emily made a sketch of an octagonal windowpane on a 27 cm square piece of paper. First she cut four identical isosceles triangles from the corners of the paper. Then she cut a square from the centre of the octagon. Each leg of a cut off triangle is 1 over 3 the length of the paper. The side length of the cut out square is also 1 over 3 the length of the paper. What is the area of the sketch after she removed the triangles and the square?
Let the side length of the paper be $s$. The area of the original octagonal windowpane is given by $s^2$. Each leg of a cut-off triangle is 1/3 the length of the paper, so each leg has length $s/3$. The triangle can be divided into two 45-45-90 triangles with side lengths $s/3, s/3$, and $s/3\sqrt{2}$. So, the area of each triangle is $1/2\cdot (s/3)^2\cdot \sqrt{2}=\frac{2}{9}s^2\cdot \frac{1}{\sqrt{2}}=\frac{2s^2}{9\sqrt{2}}$. Since each corner of the paper has four copies of this triangle, the total area of the triangle sections is $\frac{8s^2}{9\sqrt{2}}$.
The side length of the cut-out square is also $s/3$, so the area of the square is $(s/3)^2=\frac{s^2}{9}$.
Thus, the total area of the sketch after removing the triangles and the square is
\[s^2-\frac{8s^2}{9\sqrt{2}}-\frac{s^2}{9}=\left(1-\frac{8}{9\sqrt{2}}-\frac{1}{9}\right)s^2=\left(\frac{2\sqrt{2}-1}{9}\right)s^2.\]
Now, we can substitute $s=27$ since the side length of the paper is given as 27 cm. Thus, the area of the sketch is
\[\left(\frac{2\sqrt{2}-1}{9}\right)(27^2)=\left(\frac{2\sqrt{2}-1}{9}\right)(729)=\boxed{170\sqrt{2}-81}.\]