1. beens

not getting through with the link

2. generallyclay

the link works you just need to copy and paste the whole link, not just what is underlined

3. DavisAshley

can someone help me with 8-12?

4. generallyclay

could you be more specific with what you are finding difficult about, for instance # 8?

5. DavisAshley

i just dont know how to do them

6. DavisAshley

like if i can just get the equation i can find the answers.. i just need the equation

7. generallyclay

well in # 8 the tringle is an equilateral triangle, so you know all the angles already. Since you are given the radius of the circle, you can use the equation A=pi*r^2

8. generallyclay

to find the area of the circle and then subtract the area of the triangle from the area of the circle.

9. generallyclay

the sides of the triangle are all 6 btw

10. DavisAshley

so the area of the triangle is 15.59

11. DavisAshley

so the circle area would be about 34.65?

12. generallyclay

one sec

13. generallyclay

first cut the triangle in half to make two right angle triangles. then you can use pythagorean to find the height of the triangle, which i got to be sqrt(27). finding the area of the triangle with half base times height formula, 1/2*6*sqrt(27) = 9sqrt(3) =15.59. Is that what you did?

14. DavisAshley

yea

15. generallyclay

for the area of the circle did you use pi*r^2?

16. DavisAshley

yeah and i got 50.24

17. generallyclay

plugging in 16 for r^2, i got the area to be 50.27

18. generallyclay

so subtract the area of the triangle from the area of the circle to find the area of the shaded region

19. DavisAshley

34.68

20. generallyclay

super

21. DavisAshley

would it be squared?

22. generallyclay

no, why would you?

23. DavisAshley

idk.. someone helped me with number 7 and they squared it

24. generallyclay

might have been part of finding the area of a circle, you need to square the radius for that

25. generallyclay

we already did that, though, so it's fine as it is

26. DavisAshley

okay.. will you help me with the other ones

27. generallyclay

well, I have a bit of hw of my own that i must do before tomorrow, but take a look at number 12... I want to try doing that one with you.

28. generallyclay

if you like

29. DavisAshley

ok

30. generallyclay

This is a really cool and beautiful one out of the lot. do you see any way of starting to think about it?

31. DavisAshley

doing A=pi*r^2

32. generallyclay

that would give you the area of the circle, it's true. but look even more closely: there are only two lines in the triangle that you do not know the length of... right?

33. DavisAshley

yea

34. generallyclay

oops... I meant to say in the circle

35. generallyclay

and do you know the equation to find a sector a circle?

36. DavisAshley

arc divided by 360* pi*r^2

37. generallyclay

A=1/2*r^2*theta

38. generallyclay

actually since each of these sectors is 1/4 of the circle, the equation is A=1/4*pi*r^2

39. generallyclay

do you see where I'm going with this?

40. DavisAshley

so a=1/4*3.14*6^2

41. DavisAshley

=28.26

42. generallyclay

well the area of the triangle is 18 right?

43. DavisAshley

yea

44. generallyclay

good, and so given that you now have the area of a quarter of a circle, and the area of the triangle, you can subtract the area of the triangle away from the area of the quarter of the circle that you have.

45. generallyclay

sorry i skipped a step before. have you followed me up to this point or are you a bit lost?

46. generallyclay

So, here are the steps. 1) we found the area of a quarter of a circle. 2) we found the area of the triangle. 3) we subtract the area of the triangle from the area of the quarter of the circle in order to find the area of one of the shaded regions.

47. generallyclay

but since both of the triangles are the same, we also now know the area of the other shaded region. does that make sense?

48. DavisAshley

yea

49. generallyclay

Awesome. I'm glad I could help :) This is my first time on this site, but this is a pretty cool idea for a site.

50. DavisAshley

yeah

51. generallyclay

well I have to go do HW now, good luck!

52. DavisAshley