jmc

Let $\angle A{B}^{\prime }{C}^{\prime }=\theta$. By some angle chasing in $△A{B}^{\prime }E$, we find that $\angle EA{B}^{\prime }=90^{\circ }-2\theta$. Before we apply the law of sines, we're going to want to get everything in terms of $\sin \theta$, so note that $\sin \angle EA{B}^{\prime }=\sin (90^{\circ }-2\theta )=\cos 2\theta =1-2{\sin }^2\theta$. Now, we use law of sines, which gives us the following: $\frac{\sin \theta }{5}=\frac{1-2{\sin }^2\theta }{23}⟹\sin \theta =\frac{-23\pm 27}{20}$, but since $\theta <180^{\circ }$, we go with the positive solution. Thus, $\sin \theta =\frac{1}{5}$. Denote the intersection of $B^{\prime }{C}^{\prime }$ and $AE$ with $G$. By another application of the law of sines, $B^{\prime }G=\frac{23}{\sqrt{24}}$ and $AE=10\sqrt{6}$. Since $\sin \theta =\frac{1}{5},GE=\frac{115}{\sqrt{24}}$, and $AG=AE-GE=10\sqrt{6}-\frac{115}{\sqrt{24}}=\frac{5}{\sqrt{24}}$. Note that $△E{B}^{\prime }G\sim △C^{\prime }AG$, so $\frac{EG}{B^{\prime }G}=\frac{C^{\prime }G}{AG}⟹C^{\prime }G=\frac{25}{\sqrt{24}}$. Now we have that $AB=AE+EB=10\sqrt{6}+23$, and $B^{\prime }{C}^{\prime }=BC=B^{\prime }G+C^{\prime }G=\frac{23}{\sqrt{24}}+\frac{25}{\sqrt{24}}=\frac{48}{\sqrt{24}}=4\sqrt{6}$. Thus, the area of $ABCD$ is $(10\sqrt{6}+23)(4\sqrt{6})=92\sqrt{6}+240$, and our final answer is $92+6+240=\boxed{338}$.