The diagram shows a chocolate bar in the form of a triangular prism and the cross-section of the chocolate bar is an isosceles triangle with $AB = AC$.
$MC = \left(\sqrt{2} + \frac{1}{2}\right)\text{ cm}$ and $\angle ACB = 45^\circ$.

1. Identify the right-angled triangle and the relevant trigonometric ratio

In $\triangle AMC$, $\angle AMC = 90^\circ$.
$\cos(\angle ACM) = \frac{MC}{AC}$

2. Substitute known values and exact trigonometric value

$\cos(45^\circ) = \frac{\sqrt{2} + \frac{1}{2}}{AC}$
$\frac{\sqrt{2}}{2} = \frac{\sqrt{2} + \frac{1}{2}}{AC}$

3. Solve for AC and simplify the surd expression

$AC = \frac{\sqrt{2} + \frac{1}{2}}{\frac{\sqrt{2}}{2}}$
$AC = \frac{2\sqrt{2} + 1}{\sqrt{2}}$
$AC = \frac{(2\sqrt{2} + 1)\sqrt{2}}{2}$
$AC = \frac{4 + \sqrt{2}}{2} = 2 + \frac{\sqrt{2}}{2}$

1. Find the area of the triangular cross-section $ABC$.

In $\triangle AMC$, $\angle AMC = 90^\circ$ and $\angle ACM = 45^\circ$, so $\triangle AMC$ is an isosceles right-angled triangle.
$AM = MC = \sqrt{2} + \frac{1}{2}$
Area of $\triangle ABC = \frac{1}{2} \times BC \times AM = \frac{1}{2} \times (2MC) \times MC = MC^2$
Area $= \left(\sqrt{2} + \frac{1}{2}\right)^2 = 2 + \sqrt{2} + \frac{1}{4} = \frac{9}{4} + \sqrt{2} = \frac{9 + 4\sqrt{2}}{4}\text{ cm}^2$

2. Set up the equation for the volume of the prism to find $AD$.

Volume of prism = Area of cross-section $\times$ length $AD$
$25 + 22\sqrt{2} = \left(\frac{9 + 4\sqrt{2}}{4}\right) \times AD$
$AD = \frac{4(25 + 22\sqrt{2})}{9 + 4\sqrt{2}} = \frac{100 + 88\sqrt{2}}{9 + 4\sqrt{2}}$

3. Rationalise the denominator.

$AD = \frac{100 + 88\sqrt{2}}{9 + 4\sqrt{2}} \times \frac{9 - 4\sqrt{2}}{9 - 4\sqrt{2}}$
Denominator $= 9^2 - (4\sqrt{2})^2 = 81 - 32 = 49$
Numerator $= 100(9) - 100(4\sqrt{2}) + 88\sqrt{2}(9) - 88\sqrt{2}(4\sqrt{2})$
$= 900 - 400\sqrt{2} + 792\sqrt{2} - 352(2)$
$= 900 - 704 + 392\sqrt{2}$
$= 196 + 392\sqrt{2}$

4. Simplify the fraction to get the final answer.

$AD = \frac{196 + 392\sqrt{2}}{49} = 4 + 8\sqrt{2}$