A triangular pyramid has a surface area of 100, the base equals 6, and a slant height equals 8. What is the height of the base to the nearest tenth?
If the base of the triangular pyramid is a triangle, then we can use Heron's formula to find the area of the base. Let $s$ be the semiperimeter of the base, so $s=\dfrac{6+6+8}{2}=10$. The area of the base $\triangle ABC$ is $\sqrt{s(s-6)(s-6)(s-8)}=\sqrt{10(10-6)(10-6)(10-8)}=12$. If the height of the triangular pyramid is $h$, then the area of each face is $\dfrac{bh}{2}=24$.
[asy] size(60); pair A, B, C, D, F; A=(0,0); B=(6,0); C=(4,4); D=(5,2); F=(4.8,2.6); draw(A--B--C--cycle); draw(A--D); draw(B--D); draw(C--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NW); dot(D); label("$D$", D, NE); [/asy]
Also, each face is a right triangle with legs 6 and 8. The area of each triangular face given a leg and the hypotenuse is $\dfrac{bc}{2}$, so $24=\dfrac{6h}{2}$, which gives us $12=3h$. Thus, the height is $\boxed{4}$.