Is Heron necessary?

We can divide the $29/29/40$ triangle into two congruent right triangles with base length $20,$ height $x$ and hypotenuse length $29.$ By Pythagoras, we then have that $x^{2} + 20^{2} = 29^{2} \Longrightarrow x = 21.$ The area of the original triangle is then $2*\dfrac{20*21}{2} = 420.$

Similarly dividing the "mystery" triangle into two congruent right triangles with base length $b$ and height $h$ we have that $b^{2} + h^{2} = 29^{2}$ and $bh = 420,$ (due to the "equal area" condition). We then have that

$(b + h)^{2} = b^{2} + h^{2} + 2bh = 29^{2} + 2*420 = 1681 \Longrightarrow b + h = 41 \Longrightarrow bh = b(41 - b) = 420$

$\Longrightarrow b^{2} - 41b + 420 = 0 \Longrightarrow (b - 20)(b - 21) = 0.$

Thus $b = 20$ or $b = 21.$ Since we have already been given the $b = 20$ triangle, the unknown side length must be $2*21 = \boxed{42}.$

The altitude to the unequal side(base) in the given ∆ (29,29,40) can easily be found since it will bisect the base.

If comes out to be 21.

Similarly, in the unknown ∆, the altitude to the base bisects it. In the two congruent right triangles we get, legs will be 21 and 20.

Since the initial two triangles are given to be dissimilar, so altitude of the unknown ∆ is 20, so side length is 21*2=42. :)