In the diagram above, all the points are on a line, and the number of

Answer = B. \(21\pi\)

Radii of all green shaded circles is even i.e 2, 4, 6

Radii of all white circles is odd i.e 1, 3, 5

Area of required shaded region \(= \pi(2^2 + 4^2 + 6^2) - \pi(1^2 + 3^2 + 5^2)\)

\(= \pi(4+16+36 - 1 - 9 - 25) = 21\pi\)

MAGOOSH OFFICIAL SOLUTION:

First, let’s look at the outer “lobe,” the one between 10 and 12. The circle through point 12 has a center a 6 and radius of 6, so its area is 36pi. The circle through point 10 has a center a 5 and radius of 5, so its area is 25pi. If we subtract the latter from the former, we an area of 11pi for this lobe.

Now, let’s look the middle lobe, the one between 6 and 8. The circle through point 6 has a center a 3 and radius of 3, so its area is 9pi. The circle through point 8 has a center a 4 and radius of 4, so its area is 16pi. If we subtract the latter from the former, we an area of 7pi for this lobe.

Now, let’s look the smallest lobe, the one between 2 and 4. The circle through point 2 has a center a 1 and radius of 1, so its area is pi. The circle through point 4 has a center a 2 and radius of 2, so its area is 4pi. If we subtract the latter from the former, we an area of 3pi for this lobe.

Add the areas of the three separate lobes: 3pi +7pi + 11pi = 21pi.

Answer = (B)

We divide the it into 3 parts with 2 circles each then add the 3 parts to get the total area as follows:
[pi(2)^2-pi(1)^2] + [pi(4)^2-pi(3)^2]+[pi(6)^2-pi(5)^2]
=3pi + 7pi + 11pi
=21 pi

36pi-25pi+16pi-9pi+4pi-pi

we are subtracting (25+9+1) i.e. an odd number from even number. Answer will be odd and there is only only odd option in the answer choices.

B is the answer

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