Octagon Areas... XD

The area of regular polygon having $n$ sides with side length $a$ is : $\huge A=\frac{1}{4} n a^2 \cot (\frac{π}{n})$ Plugging the values gives the result.

proving of getting the area of the Octagon.

By Letting S = Side,

Draw a square and name it as ABCD and Construct a Octagon, GHIJKLEF, inside the square.

Area of Octagon = Area of Square - Area of 4 equal isosceles triangle.

The sides of the triangle is s/(√2) or (s√2)/2

= (s + (s√2)/2 + (s√2)/2)^2 - [4( (s√2)/2)^2]/2

= (s+s√2)^2 - [2(2s^2)/4]

= s^2 + 2s^2 + (2s^2)√2 - s^2

= 2s^2 + (2s^2)√2

Substituting, s = 4,

2s^2 + (2s^2)√2

= 2(4)^2 + (2(4)^2)√2

= 2(16) + 2(16)√2

= 32+32√2

Final Answer: 32 + 32√2 sq. dm.