The shaded area can be split into a square and four identical triangles. So, to find the shaded area, we simply need to find the area of these smaller, simpler shapes.

Let’s start with the triangles.

We know they must be right-angled triangles because an exterior angle of a regular octagon is 45°, so the angles in each triangle must be 45°, 45° and 90°.

Each triangle has a base and height of length \(a\).

\(\begin{aligned}Area &= \frac{1}{2} \times a \times a\\&=\frac{1}{2}a^2\end{aligned}\)

There are four of these triangles, so their total area is \(4\times\frac{1}{2}a^2=2a^2\).

Now, we look at the square. To find its area, we need to find the length of a side.

As we can see, the length of the sections either side of the triangle are both \(a\) and the central section is the hypotenuse (\(h\)) of our triangle.

Using Pythagoras’ theorem, we get

\(\begin{aligned}h^2 &= a^2 + a^2\\&=2a^2\\h&= a\sqrt{2}\end{aligned}\)

So, each side of the square is \((2a + a\sqrt{2})\) units long.

Multiplying this by itself will give us the area of the square.

\(\begin{aligned}(2a + a\sqrt{2})(2a + a\sqrt{2})&= 4a^2 + 2a^2\sqrt{2} +2a^2\sqrt{2} + 2a^2\\&= 6a^2 + 4a^2\sqrt{2}\end{aligned}\)

Now, we just add on the area of the 4 triangles and then factorise the result.

\(\begin{aligned}6a^2 + 4a^2\sqrt{2} + 2a^2 &= 8a^2 + 4a^2\sqrt{2} \\&= 4(2 + \sqrt{2})a^2 \end{aligned}\)