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Let $CF'G'$ be the reflection of $C'FG$. And as $CDE$ is the reflection of $C'DE$, we can say,

$F'G'||DE||AB$

So, $(CF'G')=(C'FG)=80$.

(ABC)=$\frac{1}{2}AB.CP=500$ or, $10CP=500$ or,$CP=50$

We know, $\triangle ABC~\triangle CF'G'$

$\frac {AB}{F'G'}=\frac {CP}{CQ}=k$

So, $F'G'=\frac{AB}{k}=\frac{20}{k}$, $CQ=\frac{CP}{k}=\frac{50}{k}$

$(CF'G')=\frac{CQ.F'G'}{2}=\frac{20*50}{2.k.k}=\frac{500}{k.k}=80$ or,$k=\frac{5}{2}$

$FG=F'G'=\frac{20}{k}=8$, $CQ=\frac{50}{k}=20$

Then, $CP+C'P'=50+20=70$ or,$CQ+C'P'+PR+RQ=70$ or,$2CR=70$ or,$CR=35$

We proved, $\triangle CF'G'~\triangle CDE$

So, $\frac{DE}{F'G'}=\frac{CR}{CQ}=\frac{35}{20}=\frac{7}{4}$

$DE=\frac{7F'G'}{4}=\frac{7*8}{4}=14$

So, $(CDE)=\frac{1}{2}DE.CR=\frac{14*35}{2}=245$; that we need.