## Dallasb22 Group Title Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 3-inch radius A = sq. in. one year ago one year ago

1. Dallasb22 Group Title

Area= 1/2ap

2. Dallasb22 Group Title

I draw an auxiliary triangle off of the radius, the radius being the hypotenuse

3. tcarroll010 Group Title

By 3 inch radius, do you mean that it is inscribed in a circle of radius 3 inches?

4. Dallasb22 Group Title

|dw:1348254489061:dw|

5. Dallasb22 Group Title

The apothem is half of the radius

6. tcarroll010 Group Title

|dw:1348254528658:dw| So it looks like both are drawings are similar?

7. Dallasb22 Group Title

So its 1.5

8. Dallasb22 Group Title

I multiply 1.5 times √3 to get the long side

9. Dallasb22 Group Title

Than i multiply that answer "3√3" by 2 to get the side of the big triangle

10. Dallasb22 Group Title

I multiply that by 3 to get the perimeter. So i have the perimeter and the apothem.

11. tcarroll010 Group Title

Yes, and to make matters easier, the angle and segment bisectors will meet 2/3 of the way to the other side, making the height = 1.5 x apothem.

12. Dallasb22 Group Title

The answer i got the first time is: 6 3/4√3 but they told me its wrong.

13. tcarroll010 Group Title

height = 1.5 x apothem = 1.5 x (r/2) = 3r/4.

14. Dallasb22 Group Title

I don't need the height to find the area though.

15. Dallasb22 Group Title

All i need is the apothem and the perimeter

16. tcarroll010 Group Title

You're right, you don't, but you can take advantage of some easy trigonometry to get the answer easier.

17. Dallasb22 Group Title

Ok, well i'm only in Geometry.. So i'm not suppose to be doing trig

18. Dallasb22 Group Title

*Well i don't know any

19. tcarroll010 Group Title

Ok, then I'll stop with that approach.

20. Dallasb22 Group Title

Lol, if it's fairly easy, I don't mind learning it

21. Dallasb22 Group Title

It must be easier than the book's method because everyone I talk to on here starts using trig

22. tcarroll010 Group Title

No, that's ok. I can do it the other way. That's the way you're supposed to be doing it and it will blow others away.

23. tcarroll010 Group Title

I'll explain the methodology and you can do the work. It won't be that hard, really.

24. Dallasb22 Group Title

I got most of my questions right, but this one i got wrong and i'm not sure why.

25. tcarroll010 Group Title

We'll use apothem and perimeter and derive some cool measurements from them.

26. Dallasb22 Group Title

Groovy

27. tcarroll010 Group Title

We can use one thing I said above, that height = 3r/4 because that does come from the apothem to radius relationship and the fact that angle and side bisectors meet in the middle and make that height 3r/4. Stop and draw yourself that on paper to convince yourself. Because that's going to be key. And grooviness is good here!

28. Dallasb22 Group Title

29. Dallasb22 Group Title

9/4 is the height?

30. tcarroll010 Group Title

yes.

31. tcarroll010 Group Title

Now, here's the trick, and it's really cool conceptually, but a little hard to explain. maybe I can draw a picture.

32. Dallasb22 Group Title

Ok.

33. tcarroll010 Group Title

|dw:1348255695035:dw|

34. psi9epsilon Group Title

For an equilateral triangle circumcentre, incentre , centroid are all at the same point use this property to solve the problem

35. tcarroll010 Group Title

I forgot to put a letter at the top, call it c and call the, here, I'll draw again

36. psi9epsilon Group Title

@tcarroll010 , has given you the correct diagram, just name the angles and measures of sides

37. tcarroll010 Group Title

|dw:1348255924927:dw| There.

38. tcarroll010 Group Title

We know ac = 9/4. We know oa is 3/4. We know ob is 3/4, so we can get ab and you can do 1/2 of h x b for oab. Then double for the whole triangle.

39. psi9epsilon Group Title

|dw:1348255963318:dw|

40. tcarroll010 Group Title

Sorry, I did ob wrong, but you can do it. It's 2/3 of oa.

41. tcarroll010 Group Title

ob is 2/3 of ac.

42. Dallasb22 Group Title

Ob is 2/3 of 3/4?

