To solve for the height of the triangular pyramid (the tent), we can use the formula for the surface area of a triangular pyramid, which is given by:
\[ \text{Surface Area} = \text{Base Area} + \text{Lateral Area} \]
The base of the pyramid is a triangle. The area of the triangular base (assuming it is an equilateral triangle, since the pyramid is regular) can be calculated using the formula for the area of a triangle:
\[ \text{Base Area} = \frac{1}{2} \times \text{base} \times \text{height}_\text{base} \]
We need to first determine the height of the base triangle. Given that the base is 6 ft, we can denote this height as \( h_{base} \).
Next, we also need to calculate the lateral area of the pyramid. The lateral area consists of the areas of the three triangular faces. Each triangular face will have a base of 6 ft and a slant height of 8 ft:
\[ \text{Area of one triangular face} = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 6 \times 8 = 24 \text{ ft}^2 \]
Since there are three faces:
\[ \text{Lateral Area} = 3 \times 24 = 72 \text{ ft}^2 \]
Now, we can now calculate the area of the base triangle. Let's denote the height of the base as \( h_{base} \).
The total surface area is given as 100 ft², hence we can set up the equation:
\[ 100 = \text{Base Area} + \text{Lateral Area} \]
Now we calculate:
\[ 100 = \frac{1}{2} \times 6 \times h_{base} + 72 \]
Now, simplifying the equation:
\[ 100 = 3h_{base} + 72 \]
Subtract 72 from both sides:
\[ 28 = 3h_{base} \]
Now, solve for \( h_{base} \):
\[ h_{base} = \frac{28}{3} \approx 9.3 \text{ ft} \]
Thus, the height of the base to the nearest tenth is:
9.3 ft.