin%2845%5Eo%29 = %281%2F2%29%2A%28%282%2Bsqrt%282%29%29%2F2%29%2A%28sqrt%282%29%2F2%29 = %282%2Asqrt%282%29%2B2%29%2F8 = %28sqrt%282%29%2B1%29%2F4.    (3)


For the area of the entire octagon, we should take the quantity (3)  8 (eight) times to get

    area%5Boctagon%5D = 2+%2B+2%2Asqrt%282%29  square units.    ANSWER


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