Area of the Enclosed Octagon?

• Let the length of legs of the 8 right-angled isosceles triangles be $x$ .
• Then by Pythagorean theorem, the side of octagon be $x\sqrt{2}$ .
• From the side of square it equals: $x + x\sqrt{2} + x = 3$
• $\implies \quad x \approx 0.878$

$A = 9 - 4 \times \frac{1}{2} x \cdot x$ $A = 9 - 1.54 = \boxed{7.458}$

All the Red and Green triangles are congruent and Right -isosceles due to symmetry. The squares each has an area of $9$ , thus the side of the square is $3$ . The hypotenuse of each small triangle is $\sqrt{2}a$ .

We have now $2a + \sqrt{2}a= 3 \implies a = \frac{3}{2+\sqrt{2}}$ .

The area of two Red triangles = area of one small Red square of side $a$ = $a^2 =\big(\frac {3}{2+\sqrt{2}}\big)^2$

The area of the octagon is: Area of the Big square- area of the 4 Red triangles =

Area of the Big square of side $3$ - area of two Red squares of side $a$ =

$9 - 2\Big(\frac{3}{2+\sqrt{2}}\Big)^2 = \boxed{ 7.46}$ .

Consider my figure above. The side length of the square is $\sqrt{9}=3$ , so

$2k+m=3$ $\text{[equation 1]}$

Applying pythagorean theorem on the small right triangle on the corner of the square, we have the equation

$m=k\sqrt{2}$ $\text{[equation 2]}$

The area of the regular octagon is equal to the area of the square minus the area of the four small right triangles. The area of one triangle is

$A_T = \dfrac{1}{2}k^2$

So we need to find $k^2$ , substituting the value of $m$ in equation $1$ into equation $2$ we get

$k=\dfrac{3}{2+\sqrt{2}}$

Squaring $k$ , we get

$k^2 = \dfrac{9}{6+4\sqrt{2}}$

Now, substitute

$A_T = \dfrac{1}{2}k^2=\dfrac{1}{2}\times \dfrac{9}{6+4\sqrt{2}}=\dfrac{9}{12+8\sqrt{2}}$

So the area of the $4$ small right triangles is

$4\times A_T = \dfrac{4\times 9}{12+8\sqrt{2}}=\dfrac{36}{12+8\sqrt{2}}$

Now, the area of the regular octagon is

$A_{octagon}=9-\dfrac{36}{12+8\sqrt{2}}\approx \boxed{\text{7.46 square units}}$ $\boxed{\text{answer}}$