$\frac{16}{81}$ is shaded.

Let $x$ and $y$ be the distances shown. Then the shaded area is $8y + x$. But there are a number of similar triangles and from one pair $${x\over8} = {y\over1}$$ i.e. $$x = 8y$$ So, $$\frac{\mbox{shaded area}}{\mbox{total area}} = \frac{8y + x}{9(x + y)} = \frac{8y + 8y}{9 \times 9y} = \frac{16}{81}$$

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.