how to calculate the area of an octagon?

Answer 1

Divide octagon into
upper Trapezium
Middle rectangle
Lower trapezium
You draw figure and you will get idea.
Height of trapezium = s/rt2 ---- where s is side of octagon
parallel sides of trapezium = s and other side = s + 2 × s/rt2 = s + rt2 × s
Length of middle rectangle = s + 2 × s/rt2 -------(rt2 is square root 2) = s + s/rt2
Breadth of rectangle = s
Area of rectangle = s [s + s/rt2] = s^2[1 + 1.rt2]
Total area = area of 2 trapeziums + area of rectangle.

Answer 2

if it's a regular octagon (ie all sides are the same length and all the internal angles are the same):

Call the side length L

Then join each corner of the octagon to the central point, creating eight identical triangles. Find the area of a single triangle and multiply by 8.

Wikipedia gives the formula for a regular octagon's area as

(2 x cot (pi/8)) x (L^2)

Answer 3

It depends if the octagon is regular or irregular.
If it is irregular, all you can do is divide it into triangles and find the area of each triangle using Hero's formula.

If the octagon is regular, then the area = .5Pa, where P= perimeter of octagon = 8 times the length of a side, and a is the apothem of the octagon which is the length of a line drawn from the center of the octagon perpendicular to one of its equal sides. In a regular octagon, the apothem is equal to .5s(1+sqrt(2)) where s is the length of a side.

Answer 4

Devide the octagon itno eight isosceles triangles. The area of a triangle multiplied by eight will give you the area of the octagon.

Answer 5

If it's a regular octagon, it's

2*(x^2)*[1+sqrt(2)]

Where x = length of a side

Answer 6

area of octagon ABCDEFGH of side length a is given by A = a^2 + 4a^2/root(2) + 2(a/root2)^2 = 2a^2(1 + root2) = 4.8286a^2

Answer 7

http://mathcentral.uregina.ca/QQ/database/QQ.09.01/laurie2.html

Check that out.

Basically you find the area the squares made by the 4 triangles, and subtract that from the big square.

Answer 8

break it up into smaller shapes

rectangles and triangles

Answer 9

http://mathcentral.uregina.ca/QQ/database/QQ.09.04/jeremy2.html