## chancemorris123 one year ago Find the circumference of a circle whose area is /. A. B. C. D.

1. chancemorris123

2. jdoe0001

$$\bf \textit{area of a circle}=\pi r^2\qquad area=60\pi \qquad thus \\ \quad \\ 60\pi =\pi r^2\impliedby \textit{solve for "r"}$$ what does that give you for "r"? or radius

3. chancemorris123

60=rsqure

4. jdoe0001

hmm rsquare?

5. chancemorris123

idk what r is

6. jdoe0001

$$\bf \textit{area of a circle}=\pi r^2\qquad area=60\pi \qquad thus \\ \quad \\ 60\pi =\pi r^2\implies \cfrac{60\cancel{\pi }}{\cancel{\pi }}=r^2\implies \sqrt{60}=\sqrt{r^2}\implies \sqrt{60}=r$$

7. jdoe0001

that was close though

8. chancemorris123

idk

9. chancemorris123

7.777

10. jdoe0001

so now that we know what "r" is let us use it in the circumference formula then $$\bf circumference=2\pi r\qquad \sqrt{60}=r\qquad then \\ \quad \\ circumference=2\pi \left( \sqrt{60} \right)$$

11. jdoe0001

and then you'd want to simplify that 60, see if you can squeeze something out of the radical

12. chancemorris123

30

13. chancemorris123

30 pie

14. jdoe0001

one may note that $$\huge \pi \ne pie$$ but yours is tastier though

15. chancemorris123

so is it 30pie

16. chancemorris123

so is it 30 pie?

17. jdoe0001

$$\bf circumference=2\pi r\qquad \sqrt{60}=r\qquad then \\ \quad \\ circumference=2\pi \left( \sqrt{60} \right) \\ \quad \\ {\color{brown}{ 60\to 2\cdot 2\cdot 15\to 2^2\cdot 15 }}\qquad thus \\ \quad \\ 2\pi \left( \sqrt{60} \right)\implies 2\pi \left( \sqrt{{\color{brown}{ 2^2\cdot 15}}} \right)\implies 2\pi \sqrt{2^2}\sqrt{15}$$

18. chancemorris123

so its 2square root 5