Is the question correct ?

Let $FG = x$ be parallel to $AB$ and $CD$ . and $FH$ is perpendicular to $AB$ . Then $CG=BG=6$ , $\triangle CFG$ and $\triangle FHK$ are similar, and that

$\begin{aligned} \frac 6x & = \frac {x-5}6 \\ x^2 - 5x - 36 & = 0 \\ (x+4)(x-9) & = 0 \\ \implies x & = 9 & \small \color{#3D99F6} \text{Since }x > 0 \end{aligned}$

We note that $\triangle AFH$ and $\triangle DFL$ are congruent, making the area of trapezium $ABCD$ equal to the area of rectangle $BCLH$ which is $9 \times 12 = \boxed{108}$ .

$Right~\Delta~CKB~is~5-12-13, ~CK=13.$

$Add$ an inverted copy of trapezium ABCD to the side AD..

$This$ makes $FLCK$ a $rhombus$ . $Making$ $FL=LC=CK=KF=13$

Draw $CK$ and rotate $\triangle CDF$ $180°$ about $F$ :

By Pythagorean's Theorem on $\triangle BCK$ , $KC = 13$ .

Since $FC' \cong FC$ , $\angle KFC' \cong \angle KFC$ , and $FK \cong FK$ , $\triangle KFC' \cong \triangle KFC$ by SAS congruence.

Therefore $KC' = KC = 13$ , and the area of trapezium $ABCD$ is $A_{ABCD} = A_{\triangle CBC'} = \frac{1}{2}(13 + 5)12 = \boxed{108}$ .

I used coordinate geometry to solve this problem. Let $B = (0, 0)$ , then $K = (-5, 0)$ and $C = (0, 12)$ , while point $F = (x, 6)$ , where $x$ is negative.. Now, since $\angle CFK$ is a right angle then the dot product $CF \cdot FK = 0$ . Now, $CF = F - C = (x , -6)$ and $FK = K - F = (-5 - x, -6 )$ , and therefore $CF \cdot FK = x (-5 - x) + 36 = 0 \Rightarrow x^2 + 5 x - 36 = 0 \Rightarrow (x + 9)(x - 4) = 0 \Rightarrow x = -9$ , because $x \lt 0$ . The area of the trapezium $a = \frac{1}{2} (DC + AB) CB = 12 (-x) = 12 (9) = 108$ .