Since the area of a triangle is "half the base by the height", then, in the usual notation, $2\Delta = ah_a = bh_b = ch_c$. This can be rewritten as follows
$$\frac{1}{h_a} = \frac{1}{2\Delta}a, \quad \frac{1}{h_b} = \frac{1}{2\Delta}b, \quad \frac{1}{h_c} = \frac{1}{2\Delta}c,$$
hence the triangle inequalities, $a < b + c$, etc., for the original triangle are equivalent to the triangle inequalities, $\frac{1}{h_a} < \frac{1}{h_b} + \frac{1}{h_c}$, etc., for the triangle with sides $1/h_a, 1/h_b, 1/h_c$. Thus the reciprocals are the side lengths of another triangle, which is similar to the original triangle with similarity factor $1/2\Delta$. Therefore, the perimeter of this new triangle is equal to $2s/2\Delta = s/\Delta$ and its area is equal to $\Delta/(2\Delta)^2 = 1/4\Delta$.