## anonymous one year ago HELP PLEASE!!!! Find the area of a sector with an arc length of 60 in. and a radius of 15 in. A. 117.75 in.² B. 282.60 in.² C. 450 in.² D. 6,750 in.²

1. jiteshmeghwal9

Area of a sector=$$\Large{\frac{\theta}{360} \times \pi r^2}$$

2. anonymous

thank you!

3. anonymous

4. anonymous

because it says its wrong

5. anonymous

is 60 the theta

6. jim_thompson5910

60 is not the theta 60 is the distance around the curve. It's the length of the arc this shaded distance |dw:1436594438367:dw|

7. jim_thompson5910

if the radius is 15, then what is the total circumference?

8. anonymous

30?

9. jim_thompson5910

use the formula C = 2*pi*r

10. jim_thompson5910

you're close

11. anonymous

i got 94.24

12. anonymous

when i plugged it in

13. jim_thompson5910

yeah $\Large 30\pi \approx 94.2477796$

14. jim_thompson5910

I'm going to leave it as 30pi to keep things as exact as possible

15. anonymous

okay so what i do with that

16. jim_thompson5910

30pi is the entire circumference the arc length is 60 we divide the two to get the fraction of the circle in which this arc is applied to $\Large \frac{60}{30\pi} = \frac{2}{\pi}$

17. jim_thompson5910

now what we do is multiply that by the area pi*r^2 = pi*15^2 = 225pi $\large 225\pi*\frac{2}{\pi} = 225\cancel{\pi}*\frac{2}{\cancel{\pi}} = 225*2 = 450$

18. anonymous

thank you so much!

19. jim_thompson5910

|dw:1436594886725:dw|

20. jim_thompson5910

no problem

21. anonymous

can you help me with another problem? @jim_thompson5910

22. jim_thompson5910

sure

23. jim_thompson5910

24. anonymous

Find the length of the radius of . A. 2004-05-03-07-00_files/i0360003.jpg B. 2004-05-03-07-00_files/i0360004.jpg C. 2004-05-03-07-00_files/i0360005.jpg D. 2004-05-03-07-00_files/i0360006.jpg

25. anonymous

wait sorry

26. anonymous

A. r=1 B. r= 3 C. r=12 D. r=15 i have to find the length of the radius

27. jim_thompson5910

does it give you the entire circumference?

28. jim_thompson5910

oh wait nvm, we don't need that info

29. anonymous

no it does not

30. jim_thompson5910

notice how the central angle is 4pi/9 what happens when you divide that central angle by 2pi? what do you get?

31. jiteshmeghwal9

$\frac{\theta}{360} \times 2 \pi r= s$$\frac{\frac{4}{9}\pi} {360} \times 2 \pi r=\frac{20}{3} \pi$solve for 'r'.

32. anonymous

i have to put theta over 180 pi and then pi*radius

33. jim_thompson5910

I guess a shortcut is to use the formula $\Large s = \theta*r$ this formula only works if theta is in radian mode. In this case, theta = 4pi/9 and s = 20pi/3 $\Large s = \theta*r$ $\Large \frac{20\pi}{3} = \frac{4\pi}{9}*r$ do you see how to solve for r from here?

34. anonymous

i know that the pi's cancel out right?

35. jim_thompson5910

yep

36. anonymous

simplify the fractions then?

37. jim_thompson5910

not quite

38. jim_thompson5910

we now have 20/3 = (4/9)*r how would you isolate r ?

39. anonymous

divide?

40. jim_thompson5910

divide both sides by what

41. anonymous

i'm not sure

42. jim_thompson5910

|dw:1436596805511:dw|

43. jim_thompson5910

or you can multiply both sides by the reciprocal of 4/9 the reciprocal of 4/9 is 9/4 |dw:1436596829027:dw|

44. anonymous

i got 15

45. jim_thompson5910

|dw:1436596853297:dw|

46. jim_thompson5910

yes, r = 15

47. anonymous

thank you!

48. jim_thompson5910

no problem