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# Area of Regular Octagon?

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What would the area of a regular octagon be if the only measurement known is the perimeter which in this case is 32 units.

Apr 22, 2019

#1
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Octagon has 8 sides.       32/8 = 4 units side length

Area of octgon = 2(1 + sqrt2) s^2        (s = side length)

Apr 22, 2019
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I actually need to use trig to find the area. Could you help me find the area using trig and special right triangles. I hope you understand what I am trying to do.

OwenT154  Apr 22, 2019
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Go here:   (this site even has your question as an example)

https://byjus.com/octagon-formula/

ElectricPavlov  Apr 22, 2019
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256 units

Apr 22, 2019
#4
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Explain.

OwenT154  Apr 22, 2019
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We can divide the octagon into  16 equal   right triangles

We will have one angle of 22.5°.....with an opposite side of  4/2  = 2

The other angle will =  67.5°

Using the Law of Sines....we can find the  side, s,  opposite this angle

s / sin 67.5  =  2/sin 22/5

s  =   2 sin 67.5 / sin 22.5

Note that  sin 22.5  = cos 67.5  ....so we have

s  =  2 sin 67.5 / cos 67.5

s  = 2 tan 67.5

s  = 2 tan (135 / 2)  =  2   [ csc 135  - cot 135 ]  =  2 [ √2 - -1 ]   = 2√2 + 2

So....the area of the octagon  =

16 * (1/2)  product of legs  =

8 (2)(2√2 + 2)  =     32√2  + 32    units^2   Apr 22, 2019
#7
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Non-Trig Solution:

Draw an octagon...like the diagram as shown below. Since the perimeter of the octagon is 32 cm, each of its side length are  32 / 8 = 4 centimeters.

Now, draw a square around it...forming four 45-45-90 degree triangles. The side length opposite to the 45 degree angles is $$x\sqrt{2}=4, x=\frac{4}{\sqrt{2}}=2\sqrt{2}$$.

Find the area of each triangle, and multiply that by 4, yielding...$$4*4=16.$$

The area of the square on the other hand is $$(4\sqrt{2}+4)^2=48+32\sqrt{2}.$$

Finally, (the area of the square) - (area of the four triangles = (area of the octagon).

Thus, we have $$48+32\sqrt{2}-16=\boxed{32+32\sqrt{2}}.$$ Apr 23, 2019