A triangular pyramid has a surface area of 100 ft.^2, the base equals 6 ft., and the slant height equals 8 ft.. What is the height of the base to the nearest tenth?
The triangular pyramid has a triangular base and three triangular faces. The surface area is equal to the area of the base plus the sum of the areas of the three faces.
The area of the triangular base is $\frac{1}{2}bh=\frac{1}{2}(6)(h_b)$, where $h_b$ is the height of the triangular base.
The area of each face of the pyramid is $\frac{1}{2}(bh)$ again, but this time $b=6$, the base of the triangular pyramid, and $h=8$, the slant height of each face. There are three faces, so the sum of their areas is $3\cdot \frac{1}{2}(6)(8)=72$.
Since the surface area of the pyramid is 100, we have \[\frac{1}{2}(6)(h_b)+72=100.\] Solving yields $h_b=\frac{56}{3}$. Therefore, the height is $\frac{56}{3 \cdot 6}=\frac{28}{9}$ feet. Rounding to the nearest tenth, we get $\approx \boxed{3.1}$ feet.