## kittiwitti1 one year ago A two-person tent is to be made so that the height at the center is a = 6 feet (see the figure below). If the sides of the tent are to meet the ground at an angle 60°, and the tent is to be b = 8 feet in length, how many square feet of material will be needed to make the tent? (Assume that the tent has a floor and is closed at both ends, and give your answer in exact form.) Picture reference: http://prntscr.com/8a6azy I did it two ways and got different answers, not sure which one I did wrong...

1. anonymous

|dw:1440836698456:dw|

2. kittiwitti1

I tried doing it on my own and got this:$\frac{360}{\sqrt{3}}ft^{2}$

3. kittiwitti1

I feel like that's wrong somehow...

4. anonymous

Alright, let's check.

5. anonymous

You split the area into 3 separate sections, yes? The triangle faces and the rectangular sides?

6. kittiwitti1

Yes.

7. anonymous

8. kittiwitti1

The base is the same as the sides in area, right? :o

9. anonymous

|dw:1440837218920:dw|

10. kittiwitti1

These are the formulas I got. Triangle:$h=6,b=\frac{6}{\sqrt{3}},hyp=\frac{12}{\sqrt{3}}$ Square:$w=\frac{12}{3},l=8$

11. anonymous

we need to find x first, and then double it to find the area of the front facing triangluar portion.

12. kittiwitti1

I think if I did anything wrong I probably messed up there

13. anonymous

I havent worked it out yet so one minute! :P

14. kittiwitti1

|dw:1440837311425:dw| Okay lol

15. kittiwitti1

|dw:1440837440575:dw| Do these look right to you?

16. kittiwitti1

I mean correctly done* lol that sounded rude. ^^: I found another proof online with different digits for A and B though, and got a different answer: http://mathhelpboards.com/questions-other-sites-52/aju051000s-questions-yahoo-answers-involving-trigonometry-6263.html

17. anonymous

Yeah. the $$\dfrac{12}{\sqrt{3}}$$ looks correct, and that would be the side of the triangular prism... so $$A_2 = \dfrac{12}{\sqrt{3}} \cdot 8$$

18. kittiwitti1

Yes

19. kittiwitti1

I got it like this:$2\times\frac{1}{2}\times(6\times\frac{12}{\sqrt{3}})+3(8\times\frac{12}{\sqrt{3}})$

20. kittiwitti1

Before the plus: Triangles. After the plus: Rectangles

21. kittiwitti1

I thought the link's method was too complicated and tried to do it myself...but the answers don't match. Not sure which one's wrong

22. kittiwitti1

I mean I got an answer of $\frac{360}{\sqrt{3}}$ from doing it myself but it doesn't equal the answer I got from the method in the URL I posted:$\ne\frac{240}{\sqrt{3}}$

23. anonymous

Arg... latex!

24. kittiwitti1

LOL Brb I have a mosquito in my house. -_-

25. anonymous

|dw:1440838289662:dw| $SA = \color{red}{A_1} +\color{blue}{A_2} +\color{green}{A_3}$$SA = \color{red}{2\left(\frac{ 12}{\sqrt{3}} \cdot 8\right)} +\color{blue}{2\left(\frac{1}{2} \cdot \frac{12}{\sqrt{3}} \cdot 6\right)} + \color{green}{\left(\frac{12}{\sqrt{3}} \cdot 8\right)}$

26. kittiwitti1

I used 3 instead of 2 for red because I just stuck all the rectangular areas together.

27. anonymous

yeah that works too.

28. anonymous

$SA = \color{red}{2\left(\frac{ 12}{\sqrt{3}} \cdot 8\right)} +\color{blue}{2\left(\frac{1}{2} \cdot \frac{12}{\sqrt{3}} \cdot 6\right)} + \color{green}{\left(\frac{12}{\sqrt{3}} \cdot 8\right)}$$SA = 3\left(\frac{12}{\sqrt{3}} \cdot 8 \right) +2\left(\frac{1}{2}\cdot \frac{12}{\sqrt{3}} \cdot 6\right)$

29. kittiwitti1

$3\left(8\times\frac{12}{\sqrt{3}}\right)+2\left(\frac{1}{2}\times6\times\frac{12}{\sqrt{3}}\right)=\frac{72}{\sqrt{3}}+3\left(\frac{96}{\sqrt{3}}\right)$So far so good?

