Calculate Orthohedron Dimensions: Solving √(5a²+6a+b⁴+9) Diagonal Problem

The problem involves an orthohedron with a given diagonal length expressed as $\sqrt{5a^2+6a+b^4+9}$. The known dimensions are length $2a$ and width $a+3$, and we need to calculate the unknown height.

Using the Pythagorean theorem in three dimensions: $d^2 = (\text{length})^2 + (\text{width})^2 + (\text{height})^2$.

Substitute the given values: $\left( \sqrt{5a^2 + 6a + b^4 + 9} \right)^2 = (2a)^2 + (a+3)^2 + h^2$.

Simplifying gives: $5a^2 + 6a + b^4 + 9 = 4a^2 + (a^2 + 6a + 9) + h^2$.

Combine like terms on the right: $5a^2 + 6a + b^4 + 9 = 5a^2 + 6a + 9 + h^2$.

Subtract $5a^2 + 6a + 9$ from both sides: $b^4 = h^2$.

This gives the height as $h = b^2$.

Thus, the dimensions of the orthohedron are $(2a, a+3, b^2)$.

Therefore, the solution to the problem is $\boxed{(2a, a+3, b^2)}$.