Need help with finding the area of the shaded region in a circle!

Question

rebeccaxhawaii · Answer

post

rebeccaxhawaii · Answer

AND WELCOME TO OS

anonymous · Answer

attachment

anonymous · Answer

@rebeccaxhawaii posted (:

zepdrix · Answer

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zepdrix · Answer

To find the area of this outer rim,
we have some big circle,
and we're subtracting some smaller circle from inside of it.

zepdrix · Answer

Let's call the area of the big shaded circle A,
and the area of the small circle (which we will cut out) little a,

So the area of this rim around the edge = A - a.

zepdrix · Answer

So if the radius of the big circle is R,
then $\rm A=\pi R^2$

and if the radius of the smaller circle is r,
then $\rm a=\pi r^2$

So the area of the "rim" as I'm calling it, is,$$\large\rm A-a=\pi R^2-\pi r^2$$And they gave us this area, 32pi.$$\large\rm 32\pi=\pi R^2-\pi r^2$$

zepdrix · Answer

If we somehow.... (put your thinking cap on) ...
plug in a value for R,

then we can move stuff around and solve for this unknown r, ya?

zepdrix · Answer

Following along? :)
Confusing?
What do you think?

anonymous · Answer

@zepdrix i understand, but we need to find "R" before we get "r", so do we get "R" from the diameter given?

zepdrix · Answer

Yessss, good good good :)

anonymous · Answer

@zepdrix okay so we'd have 32π=π9^2−πr^2 because 18 is diameter and a radius if half of a diameter, so it'd be 9^2 right?

zepdrix · Answer

Good good, R=9 so you plugged it in :)
Maybe start by dividing both sides of the equation by pi to give you a clear path to solving for r.

anonymous · Answer

@zepdrix now that's where i get stuck at...

zepdrix · Answer

$$\large\rm 32\pi=9^2\pi-r^2\pi$$Dividing by pi,$$\large\rm 32=9^2-r^2$$That simplifies things a bit, ya? :)

anonymous · Answer

oh because they all cancel out! so after that we're going to have 32=81-r^2

zepdrix · Answer

Mmmm k great.
Just a few more steps.

anonymous · Answer

now we need to get r alone right? so we have to get rid of the squared

zepdrix · Answer

Here's a little hint when solving equations algebraically:

Remember your PEMDAS thing?
It tells you how to prioritize operations.
You start with the most complex operations, exponents, brackets, etc...
and finish with the simplest, addition/subtraction.

When you're trying to solve for a variable,
try to think of this process in reverse.
You want to try and undo the simplest things first.

So maybe we can deal with the 81 before we worry about the square.

anonymous · Answer

32=81-r^2
32=-r^2+81 (to simplify both sides of the equation)
then add r^2 to both sides
32+r^2=81 (because the negative/positive r^2 cross each other out)
subtract 32 from both sides
r^2=49
take the the square root which would give me r=7

anonymous · Answer

is that right? @zepdrix