What is the area of an octagon with a side length of 6 cm?

An octagon is a square with the four corners removed.

These four corners are right angles triangles, with a hypotenuse of of 6 cm. So we need to find the area of these four triangles and remove(subtract) from a larger square.

Using ~Pythagoras.

6^(2) = a^(2) + a^(2)

Hence

36 = 2a^(2)

18 = a^(2)

a = 3sqrt(2) The side length of the triangles.

So the area of these triangles is 4 x 0.5 x 3sqrt(2) x 3sqrt(2( = 36 cm^(2)

Now the side length of a large square enclosing the octagon. is

6 + 3sqrt(2) + 3sqrt(2( = 6 + 6sqrt(2)

The overall area of the 'large square; is ( 6 + 6sqrt(2))^(2) = )Use FOIL).

(6 + 6sqrt(2))(6 + 6sqrt(2)) =

36 + 36sqrt(2) + 36 sqrt(2)+ 72) = 108 + 72sqrt(2) cm^(2)

So subtract from the area , the four corner area which is from above 36cm^(2)

Hence

108 + 72sqrt(2) - 36 = (72 + 72sqrt(2() cm^(2) or

72(1 + sqrt(2)) cm^(2)

or

173.8233765 ... cm^(2).