A geometry problem

Let $AB=x$ and $JB=ID=a$ . Then $DE=AE-x$ and $AJ=EF=x-a$ . $\triangle EDI$ and $\triangle EAB$ are similar, so $\frac{DE}{ID}=\frac{AE}{AB} \implies \frac{AE-x}{a}=\frac{AE}{x} \implies AE=\frac{x^2}{x-a}$ . Finally, $[AEFJ]=AE\cdot AJ=\frac{x^2}{x-a} \cdot (x-a)=x^2$ . So, the answer is $111^2=\boxed{12321}$ .

The problem gives away the answer without requiring proof by simply using JB=0

Given: $JB = ID, AB=111$ , Let $[AEFJ]$ denote the area of rectangle $AEFJ$ .

Solution: In $\triangle JBH$ & $\triangle DIE$ ,

$\angle JBH = \angle DIE$ [Corresponding angles]

$\angle BJH = \angle IDE$ [corresponding angles]

JB = ID [Given]

$\Rightarrow \triangle JBH \cong \triangle DIE$ [SAS congruence]

$\Rightarrow [JBH] = [DIE]$ .......(1)

Similarly, in $\triangle BCI$ & $\triangle HFE$

$\angle CBI$ = $\angle FHI$ [Corresponding angles]

$FE = IG+ID = IG+JB = IG+CG = CI i.e., FE = CI$

$\angle BCI$ = $\angle FHI$ [Angles of square & rectangle]

$\Rightarrow \triangle BCI \cong \triangle HFE$

$\Rightarrow [BCI] = [HFE]$ .......(2)

$[AEFJ] = [AJHID]+[IDE]+[HFE]$

$=[AJHID]+[JBH]+[BCI]$ ...........From (1) & (2)

$=[ABCD] = AB^{2}$ = $111^{2}$ = 12321

This is the construction for drawing a rectangle given one side and having the area of a given square.

I think the diagram should be redrawn as it doesn't match the sizes and proportions.