Five Similar Rectangles

Let the short side $s_c$ of $C$ be $1$ and its long side $l_c=k$ . Then, we have:

\(\begin{array} {} \Rightarrow s_b=k & \Rightarrow l_b=k^2 \\ \Rightarrow l_a=k^2+1&\Rightarrow s_a= k +\dfrac{1}{k} \\ \Rightarrow l_d=2k+\dfrac {1} {k} & \Rightarrow s_d=2+\dfrac {1} {k ^2} \\ \Rightarrow s_{Big} =2k+\dfrac {1} {k} & \Rightarrow l_{Big} =k^2+3+\dfrac {1} {k ^2} \end{array} \)

And,

\begin{aligned} \frac{l_{Big}}{s_{Big}} & =k \\ \Rightarrow k^2+3+\frac {1} {k^2} & =2k^2 +1\\ k^4-2k^2 - 1&=0 \\ \Rightarrow k^2 & =1 +\sqrt {2} \end{aligned} \end {equation}

Now,

\begin{aligned} \frac {a} {b} & =\frac {s_al_a} {s_b l_b} \\&= \frac {(k^2 +1)^2} {k^4}\\&= \left(\frac {2+\sqrt{2}}{1 +\sqrt {2}}\right) \\&= \left(\frac {(2+\sqrt {2}) (\sqrt {2} - 1)}{(1+\sqrt {2}) (\sqrt {2}-1)} \right) ^2 \\&=(\sqrt {2}) ^2 =2 \end{aligned} \end {equation}

$\Rightarrow a+b=2+1=\boxed{3}$

Let $l(X)$ and $w(X)$ denote the length and width of rectangle $X$ , where $X$ = $A$ , $B$ , $C$ or $D$ . WLOG, let $l(C) = 1$ and $w(C) = x$ . So $w(X)/l(X) = x$ . It follows that $w(B) = 1$ , $l(B) = w(B)/x = 1/x$ , $l(A) = l(B)+w(C) = 1/x+x$ , $w(A) = x\cdot l(A) = 1+x^2$ , $l(D) = w(A)+l(C) = 2+x^2$ , and $w(D) = x\cdot l(D) = 2x+x^3$ . Then $w(A\cup B\cup C\cup D) = x\cdot l(A\cup B\cup C\cup D),$ $l(D) = x\cdot [l(A)+w(D)],$ $2+x^2 = x(1/x+x+2x+x^3),$ $x^4+2x^2 = 1,$ $(1+x^2)^2 = 2.$ Hence $a/b = [w(A)/w(B)]^2 = (1+x^2)^2 = 2$ ; and so $a+b = 2+1 = 3$ .

Let short side be denoted by the capital letter, with E the containing rectangle. k times this is the longer side.
From the lengths in the fig, we get the following.
kE=kA+D...........1
kA=kB+C...........2
kC=B..................So C=B/k...................3
kD=A+B.............So D=(A+B)/k............4
E=A+B...............5
Substituting C, D, E from 3, 4, 5 into 1 and 2, we get:-
Into 1 :- k(A+B)=kA+(A+B)/k...........So kB-B/k=A/k....... $\implies\ \ \dfrac A B=k^2 - 1$ ..............................6
Into 2 :- kA=kB + B/k.................................................... $\implies\ \ \dfrac A B=1 + \dfrac 1 {k^2}$ .............7
$\therefore\ k^2 - 1 = 1 + \dfrac 1 {k^2},\ \ \ \ \implies\ \ k^4 - 2k^2 - 1 = 0. \\ Solving\ \ the\ quadratic\ in\ x^2, \ \ \ we\ get\ \ x^2 = 1 + \sqrt2,\ \ since\ x^2\ can\ only\ be\ positive. \\ \dfrac{Area\ of\ A}{Area\ of\ B}=\dfrac{kA^2}{kB^2}=(k^2 - 1)^2=(1 + \sqrt2 - 1)^2=2=\dfrac 2 1 =\ \dfrac a b.\\ So\ a + b = 2 + 1=\Large\ \ \ \color{#D61F06}{3}.$