+213 What would the area of a regular octagon be if the only measurement known is the perimeter which in this case is 32 units.
+37153 Octagon has 8 sides. 32/8 = 4 units side length
Area of octgon = 2(1 + sqrt2) s^2 (s = side length)
+213 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.
+37153 Go here: (this site even has your question as an example)
https://byjus.com/octagon-formula/
+129899 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
+4622 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 \(48+32\sqrt{2}-16=\boxed{32+32\sqrt{2}}.\)
