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$\frac{16}{81}$ 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, $$\frac{\mbox{shaded area}}{\mbox{total area}} = \frac{8y + x}{9(x + y)} = \frac{8y + 8y}{9 \times 9y} = \frac{16}{81}$$
Diagonal Touch
Age 14 to 16
ShortChallenge Level 
$\frac{16}{81}$ 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.
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