Powered Parallelogram

Let the two side lengths of the parallelogram be $a$ and $b$ and those of its diagonals be $c$ and $d$ . Let an angle be $\theta$ . Then $c^2 + d^2 = 625$ . By cosine rule , we have

$\begin{aligned} c^2 + d^2 & = 625 \\ a^2 + b^2 - 2ab\cos \theta + a^2 + b^2 - 2ab \blue{\cos (180^\circ - \theta)} & = 625 & \small \blue{\text{Note that }\cos (180^\circ - \theta) = - \cos \theta} \\ a^2 + b^2 - 2ab\cos \theta + a^2 + b^2 + 2ab \blue{\cos \theta} & = 625 \\ 2(a^2 + b^2) & = 625 \\ \implies S & = 625 \end{aligned}$

Therefore the square of sum of digits of $S$ is $(6+2+5)^2 = 13^2 = \boxed{169}$ .

In any parallelogram, $AB^2+BC^2+CD^2+DA^2 = AC^2+BD^2$ . To prove this, we can just use the cosine law. Let $\angle ABC = \angle CDA = \theta$ . Then $\angle BCD = \angle DAB = 180^{\circ} - \theta$ . $\begin{aligned} AC^2 &= AB^2+BC^2 - 2AB \cdot BC \cdot \cos{\theta} \\ BD^2 &= BC^2+CD^2 - 2BC \cdot CD \cdot \cos{\left(180^{\circ} - \theta \right)} \\ AC^2 &= CD^2+DA^2 - 2CD \cdot DA \cdot \cos{\theta} \\ BD^2 &= DA^2+AB^2 - 2DA \cdot AB \cdot \cos{\left( 180^{\circ} - \theta \right)} \end{aligned}$

Since $\cos{\theta} + \cos{\left(180^{\circ} - \theta \right)}=0$ , adding up all four of these equations gives the result.

So in this case, $AB^2+BC^2+CD^2+DA^2 = 625$ , and the answer is $(6+2+5)^2=\boxed{169}$ .

Applying appolonius theorem in triangle ABD

$AB^{2}+BD^{2}=2(CD^{2}+BC^{2})$

$625=sum$ $of$ $square$ $of$ $sides$

$\Rightarrow$ Result $=6+2+5=\boxed{13}$

Answer = $13^{2} = 169$