Question

if the height of h at the center of the tent is to be 6 feet and the length of L of the tent is to be 6 feet then how many square feet of material will be needed to make the tent?

Answer 1

The tent is in the shape of a triangular prism with one rectangular side face open.

So the total area of the material required will be equal to the area of two rectangular faces and two triangular faces of the prism.

Moreover, the two rectangular faces are identical, and the two triangular faces are also identical.

So the area A can be determined as,

`\text{TSA}=2\cdot A_t+2\cdot A_r`

Given that the height (h) of the triangular face is 6 feet, and the angle of the inclined side is 60 degrees. So base (b) of the triangle can be calculated as follows,

`\begin{gathered} \tan 60=(h)/(((b)/(2))) \\ \sqrt[]{3}=(6)/(b)\cdot2 \\ b=\frac{12}{\sqrt[]{3}} \\ b=4\sqrt[]{3} \end{gathered}`

Now, that we have the height and base, the area can be calculated directly as,

`\begin{gathered} A_t=(1)/(2)\cdot4\sqrt[]{3}\cdot6 \\ A_t=12\sqrt[]{3} \end{gathered}`

Now, we have to find the area of the rectangle. For this, the width of the rectangle (w) is to be calculated first.

This can be again calculated using the given angle of 60 degrees,

`\begin{gathered} \sin 60=(h)/(w) \\ \frac{\sqrt[]{3}}{2}=(6)/(w) \\ w=\frac{12}{\sqrt[]{3}} \\ w=4\sqrt[]{3} \end{gathered}`

The area of the rectangular face is calculated as,

`\begin{gathered} A_r=l\cdot w \\ A_r=6\cdot4\sqrt[]{3} \\ A_r=24\sqrt[]{3} \end{gathered}`

Substituting the values in the expression for total area,

`\begin{gathered} \text{TSA}=2\cdot(12\sqrt[]{3})+2\cdot(24\sqrt[]{3}) \\ \text{TSA}=24\sqrt[]{3}+48\sqrt[]{3} \\ \text{TSA}=72\sqrt[]{3} \\ \text{TSA}\approx124.7 \end{gathered}`

Thus, the total material required for the tent is 72sqrt3 sq. feet exactly, and 124.7 sq. feet approximately.