For what positive value of k is the triangle with vertices A(0, 0), B(8, 0) and C(4, k) equilateral?

AB = |8 - 0| = 8. $C$ sits directly above the midpoint of AB, so $AC$ = BC automatically and we only need $AC$ = AB: $\sqrt{(4-0{)}^2+k^2}=8$, so $16+k^2=64,k^2=48$ and $k=\sqrt{48}=4\sqrt{3}\approx 6.93$. Taking $k=8$ makes $AC=\sqrt{80}$, which is longer than AB; $4\sqrt{2}$ comes from solving $k^2=32$.