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Thread: Area of a Tent

  1. #1

    Area of a Tent

    A two-person tent is to be made so that the height at the center is a = 3 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.)

    Attached is image of tent

    [attachment=0:3r9kzfxz]1-1-052alt.gif[/attachment:3r9kzfxz]

    Here is my work.... however I may be way out in outfield

    So the center pole is 3 ft which if we make X the bottom side,

    3 = X*Sqrt(3)
    than
    3 sqrt(3)/3 = x
    than
    sqrt(3)=x

    So if we make y= Full bottom than since the bottom side is actually double that so 2 sqrt(3)=y

    To calculate the area of a triangle we do 1/2 L+H so that would be 6 sqrt(3) one side. Multiply that * 2 sides and we have 12 sqrt(3)

    NOw for the rectangles 6 sqrt(3) * 8 = 48 sqrt(3). Multiply that times 3 = 144 q=sqrt(3)

    Add those together and I get 156 sqrt(3)
    Attached Images Attached Images

  2. #2
    Elite Member
    Join Date
    Sep 2005
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    7,293

    Re: Area of a Tent

    OK, let's see.

    For the triangular ends.

    [tex]3cot(\frac{\pi}{3})=\sqrt{3}[/tex]

    The length of the hypotenuse is then [tex]\sqrt{(\sqrt{3})^{2}+3^{2}}=2\sqrt{3}[/tex]

    There are 4 of the right triangles, front and back, that make up the ends so they have area [tex]4\cdot \frac{\sqrt{3}\cdot 3}{2}=\boxed{6\sqrt{3}}[/tex]

    The slant length, hypoteneuse, has length [tex]2\sqrt{3}[/tex], so the 2 sides have area [tex]2\cdot 2\sqrt{3}\cdot 8=\boxed{32\sqrt{3}}[/tex]

    The bottom has area [tex]2\sqrt{3}\cdot 8=\boxed{16\sqrt{3}}[/tex]

    Add them all up and we have [tex]6\sqrt{3}+32\sqrt{3}+16\sqrt{3}=\boxed{54\sqrt{3}} \approx 93.53[/tex]

  3. #3
    Elite Member
    Join Date
    Jan 2005
    Location
    Lexington, MA
    Posts
    5,603

    Re: Area of a Tent

    Hello, cademan!

    A slightly different approach . . .


    A two-person tent is to be made so that the height at the center is 3 feet,
    If the sides of the tent are to meet the ground at an angle of 60, and the tent is to be 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. .Give your answer in exact form.)
    Code:
                                  *
                            *      \
                      *             \
                *                    \ x
               /|\                    \
              / | \                    \
           x /  |  \ x                  *
            /   |3  \             *
           /    |    \      *    8
          *-----*-----*
                x

    [tex]\text{Let } x\text{ = side of the equilateral triangle.}[/tex]


    [tex]\text{Area of triangle} \:=\:\tfrac{1}{2}(x)(3) \;=\; \tfrac{3}{2}x[/tex]

    . . [tex]\text{Area of front and back} \;=\;2 \cdot \tfrac{3}{2}x \;=\;3x[/tex]


    [tex]\text{Area of one side} \;=\;(x)(8) \;=\;8x[/tex]

    . . [tex]\text{Area of two sides} \;=\;2\cdot8x \;=\;16x[/tex]


    [tex]\text{Area of floor} \;=\;(x)(8) \;=\;8x[/tex]


    [tex]\text{Hence: }\;A \;=\;3x + 16x + 8x \;=\;27x\text{ ft}^2[/tex]



    We have this right triangle:


    Code:
                *
               /|
              / |
           x /  |3
            /   |
           /    |
          *-----* 
            x/2

    [tex]\text{We have: }\;\left(\tfrac{x}{2}\right)^2 + 3^2 \:=\^2 \quad\Rightarrow\quad \tfrac{3}{4}x^2 \:=\:9 \quad\Rightarrow\quad x^2 \:=\:12 \quad\Rightarrow\quad x \:=\:2\sqrt{3}[/tex]


    [tex]\text{Therefore: }\;A \;=\;27(2\sqrt{3}) \;=\;54\sqrt{3}\text{ ft}^2[/tex]

    I'm the other of the two guys who "do" homework.

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