Pyramidal Mystery

Let $s$ be the side length of the square base of the pyramid. Then the height of the each lateral triangle is the hypotenuse of the right triangle, which has $h$ and $\dfrac{s}{2}$ as the adjacent arms.

Hence, the triangle's height is $\sqrt{h^2 + \dfrac{s^2}{4}}$ .

It is given that the the surface area of the pyrimid is equal to the cube's surface area.

$\begin{aligned} 4\times \dfrac12 \times s \left (\sqrt{h^2 + \dfrac{s^2}{4}} \right) + s^2 &= &6h^2 \\ 2s \left (\sqrt{h^2 + \dfrac{s^2}{4}} \right) &= & 6h^2 - s^2 \\ 4s^2 \left (h^2 + \dfrac{s^2}{4} \right) = 36h^4 -12(h^2)(s^2) + s^4 &= & 4h^{2}s^{2} + s^4 \\ 16h^{2}s^{2} &= & 36h^4 \\ \dfrac{h}{s} &= & \sqrt{\dfrac{16}{36}} = \dfrac{2}{3} \\ s &= & \dfrac{3h}{2} \end{aligned}$

Then the height of the triangle = $\sqrt{h^2 + \dfrac{1}{4} \left(\dfrac{3h}{2} \right)^2} = \sqrt{h^2 + \dfrac{9}{16}(h^2)} = \dfrac{5h}{4}$ .

Thus, the ratio of the triangle's height to the cube's height is $5:4$ , and since these heights are co-prime, then $h = 4$ and the triangle's height = $5$ .

And so $s = 6$ .

As a result, the pyramid's surface area = $6^2 + 4\times \dfrac{1}{2}\times 5 \times 6 = \boxed{96}$ .

Alternatively, the cube's surface area = $6\times 4^2 = \boxed{96}$ .