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...

Question

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...

jhannybean · Answer

|dw:1440836698456:dw|

kittiwitti1 · Answer

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

kittiwitti1 · Answer

I feel like that's wrong somehow...

jhannybean · Answer

Alright, let's check.

jhannybean · Answer

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

kittiwitti1 · Answer

Yes.

jhannybean · Answer

And then added the base?

kittiwitti1 · Answer

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

jhannybean · Answer

|dw:1440837218920:dw|

kittiwitti1 · Answer

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

jhannybean · Answer

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

kittiwitti1 · Answer

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

jhannybean · Answer

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

kittiwitti1 · Answer

|dw:1440837311425:dw|
Okay lol

kittiwitti1 · Answer

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

kittiwitti1 · Answer

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

jhannybean · Answer

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 $

kittiwitti1 · Answer

Yes

kittiwitti1 · Answer

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

kittiwitti1 · Answer

Before the plus: Triangles.
After the plus: Rectangles

kittiwitti1 · Answer

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

kittiwitti1 · Answer

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:$$\ne\frac{240}{\sqrt{3}}$$

jhannybean · Answer

Arg... latex!