Inspired by Nihar Mahajan

Let:

Equation of DE: $y=x$

Equation of DB: $y=\frac{2}{3}x$

Equation of AF: $y=-\frac{1}{3}x+4$

Equation of AC: $y=-\frac{2}{3}x+4$

Then, we get $P=(2.4,2.4),Q=(3,3),R=(4,\frac{8}{3}),S=(3,2)$

Now, we can use the shoelace method: $[PQRS]=\frac{1}{2}\begin{vmatrix}2.4&3&4&3&2.4 \\2.4&3&\frac{8}{3}&2&2.4 \end{vmatrix}=0.8$

Construction : - Draw PL || BC and QN || BC .

Clearly , PL = LE and QN = NE .............. (1)

Consider △ ANQ and △ ABF,

Clearly they are similar .

.’. $\frac{AN}{AB}$ = $\frac{NQ}{BF}$

=> $\frac{AE - NE }{AB}$ = $\frac{NQ}{BF}$

=> $\frac{AN - NQ}{AB}$ = $\frac{NQ}{BF}$ (using 1 )

=> $\frac{4 - NQ}{6}$ = $\frac{NQ}{2}$

=> NQ = 1

.’. Ar . △ AQE = ½ * NQ *AE = ½ * 1 *4 = 2

Consider △ ALP and △ ABC,

Clearly they are similar .

.’. $\frac{AL}{AB}$ = $\frac{LP}{BC}$

=> $\frac{AE - LE }{AB}$ = $\frac{LP}{BC}$

=> $\frac{AE - LP }{AB}$ = $\frac{LP}{BC}$ (using 1 )

=> $\frac{4 - LP }{6}$ = $\frac{LP}{4}$

=> LP = 1.6

.’. Ar . △ APE = ½ * LP *AE = ½ * 1.6 *4 = 3.2

Again , Consider △ RMF and △ ABF,

Clearly they are similar .

.’. $\frac{RM}{AB}$ = $\frac{MF}{BF}$

=> $\frac{RM}{6}$ = $\frac{MF}{2}$

=> MF = $\frac{RM}{3}$ ……………….. (2)

Again △ BRM and △ BDC are similar .

.’. $\frac{RM}{DC}$ = $\frac{BM}{BC}$

=> $\frac{RM}{6}$ = $\frac{BF - MF}{4}$

=> $\frac{RM}{6}$ = $\frac{2 - MF}{4}$

=> MF = 2- $\frac{2RM}{3}$ ………….(3)

From EQ. 2 and EQ . 3 we get

RM = 2

.’. Ar . △BRF = ½ * RM * BF = ½ * 2 * 2 = 2

Also, Ar. △ BSC = ¼ * Ar . rectangle = ¼ * 24 = 6

Again , Ar. △ AFC = ½ * CF * AB = ½ * 2 * 6 = 6

Required area = Ar. △ AFC - Ar . △ APQ - Ar. Quad . SRFC

= 6 - (Ar . △ APE - Ar . △ AQE ) – (Ar. △ BSC- Ar . △BRF )

= 6 – (3.2 -2) –(6-2)

= 6-3.2+2-6+2 = 0.8

Found the coordinates by similar triangles. P(2.4,2.4), R(4,8\3) and Q(3,3), S(3,2).
Since Diagonal QS is vertical base of the two triangles PQS and RQS and Horizontal distance between PQ=1.6, Vertical distance between QS=1.
Area PQRS= $\frac 1 2 *1.6*1= \Large \color{#D61F06}{0.8}.$
However the method of Jerry Han Jia Tao is far better for finding coordinate. It is simple and straight.