Constrained Triangle

Let $\angle PCD=\theta$ so that after some angle chasing we get $\angle DEQ=60-\theta$ .

Since triangle is equilateral, $\implies CD=DE$ $\implies \dfrac{5}{\sin \theta}=\dfrac{2}{\sin(60-\theta)}$ $\implies 5\left(\dfrac{\sqrt{3}}2\cos \theta-\dfrac{1}2\sin \theta\right)=2\sin \theta$ $\implies 9\sin \theta=5\sqrt 3\cos \theta$ $\implies \tan \theta=\dfrac{5\sqrt 3}{9}\implies \sin \theta=\dfrac{5}{2\sqrt{13}}$

Hence, $CD=\dfrac{5}{\sin \theta}=\large 2\sqrt{13}$ .

$\huge \therefore 2+13=\boxed{\color{#007fff}{15}}$

Let the side length be $a$ .
Now draw altitudes $AF,BE$ and $CE$ from $F, E$ and $E$ on $l,m$ and $n$ respectively.
So, $AD=\sqrt{a^{2}-49}, BD=\sqrt{a^{2}-25},CE=\sqrt{a^{2}-4}$
But $\square ABCF$ is a rectangle.
$\therefore AD+BD=CF$ $\therefore \sqrt{a^{2}-49}+\sqrt{a^{2}-25}=\sqrt{a^{2}-4}$ Squaring both sides and rearranging - $a^2-70=2\sqrt{(a^2-25)(a^2-49)}$ Squaring both sides and rearranging - $3a^4=156a^2$ $\boxed{a=2\sqrt{13}}$

Let CE and m intersect at Y,
EF be perpendicular to n (also m and l ) from E to meet l at F,
X the angle CEF,
and S be the sides.
So the angle DYC = 90 - X,
angle CDY=180 - (90 - X) - 60 = 30+X.
DC * SinCDY = 5, so S * Sin(30 +X) = 5.
So S * {Sin30 * CosX + Cos30 * SinX) = 5.
S * {1/2} * CosX * {1 + $\sqrt3$ * TanX.} = 5 ...( * )
In right triangle CEF, S * CosX= 7. ....( * * )
{ * } divide by { * * } and simplifying, TanX= $\dfrac{\sqrt3}7.\\ \therefore\ S\ =\ \sqrt{CF^2+FE^2}= \sqrt{(7*TanX)^2+7^2}\\ =\sqrt{3+49}=2\sqrt{13}=A*\sqrt{B}.\\ \therefore\ \ A+B=2+13=\Large\ \ \ \ \color{#D61F06}{15}.$

Consider the circumcircle of $\triangle CDE$ . Let the other point of intersection between the circle and line $m$ be called $P$ . $\angle CPD=\angle CED=60^{\circ}$ , so $CP=\frac{5}{\sin 60^{\circ}}=\frac{10}{\sqrt{3}}$ . Similarly, $\angle DPE=\angle DCE=60^{\circ}$ , so $CP=\frac{2}{\sin 60^{\circ}}=\frac{4}{\sqrt{3}}$ . $\angle CPE=120^{\circ}$ . By law of cosines on $ADC$ , $s^2=CE^2=CP^2+EP^2+CP\cdot EP=\frac{100}{3}+\frac{16}{3}+\frac{40}{3}=52$ , so $s=2\sqrt{13}$ , and our answer is $\fbox{15}$ .