33 of 100: The First Rule of Octagon Club Is Don't Talk About Octagon Club

Triangle $ABO$ , where $O$ is the center of the octagon, has an area equal to $\frac18$ of the area of the octagon.

Triangle $ABC$ has double the height of triangle $ABO$ and the same base. It therefore has double the area, that is $\frac14$ of the area of the octagon.

Just an another way to see it! There are $\color{#CEBB00}8$ equal triangles (eight isosceles right triangles with equal hypotenuse, which is indeed the side of octagon) and $\color{cyan}4$ equal rectangles (with a length equal to side of octagon and a width equal to leg of isosceles right triangle) in the picture. Thus area $\color{#D61F06}1$ is equal to area $\color{#D61F06}2$ . Shaded region (red color in the original figure) is half of the area $\color{#D61F06}1$ , and, thus, its $\color{#D61F06}\frac{1}{4}$ of the total area.

The area of a regular polygon is 1/2ap and p=8s. Substitute and reduce to get A=4as.

The area of the red triangle is 1/2bh, b=s, and h=2a. Substitute and reduce to get A=as

Therefore the area of the entire octogon is 4 times the area of the red triangle. In other words the area of the triangle is 1/4 of the octogon.

The triangle connecting the base of the red triangle to the center of the octagon is 1/8 of the octagon. The red triangle has double its height, so it must have twice the area. It must cover 1/4 of the octagon.

Area of octagon = 1/2 (apothem) (perimeter)

Perimeter = 8s

So area = 1/2 (apothem) (8s) = 4as

Area of triangle = 1/2 bh

b = s

h = 2a

So, area = as

Therefore the area of the triangle is 1/4 of the area of the octagon.

When $a$ = the apothem, and $P$ = perimeter the Area of a polygon is

$A_{ polygon}=\frac{1}{2}aP$

The perimeter of a polygon when $n$ = number of sides and $s$ = side length is

$P_{ polygon}=ns$

Substitution yields

$A_{ polygon}=\frac{1}{2}ans$

The area formula for a triangle is

$A_{ triangle}=\frac{1}{2}bh$

Given that $b$ = a side (/(s/) and /(h/) = two apothems /(a/) the equation becomes

$A_{ triangle}=\frac{1}{2}(s)(2a)$

Rearranging using the commutative property and removing the parentheses yields

$A_{ triangle}=\frac{1}{2}2as$

So the fraction of the polygon that the triangle yields is

$\frac{A_{ triangle}}{A_{ polygon}}=\frac{\frac{1}{2}2as}{\frac{1}{2}ans}$

Reducing by canceling yields

$\frac{A_{ triangle}}{A_{ polygon}}=\frac{2}{n}$

Since we are dealing with an octagon $n$ = 8 and the fraction becomes

$\frac{A_{ triangle}}{A_{ octagon}}=\frac{2}{8}$

Reducing yields

$\frac{A_{ triangle}}{A_{ octagon}}=\frac{1}{4}$

There are 8 triangles in a octagon. The red triangle takes up two triangles when cut, therefor is a 1/4 of the octagon

Given a side = 1 and the standard formula for calculation of the area of an octagon, extend a line completing the base of the octagon, then vertically to compute the height of the red triangle by solving for the sides of the 45 degree right triangle with hypotenuse 1, then after computing the area of the red triangle, dividing by the area of the octagon to determine fraction, thus =((1+2*0.5^0.5)/2)/4.83=0.249

Given the area of regular octagon as 2(1+ $\sqrt{2}$ ) $a^2$ , and using the angle formula for regular polygons = $180 - (\frac {360}{n}$ ) , a regular octagon has $145 ^ \circ$ inner angle, so that the when broken into 3 parts from the side edges, the height of the octagon or the triangle turns out to be $a + \frac {a}{\sqrt{2}} + \frac {a}{\sqrt{2}}$ which is $(1 + \sqrt{2})a$ .... $$ Calculating the area of the triangle using 1/2 X Base X Altitude, Triangle area = $\frac{(1 + \sqrt{2})a^2}{2}$ .... Thus the given area is $\frac{1}{4}$ the original octagon area

LET THE LEGS OF THE ISOSCELES RIGHT TRIANGLE ON THE LOWER RIDHT CORNER OF THE FIGURE

BE Y, AND THE LENGTH OF EACH SIDE OF THE FIDURE BE X, THEN

X^2 = 2Y^2, X = SQRT(2)Y

THE AREA OF THE FIGURE IS

A = 2(2X + 2Y)Y/2 + X(X + 2Y) = 2XY + 2Y^2 + X2 + 2XY = 4SQRT(2)Y2 + 2Y^2 + 2Y^2 = 4( SQRT(2) + 1)Y2

THE AREA OF THE PINK TIANGLE IS:

B = X(X + 2Y)/2 = Y^2 + SQRT(2)Y^2 = (1 + SQRT(2))Y2

THEN

B/A = (1 + SQRT(2))/(4(SQRT(2) + 1) )= 1/4.

I am presenting a generalized solution which is not as elegant as the previous ones but could prove useful when you do not get any special insights to simplify the problem (Apologies if the presentation is a little clumsy in terms of mathematical notation)