Answer: B

Small Hint:

Equal tangent segments from a vertex make the three sides $14,$ $6+z,$ and $8+z$ for some $z$

Big Hint:

Use both $K=rs$ and Heron’s formula with semiperimeter $s=14+z$

Solution:

Let the two tangent segments from the third vertex each have length $z.$ Equal tangents from a common vertex make the side lengths $$14,\qquad 6+z,\qquad 8+z,$$ with semiperimeter $s=14+z.$ Since the inradius is $4,$ $$K=rs=4(14+z).$$ Heron’s formula gives $$\begin{aligned} K^2 &=(14+z)z\cdot8\cdot6\\ &=48z(14+z). \end{aligned}$$ Equating the squares yields $16(14+z)=48z,$ so $z=7.$ The sides are $14,$ $13,$ and $15,$ and the shortest is $13.$

Thus, the correct answer is B.