## mckenzieandjesus one year ago A point in the figure is selected at random. Find the probability that the point will be in the shaded region.

1. mckenzieandjesus

2. mckenzieandjesus

3. mckenzieandjesus

@jim_thompson5910

4. mckenzieandjesus

How do I find the probability?

5. SolomonZelman

basically, you need the ratio between the square, and a circle that is subscribed in it.

6. SolomonZelman

lets define the side of a square using s.

7. SolomonZelman

The side of a the square is equivalent to the diameter of the circle. The area of the square is just A=s². The area of the circle however is a little bit harder.

8. mckenzieandjesus

ok

9. SolomonZelman

The radius is s/2 (considering the fact that diameter =s) Now, the formula for the area of the circle is A=π • r² A=π • (s/2)² = πs² /4  or alternatively = (π/4) • s²

10. mckenzieandjesus

are of a circle is A=pi r^2

11. mckenzieandjesus

ok

12. SolomonZelman

the probability that the point in the whole square that you randomly choose is $$\displaystyle \LARGE {\rm P}=\frac{\rm A_{circle}}{\rm A_{square}}$$

13. SolomonZelman

good luck....

14. SolomonZelman

if you have questions, you are always welcome to ask

15. mckenzieandjesus

There is no lengths or anything so im kinda confused

16. SolomonZelman

that sentence is supposed to say the probability that the point in the whole square that you randomly will lay on the shaded part  choose is (the part in gray I left out. apologize)

17. SolomonZelman

there is length. length for what don't you see?

18. SolomonZelman

radius = s/2 side of the square = diameter of the circle = s

19. mckenzieandjesus

but i dont know the side of the square or diameter of the circle to figure it out. Sorry im lost

20. SolomonZelman

(this probability that they ask you for, and that we will find, is true for any side length of the square, this is why the side of the square isn't given to us.)

21. mckenzieandjesus

So i make up a side length and a diameter of the circle?

22. SolomonZelman

Yes, that is what I showed

23. SolomonZelman

I denoted the length with letter s.

24. SolomonZelman

And now the probability of a randomly selected point, to lay on a circle is A(circle) ÷ A(square)

25. SolomonZelman

our area of the circle is πs²/4 area of the square is s²

26. SolomonZelman

still confused?

27. mckenzieandjesus

so 5^2=25 and 3.14*5^2 = 78 1/2

28. SolomonZelman

no, you don't make up a length. Not that it would matter, but you are not asked or meant to do this. I am sorry.

29. mckenzieandjesus

i thought u said i do

30. SolomonZelman

We are just using any side-length "s" (regardless of what "s" is - of course, as long as s>0)

31. SolomonZelman

I said that we use any sidelength 's'. we show or work, for why is it so, that this probability is blank in this case? if they wanted to, they would have given you the side. but they did not- for a reason.

32. SolomonZelman

But, do you understand my previous replies, how I found the area of a square and a circle in terms of s, or should I go over that again?

33. mckenzieandjesus

i know the formulas

34. SolomonZelman

ok, now divide the area of circle, by area of the square.

35. SolomonZelman

$$\displaystyle \LARGE {\rm P}=\frac{\rm A_{circle}}{\rm A_{square}}=\frac{\frac{\pi}{4}{\rm s}^2}{{\rm s}^2}=~...?$$

36. SolomonZelman

(the s² cancels on top and bottom, and you remain with ?)

37. mckenzieandjesus

pi/4?

38. mckenzieandjesus

3.14/4?

39. SolomonZelman

yes π/4 :)

40. SolomonZelman

(and again, that is regardless of the value of s, for all real values of s that are greater than 0)

41. mckenzieandjesus

okay

42. SolomonZelman

3.14/4 (if you use that approximation, then you might want to re-write the fraction, reduce it... or you know...) i would perhaps go: 3.14/4= 314/500= 157/250

43. mckenzieandjesus

i did 3.14/4 = 157/200

44. mckenzieandjesus

ok so 157/250

45. jim_thompson5910

|dw:1435286362761:dw|

46. jim_thompson5910

An alternative is to think of the square with side length 2r so the radius of each circle is r |dw:1435286688332:dw|

47. mckenzieandjesus

ok

48. jim_thompson5910

area of square = (2r)*(2r) = 4r^2 area of circle = pi*r^2 divide the area of the circle by the area of the square = pi*r^2/4r^2 = pi/4 either way you get the same answer

49. mckenzieandjesus

what do i put for r

50. jim_thompson5910

r can be any number you want the 'r's will cancel, so it really doesn't matter

51. jim_thompson5910

same with the 's's following SolomonZelman's method

52. mckenzieandjesus

so i could do 3.14*5^2/4*5^2?

53. SolomonZelman

yes, I guess to find the answer you can of course give the side of the square any length like 3.

54. SolomonZelman

I was just thinking that it asked for a complete prove that shows the probability of π/4 for all positive s.

55. mckenzieandjesus

or 3.14*5^2/4*2^2= 78 1/2 now what?

56. jim_thompson5910

once you pick a number for r, you have to stick with it

57. jim_thompson5910

and I should have used parenthesis (pi*r^2)/(4r^2)

58. mckenzieandjesus

so (3.14*5^2)/(4*5^2)?

59. mckenzieandjesus

= 157/200

60. mckenzieandjesus

now what?

61. jim_thompson5910

what format do they want the answer? as a fraction? or decimal?

62. mckenzieandjesus

63. jim_thompson5910

oh ok

64. jim_thompson5910

convert the fraction you have to a decimal then convert that decimal to a percent

65. mckenzieandjesus

0.785

66. jim_thompson5910

convert 0.785 to a percent

67. mckenzieandjesus

78.5%

68. mckenzieandjesus

69. jim_thompson5910

correct

70. mckenzieandjesus

well that was confusing lol thanku

71. mckenzieandjesus

whats a minor arc in a circle?

72. jim_thompson5910

http://mathworld.wolfram.com/MinorArc.html any arc that is smaller than 180 degrees

73. jim_thompson5910

an arc is just a piece of the whole circumference

74. mckenzieandjesus

ok so it be PS?

75. jim_thompson5910

that's one minor arc

76. mckenzieandjesus

MY CHOICES: PS, SO, SQ, PSR

77. mckenzieandjesus

the others didnt look right

78. jim_thompson5910

yeah PS is the only minor arc SO isn't even an arc (it's a radius)

79. mckenzieandjesus

thats what i thought