What is a simple (or not) way of finding the lengths of the diagonals of this rhombus? I recently taught a group of geometry students about properties of rhombuses, and gave them a set of homework exercises created by a previous instructor which included the following problem.

If a rhombus has sides of length $25$ mm and a $45^{\circ}$ angle, determine the lengths of the diagonals.

Here is an image for reference.

I ultimately removed this problem because I haven't found a simple solution at a high school geometry level (i.e. some algebra, no trig).  I've solved the problem two ways, which I will post as an answer, one way involved simple trig and the other way was geometrical but messy (involved solving a system of nonlinear equations and a fourth degree polynomial).
I'm curious if anyone can solve this problem a different way.  I'd like to know if there is nice solution involving only geometry, but I would also be interested in seeing any novel approaches anyone comes up with.

As I mentioned in the question, here are the two ways that I have found to solve this problem.
Method 1: Trigonometry
If we draw in the diagonals, we get the following image, and we know that each diagonal bisects the other as well as the angles of the rhombus:
That means that we can say
\begin{align*}
a &= 25\sin(22.5^{\circ})\text{ mm}\\
b &= 25\cos(22.5^{\circ})\text{ mm},
\end{align*}
meaning that the diagonals are 
\begin{align*}
2a &= 50\sin(22.5^{\circ})\text{ mm}\approx 19.13 \text{ mm}\\
2b &= 50\cos(22.5^{\circ})\text{ mm} \approx 46.19 \text{ mm}.
\end{align*}
Method 2: (Messy) Geometry
Here I started by drawing a line perpendicular to the top connecting to the lower right angle.  Since the upper right angle is $45^{\circ}$ we know that this line makes a $45^{\circ}$ angle as well, meaning we have an isosceles triangle: 
By the Pythagorean Theorem, we know that 
$$2x^{2} = 25^{2} \implies x = \frac{25}{\sqrt{2}} \text{ mm}.$$  Therefore the area of the rhombus is given by
$$\text{Area} = 25 \text{ mm } \times \frac{25}{\sqrt{2}} \text{ mm} = \frac{25^{2}}{\sqrt{2}} \text{ mm}^{2}.$$ 
Furthermore, going back the the variables introduced in the first image, we can say that 
$$a^{2} + b^{2} = 25^{2} \quad \text{and} \quad 4\times \frac{1}{2}ab = \frac{25^{2}}{\sqrt{2}}.$$
Solving for $a$ in the second equation yields
$$ a = \frac{25^{2}}{2b\sqrt{2}},$$
which when substituted into the first equation gives us
$$b^{2} + \frac{25^{4}}{8b^{2}} = 25^{2},$$
which upon multiplying by $b^{2}$ can be written as
$$b^{4} - 25^{2}b^{2} + \frac{25^{4}}{8} = 0.$$
Now, this equation is quadratic in $b^{2}$, so we can just apply the quadratic formula to obtain
$$b^{2} = \frac{25^{2}\pm\sqrt{25^{4} - \frac{25^{4}}{2}}}{2} = \frac{25^{2}\pm\sqrt{\frac{25^{4}}{2}}}{2} = \frac{25^{2}\pm\frac{25^{2}}{\sqrt{2}}}{2} = \frac{25^{2}}{2}\left(1 \pm \frac{\sqrt{2}}{2}\right),$$
and so $$b = \frac{25\sqrt{2\pm\sqrt{2}}}{2} \approx 23.096, 9.567.$$
Back substituting then we obtain
$$a = \frac{25^{2}}{2\left(\frac{25\sqrt{2\pm\sqrt{2}}}{2}\right)\sqrt{2}} = \frac{25}{\sqrt{4\pm 2\sqrt{2}}}\approx 9.567, 23.096.$$
Since we chose $b$ to be the longer diagonal then we can say
\begin{align*}
2a &= \frac{25}{\sqrt{4+ 2\sqrt{2}}}\text{ mm} \approx 19.13 \text{ mm}\\
2b &= \frac{50\sqrt{2+\sqrt{2}}}{2}\text{ mm} \approx  46.19 \text{ mm}.
\end{align*}
