Due to symmetry, the green and pink isosceles right triangles are all identical. Let each of the two equal side lengths be $a$ . Then the hypotenuse of the isosceles right triangle is $\sqrt 2a$ . So we have $2a+\sqrt 2a= 3$ $\implies a = \dfrac 3{2+\sqrt 2}$ .

We note that the area of the yellow region is the area of the square minus the area of four pink triangles. Since two pink triangles form a square of side length $a$ . Then, $A_{\color{#CEBB00}\text{yellow}} = 9 - 2a^2 = 9 - 2\left(\dfrac 3{2+\sqrt 2}\right)^2 \approx \boxed{7.46}$ .

A square with an area of 99 has a side of 33 and an apothem of a = \frac{3}{2}a= 2 3 ​
. The yellow part is a regular octagon with the same apothem as the square, with an area of A = na^2 \tan \frac{180°}{n} = 8(\frac{3}{2})^2 \tan \frac{180°}{8} \approx \boxed{7.46}A=na 2 tan n 180° ​
=8( 2 3 ​
) 2 tan 8 180° ​
≈ 7.46

A square with an area of $9$ has a side of $3$ and an apothem of $a = \frac{3}{2}$ . The yellow part is a regular octagon with the same apothem as the square, with an area of $A = na^2 \tan \frac{180°}{n} = 8(\frac{3}{2})^2 \tan \frac{180°}{8} \approx \boxed{7.46}$ .

A square with an area of 9 9 has a side of 3 3 and an apothem of

a

3 2 a= 2 3 ​
. The yellow part is a regular octagon with the same apothem as the square, with an area of

A

n a 2 tan ⁡ 180 °

n

8 ( 3 2 ) 2 tan ⁡ 180 ° 8 ≈ 7.46 A=na 2 tan n 180° ​
=8( 2 3 ​
) 2 tan 8 180° ​
≈ 7.46 ​
.