Diagonal Triangle of a Cuboid

The dimensions of this cuboid is $34 \times W \times (34 + W)$ , where $W\geq 34$ .

By Pythagorean theorem , the side lengths of the triangle in question are $\sqrt{34^2 + W^2}, \sqrt{34^2 + (34 + W)^2} , \sqrt{W^2 + (34 + W)^2} .$ Let $a,b,c$ denote the these side lengths, respectively. Then, using one of the formula of Heron's formula , we have $2418 = \dfrac14 \sqrt{(a^2 +b^2 + c^2)^2 - 2(a^4 + b^4 + c^4) } .$ With $\begin{cases} a^2= 34^2 + W^2 \\ b^2 = 34^2 + (34 + W)^2\\ c^2 = W^2 + (34 + W)^2 , \end{cases}$ the equation simplifies to $(2418\cdot 4)^2 = 4(W^2 + 34W+ 1156)^2 \implies W = 46.$ The volume of the cuboid is simply $34W(34 + W) = \boxed{125120}$ .

I took somewhat of a brute force approach, solving the equation $||(L,0,H) \times (0,W,H)||=||(34+W,0,34) \times (0,W,34)||=2\times 2418$ . After a bit of computation we find $W(W+34)+34^2=4836$ so $W(W+34)=3680$ and the volume of the cuboid is $34W(W+34)=\boxed{125120}$ .