Calculate C¹B¹ in a Rectangular Prism with Diagonal √(6x²-12x+41)

To solve this problem, we'll use the expressions for the diagonals given:

• The space diagonal of the prism: $\sqrt{6x^2 - 12x + 41}$.
• The face diagonal $DC^1 = \sqrt{5x^2 - 4x + 25}$.
• We need to find the edge $C^1B^1$.

Since $DC^1 = \sqrt{a^2 + b^2}$, we will denote one dimension as $a$ and another as $b$. Therefore:

$DC^1 = \sqrt{5x^2 - 4x + 25} = \sqrt{(x-2)^2 + (b)^2}$.

Assuming that $x-2$ is one of the lengths, we have:

$5x^2 - 4x + 25 = (x-2)^2 + b^2$.

On expanding $(x-2)^2$, we get $x^2 - 4x + 4$. Thus:

$5x^2 - 4x + 25 = x^2 - 4x + 4 + b^2$.

Cancelling the $x^2 - 4x$ terms on both sides gives,

$4x^2 + 21 = b^2$.

For the diagonal of the prism, we assume if $x-2$ is one dimension and $b = \sqrt{4x^2 + 21}$, we solve:

$\left(\sqrt{6x^2 - 12x + 41}\right)^2 = (x-2)^2 + (b)^2 + ( C^1B^1 )^2$.

$6x^2 - 12x + 41 = (x-2)^2 + 4x^2 + 21 + (C^1B^1)^2$.

Substituting $( x-2 )^2 = x^2 - 4x + 4$:

$6x^2 - 12x + 41 = x^2 - 4x + 4 + 4x^2 + 21 + (C^1B^1)^2$.

Simplify and solve for $C^1B^1$:

$6x^2 - 12x + 41 = 5x^2 + 25 + (C^1B^1)^2$.

Thus, $x^2 - 12x + 16 = (C^1B^1)^2$.

This implies $C^1B^1 = x - 4$.

Therefore, the solution to the problem is $C^1B^1 = x-4$.