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cademan
08-19-2010, 03:57 PM
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)

galactus
08-19-2010, 04:19 PM
OK, let's see.

For the triangular ends.

3cot(\frac{\pi}{3})=\sqrt{3}

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

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

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

The bottom has area 2\sqrt{3}\cdot 8=\boxed{16\sqrt{3}}

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

soroban
08-19-2010, 05:05 PM
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.)

*
* \
* \
* \ x
/|\ \
/ | \ \
x / | \ x *
/ |3 \ *
/ | \ * 8
*-----*-----*
x
\text{Let } x\text{ = side of the equilateral triangle.}


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

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


\text{Area of one side} \;=\;(x)(8) \;=\;8x

. . \text{Area of two sides} \;=\;2\cdot8x \;=\;16x


\text{Area of floor} \;=\;(x)(8) \;=\;8x


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



We have this right triangle:



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


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