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Calculating square feet of a tent/triangle

  1. Jun 13, 2005 #1
    The question is this:

    A 2 person tent is to be made so that the height at the center is 4 ft. if the sides of the tent meet at the ground at an angle of 60(degrees), and the tent is to be 6 feet in length, how many square feet of material will be needed to make the tent?

    My answer:
    Since it's a 60 degree angle, then it's a 30-60-90 triangle in which the smallest side is t, the side opposite of the largest side is 2t and the h is t x sq root of 3.

    So t ends up being 4 x sq root 3 divided by 3

    which means the width is 4.619 ft ( 4x1.732)/3 = 2.309 ft ~4.618 (since 2.309 is only half the length)

    and 2t= 8(1.732)/3 = 4.619 ft

    Now i just need help figuring out the sq feet. The answer in the book says 101.6 ft squard.

    Any help is appriciated
  2. jcsd
  3. Jun 13, 2005 #2
    Looks ok. If you find the other side, pythagorean, then you can use that for the other sides of the tent. For the material, think of breaking the tent into pieces. You will have 2 triangles, the ones on the front and back, you will have 2 sides (the left and right), and one bottom (thinking in 3d). Now compute up the areas, keeping in mind that the tent is supposed to be 6ft long, and you will get 101.6ft^2
  4. Jun 13, 2005 #3


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    If you know the long side (4'), and you know the short side (2'), what is the hypotenuse?

    The hypotensue then forms the side of one of two rectangles that are the tent sides, which is your area.

    But I don't figure anywhere near 100 sq. ft. Barely half of that.
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