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...

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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...

Mathematics
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|dw:1440836698456:dw|
I tried doing it on my own and got this:\[\frac{360}{\sqrt{3}}ft^{2}\]
I feel like that's wrong somehow...

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Alright, let's check.
You split the area into 3 separate sections, yes? The triangle faces and the rectangular sides?
Yes.
And then added the base?
The base is the same as the sides in area, right? :o
|dw:1440837218920:dw|
These are the formulas I got. Triangle:\[h=6,b=\frac{6}{\sqrt{3}},hyp=\frac{12}{\sqrt{3}}\] Square:\[w=\frac{12}{3},l=8\]
we need to find x first, and then double it to find the area of the front facing triangluar portion.
I think if I did anything wrong I probably messed up there
I havent worked it out yet so one minute! :P
|dw:1440837311425:dw| Okay lol
|dw:1440837440575:dw| Do these look right to you?
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
Yeah. the \(\dfrac{12}{\sqrt{3}}\) looks correct, and that would be the side of the triangular prism... so \(A_2 = \dfrac{12}{\sqrt{3}} \cdot 8 \)
Yes
I got it like this:\[2\times\frac{1}{2}\times(6\times\frac{12}{\sqrt{3}})+3(8\times\frac{12}{\sqrt{3}})\]
Before the plus: Triangles. After the plus: Rectangles
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
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:\[\ne\frac{240}{\sqrt{3}}\]
Arg... latex!
LOL Brb I have a mosquito in my house. -_-
|dw:1440838289662:dw| \[SA = \color{red}{A_1} +\color{blue}{A_2} +\color{green}{A_3}\]\[SA = \color{red}{2\left(\frac{ 12}{\sqrt{3}} \cdot 8\right)} +\color{blue}{2\left(\frac{1}{2} \cdot \frac{12}{\sqrt{3}} \cdot 6\right)} + \color{green}{\left(\frac{12}{\sqrt{3}} \cdot 8\right)} \]
I used 3 instead of 2 for red because I just stuck all the rectangular areas together.
yeah that works too.
\[SA = \color{red}{2\left(\frac{ 12}{\sqrt{3}} \cdot 8\right)} +\color{blue}{2\left(\frac{1}{2} \cdot \frac{12}{\sqrt{3}} \cdot 6\right)} + \color{green}{\left(\frac{12}{\sqrt{3}} \cdot 8\right)}\]\[SA = 3\left(\frac{12}{\sqrt{3}} \cdot 8 \right) +2\left(\frac{1}{2}\cdot \frac{12}{\sqrt{3}} \cdot 6\right)\]
\[3\left(8\times\frac{12}{\sqrt{3}}\right)+2\left(\frac{1}{2}\times6\times\frac{12}{\sqrt{3}}\right)=\frac{72}{\sqrt{3}}+3\left(\frac{96}{\sqrt{3}}\right)\]So far so good?
I mean is it good so far* (what is wrong with my grammar)
\[SA = \frac{288}{\sqrt{3}} +\frac{72}{\sqrt{3}}\]
Then simplify it some more, and you get \[SA= \frac{360\sqrt{3}}{3} = 120\sqrt{3}\]
Yes. I did get that... But what's up with the other one giving me\[\frac{240}{\sqrt{3}}\]
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 \(SA = \dfrac{244}{\sqrt{3}}\).
a = 6ft b = 8ft θ = 60° \[\tan60°=\frac{a}{\frac{w}{2}}=\frac{2a}{w}\]\[w=2a\cot60°=\frac{8}{\sqrt{3}}ft\]\[\sin60°=\frac{a}{s}\]\[s=a\csc60°=\frac{8}{\sqrt{3}}ft\]
I did redo it with my own numerical values from my problem, but I still got a different answer.
OK let me work it out first, one min.
So they started ff by solving for half the base first, given the height of the triangle. |dw:1440839311853:dw|
Sorry if im a ltitle slow in my responses right now, Kind of getting drowsy.
Okay, I'll type out the rest in case you guys can spot an error\[A_{T}=2\times\frac{1}{2}\times\frac{8}{\sqrt{3}}ft\times6ft=\frac{48}{\sqrt{3}}ft^{2}\]\[A_{R}=3\times\frac{8}{\sqrt{3}}\times8ft=64\sqrt{3}ft^{2}\] \[A=A_{T}+A_{R}=64\sqrt{3}ft^{2}+\frac{48}{\sqrt{3}}ft^{2}=\]\[64\sqrt{3}+\frac{48}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}}\right)=\frac{48\sqrt{3}}{3}+64\sqrt{3}\]
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 -\[\frac{240}{3}\]
*It wasn't 240/root 3
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?
So then for the next equation (which just has different values) I should do the same? http://prntscr.com/8a6nx4
Yeah I think it was too long-winded and complex as well xD
|dw:1440839957574:dw| \[\tan(60) = \frac{6}{\dfrac{w}{2}} = \frac{12}{w} \iff w = 12\cot(60^\circ)\] etc...
Okay, now you just solve it with \(a=5\) instead of \(a=6\)
Yeah, so I don't need to go the long-winded method... and just change the values in the equation right? :o
|dw:1440840212338:dw| and yes. I'll start you off here by helping you find the area of the triangular face.
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. ^^
But thank you :D
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.
What o-o
So what I would do is... |dw:1440840514569:dw|
I think I did that subconsciously xD
I got 6/sqrt 3 then x2 = 12/sqrt3
\[\tan(60) = \frac{5}{x} \iff x = \frac{5}{\tan(60)}\]\[y=2\left(\frac{5}{\tan(60)}\right)\]
YEah, then you're good. no need for a long fancy explanation like that guy did.
What is the weird arrow you put there lol Is there a latex code?
That means its related. I could write either or interchangeably.
`\[\iff\]`
Oh. Is there a latex code x_x?|dw:1440840784961:dw|
Alright so far? (got the code thanks)
Yep, you gt it.
Thank you! :D I can handle it from here. Sorry to make you stay up to help me ^^;
just remember that the triangle is equilateral, therefore the base will be the same as the hypotenuse.
No its fine :)
Good luck~

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