Area Of Triangle Problem

The question can easily be solved using the similarity between the triangles. It can easily be deduced that $FPCL\quad and\quad EPBK\quad are\quad parallelograms\\ \Longrightarrow FP=LC.......(1)\\ \quad \quad \quad \quad PE=BK.......(2)\\ \triangle AFP\sim \triangle PCB\\ \triangle DPE\sim \triangle PCB\\ \\ therefore,\quad \frac { ar(\triangle AFP) }{ ar(\triangle PCB) } ={ (\frac { FP }{ CB } })^{ 2 }=\frac { 4 }{ 49 } .......(3)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { ar(\triangle DPE) }{ ar(\triangle PCB) } ={ (\frac { PE }{ CB } })^{ 2 }=\frac { 9 }{ 49 } ........(4)\\ \Longrightarrow \frac { FP }{ PE } =\frac { 2 }{ 3 } ........(5)\\ Using\quad (1),(2),(3),(4)\quad and\quad (5)\quad we\quad get\quad \\ LK=6(FP)\quad \\ Since,\quad \triangle AFP\sim \triangle JLK\quad \quad \quad \quad \quad \\ \quad \quad \frac { ar(\triangle JLK) }{ ar(\triangle AFP) } ={ (\frac { LK }{ FP } })^{ 2 }=36\\ Therefore,\quad ar(\triangle JLK)=\quad 36\quad \times \quad 4\quad =\boxed{144}$

Triangles where the areas are given are all similar. $....Note; \sqrt{\dfrac{49}{4}} =\dfrac 7 2. ......\sqrt{\dfrac{9}{4} }=\dfrac 3 2.$
WLOG AP=2x, FP=4.......Then because of similarity, $PB=\dfrac 7 2 *2x=7x, CB=\dfrac 7 2 *4=14,......DE=\dfrac 3 2 * 2x=3x, PE=\dfrac 3 2 * 4=6.$
FLCP, JAPD, and EPBK are | |grams. Implies LC=FP=4, .. BK=PE=6, ..JD=AP=2x, ..EK=PB=7x.
So LK=LC+CB+BK=24. ...... JK=JD+DE+EK=12x.
Triangle PCB=1/2 * PB * CB=49x=49 square units. So x=1.
So area JLK=1/2 * LK * JK=1/2 * 24 * 12 * 1= $\color{#D61F06}{144}.$

@Abhijeet Verma thanx for posting a solution