To determine the height of the base of Niko's triangular pyramid tent, we can use the formula for the surface area of a triangular pyramid (which is a tetrahedron with a triangular base). The surface area \( S \) is given by:
\[ S = B + L \]
where \( B \) is the area of the base, and \( L \) is the total lateral area.
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Calculate the area of the base: The base is a triangle with a base length of \( b = 6 , \text{ft} \). We need to find the height \( h_b \) of this triangle. The area of the triangle (base area) is given by:
\[ B = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h_b \]
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Calculate the lateral area: The lateral area \( L \) consists of three identical triangular faces, and the area \( A_t \) of one triangular face is given by:
\[ A_t = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 6 \times 8 = 24, \text{ft}^2 \]
Therefore, the total lateral area \( L \) is:
\[ L = 3 \times A_t = 3 \times 24 = 72, \text{ft}^2 \]
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Set up the equation for total surface area: The total surface area is given as 100 ft\(^2\), so we can set up the equation:
\[ S = B + L \Rightarrow 100 = \frac{1}{2} \times 6 \times h_b + 72 \]
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Solve for the height of the base \( h_b\): Rearranging the equation, we get:
\[ 100 - 72 = \frac{1}{2} \times 6 \times h_b \] \[ 28 = 3h_b \]
Dividing both sides by 3:
\[ h_b = \frac{28}{3} \approx 9.3, \text{ft} \]
Thus, the height of the base, rounded to the nearest tenth, is \( \boxed{9.3} , \text{ft} \).