SOLUTION: An archery target is constructed of five concentric circles such that the area of the inner circle is equal to the area of each of the four rings. If the radius of the outer circle

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Question 1148930: An archery target is constructed of five concentric circles such that the area of the inner circle is equal to the area of each of the four rings. If the radius of the outer circle is 9m, find the width of band a, in m.
Diagram: https://imgur.com/a/YRGUBMp
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The area of a circle is %28pi%29%28r%5E2%29%5D where r is the radius.

The area of the whole target is %28pi%29%289%5E2%29+=+81%28pi%29.

The area of the circle forming the outer edge of band a is 4/5 of the total area: %28324%2F5%29pi. The outer radius of band a is sqrt%28324%2F5%29.

The area of the circle forming the inner edge of band a is 3/5 of the total area: %28243%2F5%29pi. The inner radius of band a is sqrt%28243%2F5%29.

The width of band a is the difference between those two radii: sqrt%28324%2F5%29-sqrt%28243%2F5%29

Use a calculator if you need the answer in a different form.