43. tcarroll010 Group Title

Sorry, I'm trying to do 3 things at once, but it's right there in the diagram and just use pythagorean on the lengths.

44. Dallasb22 Group Title

I don't follow.

45. tcarroll010 Group Title

2/3 of 9/4 so its 3/2

46. Dallasb22 Group Title

1.5

47. tcarroll010 Group Title

ob is 3/2 oa is 3/4, so you can get ab.

48. tcarroll010 Group Title

ab will be your base. ac your height. Then you can easily get area of triangle.

49. Dallasb22 Group Title

Ab being 1.5√3

50. Dallasb22 Group Title

ac: 3

51. tcarroll010 Group Title

ac is 9/2 or (4 and 1/2)

52. tcarroll010 Group Title

oa is apothem which is 1/2 of r, so add 3 (which is oc) to 1 and 1/2 which is oa. so ac is 9/2

53. Dallasb22 Group Title

Ahhh

54. tcarroll010 Group Title

ob = oc = 3. So, you have a right triangle where you can figure out ab. ab^2 + oa^2 = ob^2

55. tcarroll010 Group Title

Once you get ab, it's a piece of cake.

56. tcarroll010 Group Title

Are you all set now?

57. tcarroll010 Group Title

I gotta go. I hope you are all set.

58. Dallasb22 Group Title

No.. Lol, ab^2 + 3/4^2 =3^2?

59. tcarroll010 Group Title

I just got back to my computer after a half hour. You don't happen to still be there are you?

60. Dallasb22 Group Title

61. tcarroll010 Group Title

My condolences. I'm thinking that you're not liking this problem anymore. I'm going to try to simplify it all in one post and you can just work from info in that and the diagram.

62. Dallasb22 Group Title

OK

63. tcarroll010 Group Title

oa is apothem and = r/2. oc=ob=r. ac=oc+oa. oa^2 + ab^2 = ob^2. Triangle area = ab x ac. That's all there is to it. My guess is that you're having trouble working with oa^2 + ab^2 = ob^2. So, for that ab: ab = sqrt(ob^2 - oa^2).

64. Dallasb22 Group Title

oa= 3/2 right?

65. tcarroll010 Group Title

yes

66. tcarroll010 Group Title

Are you having trouble with the square root and getting ab? Is that the hangup?

67. Dallasb22 Group Title

I don't think so, let me work out your equation above ^^

68. Dallasb22 Group Title

OK, so i got: √6.75 So √6.75 times 4.5 = area? √6.75 = ab 4.5 = ac

69. tcarroll010 Group Title

yes, you're getting it, you really are. It might be a little better to think of that ab length, which is sqrt(6.75), as(3 x sqrt(3)) / 2. But either way works. Whatever is easier for you.

70. tcarroll010 Group Title

Because when you're done, and you really almost are, enough that I can give you the answer, the area is (27 x sqrt(3)) / 4.

71. Dallasb22 Group Title

OKay.

72. tcarroll010 Group Title

You should be able to get to that last number with what should be your upcoming and last step.

73. Dallasb22 Group Title

So... 27√3 ------ 4 ???

74. Dallasb22 Group Title

$27√3\div4$

75. tcarroll010 Group Title

yes, but more importantly, do you see the flow of the steps? That's the important thing.

76. Dallasb22 Group Title

I can't carry that out?

77. tcarroll010 Group Title

You can put it in decimal from or any other representation if you want. Sqrt(3) is irrational, so any decimal representation will truncate digits, but that's ok.

78. tcarroll010 Group Title

Decimal-wise, it can be truncated to 11.69134.

79. Dallasb22 Group Title

Oh wait sorry, 27√3/4 is the area!! Okay

80. tcarroll010 Group Title

yes, and area is ac x ab.

81. Dallasb22 Group Title

I put the answer as: 27/4√3

82. Dallasb22 Group Title

I'll let you know if they accept it in 5 minutes.

83. tcarroll010 Group Title

Not 27/4√3. It's $27\sqrt{3}/4$

84. tcarroll010 Group Title

The order of those multiplicands is important.

85. Dallasb22 Group Title

Ok

86. Dallasb22 Group Title

Correct Score: 100 Thank you!!