30. kittiwitti1

I mean is it good so far* (what is wrong with my grammar)

31. anonymous

$SA = \frac{288}{\sqrt{3}} +\frac{72}{\sqrt{3}}$

32. anonymous

Then simplify it some more, and you get $SA= \frac{360\sqrt{3}}{3} = 120\sqrt{3}$

33. kittiwitti1

Yes. I did get that... But what's up with the other one giving me$\frac{240}{\sqrt{3}}$

34. anonymous

Just saw the link you posted earlier, the height being different would result in a different area for the triangle..hence different answer? that's why this guy is getting his $$SA = \dfrac{244}{\sqrt{3}}$$.

35. kittiwitti1

a = 6ft b = 8ft θ = 60° $\tan60°=\frac{a}{\frac{w}{2}}=\frac{2a}{w}$$w=2a\cot60°=\frac{8}{\sqrt{3}}ft$$\sin60°=\frac{a}{s}$$s=a\csc60°=\frac{8}{\sqrt{3}}ft$

36. kittiwitti1

I did redo it with my own numerical values from my problem, but I still got a different answer.

37. anonymous

OK let me work it out first, one min.

38. anonymous

So they started ff by solving for half the base first, given the height of the triangle. |dw:1440839311853:dw|

39. anonymous

Sorry if im a ltitle slow in my responses right now, Kind of getting drowsy.

40. kittiwitti1

Okay, I'll type out the rest in case you guys can spot an error$A_{T}=2\times\frac{1}{2}\times\frac{8}{\sqrt{3}}ft\times6ft=\frac{48}{\sqrt{3}}ft^{2}$$A_{R}=3\times\frac{8}{\sqrt{3}}\times8ft=64\sqrt{3}ft^{2}$ $A=A_{T}+A_{R}=64\sqrt{3}ft^{2}+\frac{48}{\sqrt{3}}ft^{2}=$$64\sqrt{3}+\frac{48}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}}\right)=\frac{48\sqrt{3}}{3}+64\sqrt{3}$

41. kittiwitti1

BLEH that took too long...$3\left(\frac{48\sqrt{3}}{3}\right)+3(64\sqrt{3})=48\sqrt{3}+192\sqrt{3}=240\sqrt{3}$Not sure what I did wrong... And sorry it wasn't this -$\frac{240}{3}$

42. kittiwitti1

*It wasn't 240/root 3

43. anonymous

I think however the guy solved it is a little too complicated and the way we solved it earlier is just easier. Don't you think?

44. kittiwitti1

So then for the next equation (which just has different values) I should do the same? http://prntscr.com/8a6nx4

45. kittiwitti1

Yeah I think it was too long-winded and complex as well xD

46. anonymous

|dw:1440839957574:dw| $\tan(60) = \frac{6}{\dfrac{w}{2}} = \frac{12}{w} \iff w = 12\cot(60^\circ)$ etc...

47. anonymous

Okay, now you just solve it with $$a=5$$ instead of $$a=6$$

48. kittiwitti1

Yeah, so I don't need to go the long-winded method... and just change the values in the equation right? :o

49. anonymous

|dw:1440840212338:dw| and yes. I'll start you off here by helping you find the area of the triangular face.

50. kittiwitti1

It's okay, you're tired. I can handle this one :) I was afraid I had to do the long method but this one I can do. ^^

51. kittiwitti1

But thank you :D

52. anonymous

Oh here's an easier way, if it helps at all. split the base of the triangle in half, and label it x. Then you have 2 x's. label the entire base y. So first you'll solve for x, and when you DOUBLE it, you'll find y :) Then you just use the y-value in the rest of your calculations.

53. kittiwitti1

What o-o

54. anonymous

So what I would do is... |dw:1440840514569:dw|

55. kittiwitti1

I think I did that subconsciously xD

56. kittiwitti1

I got 6/sqrt 3 then x2 = 12/sqrt3

57. anonymous

$\tan(60) = \frac{5}{x} \iff x = \frac{5}{\tan(60)}$$y=2\left(\frac{5}{\tan(60)}\right)$

58. anonymous

YEah, then you're good. no need for a long fancy explanation like that guy did.

59. kittiwitti1

What is the weird arrow you put there lol Is there a latex code?

60. anonymous

That means its related. I could write either or interchangeably.

61. anonymous

$\iff$

62. kittiwitti1

Oh. Is there a latex code x_x?|dw:1440840784961:dw|

63. kittiwitti1

Alright so far? (got the code thanks)

64. anonymous

Yep, you gt it.

65. kittiwitti1

Thank you! :D I can handle it from here. Sorry to make you stay up to help me ^^;

66. anonymous

just remember that the triangle is equilateral, therefore the base will be the same as the hypotenuse.

67. anonymous

No its fine :)

68. anonymous

Good luck~