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The three angles of the triangle add to $180^{\circ}$, so the combined area of the three sectors of the circles that are inside the triangle add up to half a circle with area:
$\frac{1}{2} \times \pi \times 2^2 = \frac {4\pi}{2}= 2\pi$. 
So the grey area is $(80- 2\pi) cm^2$.

This problem is taken from the UKMT Mathematical Challenges.