## ksaimouli Group Title circle one year ago one year ago

1. ksaimouli

|dw:1358806909531:dw|

2. ksaimouli

area of circle

3. ksaimouli

@zepdrix

4. Outkast3r09

what are you trying to find

5. ksaimouli

area of that

6. zepdrix

|dw:1358880151258:dw|Your picture is so sloppy :D Lemme see if this looks correct.

7. zepdrix

Theta is the angle. s is the letter we use for arc length.

8. zepdrix

Umm let's first find the radius. I think we're going to need that. Here is a helpful formula:$\large s=r \theta$We want to solve for r so let's divide both sides by $$\theta$$.$\large \frac{s}{\theta}=r$

9. zepdrix

Our theta is given in degrees, so we'll need to convert it to radians.$\large 30^o \left(\frac{\pi}{180^o}\right)=\frac{\pi}{6}$ So our radius is given by this,$\large r=\frac{6\pi}{\left(\dfrac{\pi}{6}\right)}$ Which tells us that we have a radius of 36.

10. zepdrix

So now we need the area of this sector. Hmm. So it's a percentage of the area of a circle. So it will end up being the same formula we use for area of a circle, just with a fraction in front of it to show that we only want a fraction of that area.$\large A=\left(C\right)\pi r^2$From here we just need to find out what C is. You actually probably have a formula on a sheet somewhere, so I likely don't need to go into this much detail :) lol my bad. Anyway, our fraction will be the number of degrees we care about DIVIDED BY the total degrees in a circle.$\large A=\left(\frac{30^o}{180^o}\right)\pi r^2$The degree bubbles will cancel out, giving us a unit-less quantity (which is what we want).$\large A=\left(\frac{1}{6}\right)\pi r^2$

11. zepdrix

And we determined earlier that our radius was $$\large r=36$$. Plugging this in gives us,$\large A=\left(\frac{1}{6}\right)\pi (36)^2 \qquad = \qquad \left(\frac{1}{\cancel{6}}\right)\pi (6)^{\cancel{4}3}$

12. zepdrix

$\large A=216\pi$