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

Algebra ->  Polygons -> SOLUTION: We cut a regular octagon ABCDEFGH out of a piece of cardboard. If $AB = 1$, then what is the area of the octagon?       Log On


   

Question 1209347: We cut a regular octagon ABCDEFGH out of a piece of cardboard. If $AB = 1$, then what is the area of the octagon?
Found 3 solutions by math_tutor2020, greenestamps, ikleyn:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Draw a segment from the center to each vertex of the octagon.
It will produce 8 pizza slices so to speak.

Each pizza slice is an isosceles triangle with an apex angle of 360/8 = 45 degrees

Let's zoom in on one of the pizza slices.
I'll draw a line to split this slice into two mirror halves.
This splits the 45 degree angle into 45/2 = 22.5 degrees.

h = unknown height of this pizza slice
h = apothem of the octagon
The side length of the octagon has been split in half to 1/2 = 0.5

tan(angle) = opposite/adjacent
tan(22.5) = 0.5/h
h*tan(22.5) = 0.5
h = 0.5/tan(22.5)
h = 1.207106781187 approximately
Please make sure that your calculator is set to degrees mode.

The area of each of the original slices would be
area = 0.5*base*height
area = 0.5*1*1.207106781187
area = 0.603553390593
8 times of which leads to a total area of 8*0.603553390593 = 4.828427125 which is approximate.

--------------------------------------------------------------------------

Another approach would be to use this formula
Area = 0.25*s^2*n*cot(180/n)
where "cot" represents "cotangent". Recall that cot = 1/tan
s = side length
n = number of sides

If we use this formula then we'll have s = 1 and n = 8
Area = 0.25*s^2*n*cot(180/n)
Area = 0.25*1^2*8*cot(180/8)
Area = 2*cot(22.5)
Area = 2*1/tan(22.5)
Area = 4.828427125

--------------------------------------------------------------------------

Edit: after greenestamps made his post I realize that tan(22.5) can be determined exactly in terms of radicals.

Use the identity tan%28theta%2F2%29+=+%22%22%2B-sqrt%28+%281-cos%28theta%29%29%2F%281%2Bcos%28theta%29%29%29 to determine that tan%2845%2F2%29+=+sqrt%28+%281-cos%2845%29%29%2F%281%2Bcos%2845%29%29%29
Note that tan(22.5) is positive so we ignore the minus outside the root.
After simplification you should get tan%2822.5%29+=+-1%2Bsqrt%282%29 which leads to cot%2822.5%29+=+1%2Bsqrt%282%29

So, area = 2*cot(22.5) = 2*(1+sqrt(2)) = 1+2*sqrt(2)

--------------------------------------------------------------------------

Exact Answer: 2+2*sqrt(2) square units
Approximate Answer: 4.828427125 square units

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Here is another way to solve this problem, this time getting an exact answer.

Add 45-45-90 right triangles to alternating sides of the regular octagon to form a square.

The hypotenuses of those triangles are edges of the regular octagon, so they each have edge length 1; and so their legs all have length sqrt%282%29%2F2

The side length of the square is then 1%2B2%28sqrt%282%29%2F2%29=1%2Bsqrt%282%29.

Put the four added triangles together with their right angles at a common point to see that the combined area of the four triangles is the area of a square with side length 1.

Then the area of the regular octagon is the area of a square with side length 1%2Bsqrt%282%29, minus the are of a square with side length 1:

%281%2Bsqrt%282%29%29%5E2-1%5E2=%281%2B2sqrt%282%29%2B2%29-1=2%2B2sqrt%282%29

ANSWER: 2%2B2sqrt%282%29 <<== typo corrected thanks to note from tutor @ikleyn

(which to several decimal places is equal to the answer obtained by the other tutor, 4.828417...)

NOTE: This problem shows a formula that is familiar to many geometry students who participate in math contests: the area of a regular octagon with side length s is

A=s%5E2%282%2B2sqrt%282%29%29 <<== typo corrected thanks to note from tutor @ikleyn

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
We cut a regular octagon ABCDEFGH out of a piece of cardboard.
If AB = 1, then what is the area of the octagon?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


            Here is another way to solve the problem. 


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

Solved.