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Answer  


The large circles could have any radius and the answer will still be the same.

Choose a nice radius for the large circles, e.g. 1 unit

A:    =     $-$       

         Area          =   $\pi\times1^2$   $ - $   $2\times\left(\pi\times\left(\frac12\right)^2\right)$
                          =     $\pi$   $-$   $\frac12\pi$    =    $\frac12\pi = 1.571$

B:    =     $\times3$

Area of a triangle is $\frac12ab\sin{C} = \frac12\times1\times1\times\sin{120} = \frac{\sqrt3} 4 = 0.433$
Total area is $\frac{3\sqrt3}4 = 1.299$

C:      =      $-$     Area $=4 - \pi = 0.858$


D:      =       Area = 1


E:       =      $-$  $\times4$ 
                             =    $\pi$   $-$   $\frac12(1\times1 )\times4 =\pi-2 = 1.142$


This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.