So the whole circle has an area of 27^2π = 729π
which means that each part has an area of 81π cm^2
The radius of the inner circle must be 9 cm -----> (9^2π = 81π)
the area of the 2nd circle must be 2(81π) = 162π , inner circle + first ring
πr2^2 = 162π
r2 = √162 = 9√2
the area of the third circle must be 3(81π) = 243π , inner circle + 2 rings
πr3^2 = 243π
r3 = √243 = 9√3
...
the area of the 7th circle = 7(81π) = 567π
r7^2π = 567π
r7 = √567 = 9√7
ahhh, nice pattern
so the radius of the 9th circle = 9√9 = 27 , as given
What is the ratio of the dimensions of the inner circle to the dimensions of the whole target
---> 9 : 27 = 1:3
What is the ratio of the area of the inner circle to the area of the whole target?
----> 81π : 729π = 1 : 9
I expected this result!, It is one of the properties of areas of similar shapes
A circular target is divided into nine parts of equal area by eight concentric rings. If the radius of the target is 27 cm, find the radius of the inner circle.
What is the ratio of the area of the inner circle to the area of the whole target?
What is the ratio of the dimensions of the inner circle to the dimensions of the whole target?
2 answers
Wow, thank you! I now better understand this topic. I think I'm ready for my test. Thanks!