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The bull's-eye on a target has a diameter of 3 inches. The whole target has a diameter of 15 inches. What part of the whole target is the bull's-eye? Explain

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Hi Trillstone!

Hoping I'm looking at this correctly, but I'll do my best!

So we have a target with a diameter of 15 in., and a bull's-eye with a diameter of 3 in. We're basically looking for what 3 in. is to 15 in. in terms of size. So here we go:

First we're going to need to know the area of a circle, which is: Area = πr².
We're going to take the area of both the whole target, and then just the bullseye.

Area of Whole Target: π(15/2)² ⇒ π(7.5)² = 176.71 in³.
Area of Bullseye: π(3/2)² ⇒ π(1.5)² = 7.07 in³.

So far so good? (It's late around here so I'm sorry if I'm missing steps!)

Now from here, all we have to do is divide the area of the bullseye by the area of the target to find our answer.

7.07 / 176.71 = .04

In other terms, the bullseye takes up around 4% of the target.

-Hope this helps and please correct me if this turns out to be incorrect!


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