Re: Four concentric circles share the same center. The smallest circle has
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19 May 2021, 01:04
First, we figure out the area of the smallest circle. \(A_1 = \pi r^2 = \pi 1^2 = \pi\).
Now, we find the area of the second smallest circle \((n = 2)\). \(A_2 = A_1 + (2(2) - 1) \pi = \pi + 3 \pi = 4 \pi\). This means that the radius of the second smallest circle is 2 (since the area is \(\pi r^2\)).
The third smallest circle has area \(A_3 = A_2 + (2(3) - 1) \pi = 4 \pi + 5 \pi = 9 \pi\). This means that the radius of this circle is 3.
Finally, the fourth smallest circle (that is, the largest circle) has area \(A_4 = A_3 + (2(4) - 1) \pi = 9 \pi + 7 \pi = 16 \pi\). This means that the radius of this circle is 4.
The sum of all the areas is \(\pi + 4\pi + 9\pi + 16\pi = 30\pi\).
The sum of all the circumferences is \(2\pi\) times the sum of all the radii. The sum of all the radii is \(1 + 2 + 3 + 4 = 10\), so the circumferences sum up to \(20\pi\).
Thus, the sum of all the areas, divided by the sum of all the circumferences, is \(\frac{30\pi}{20\pi} = 1\frac{1}{2}\).
Answer: B