Think Ouside the Box

We can find the length of $PQ$ by subtracting the length of $DP$ from $DQ$ .

We can extend $EG$ , $FH$ , and $AB$ to meet at point $R$ . We can also extend $EG$ , $FH$ , and $DC$ to points $T$ and $S$ .

Since $\triangle AER$ and $\triangle BGR$ are similar, this proportion becomes true:

$\dfrac{AR}{AE}=\dfrac{BR}{BG}$

Substituting in known lengths,

$\dfrac{AR}{3}=\dfrac{AR+5}{6}$

$AR=5$

$\implies BR=10$

Since $\triangle CGS$ and $\triangle DES$ are similar, this proportion becomes true:

$\dfrac{CS}{CG}=\dfrac{DS}{DE}$

Substituting in known lengths,

$\dfrac{CS}{4}=\dfrac{CS+5}{7}$

$CS=\dfrac{20}{3}$

$\implies DS=\dfrac{35}{3}$

We also know that $\triangle BQR$ and $\triangle DQS$ are similar, where the ratio from $\triangle BQR$ to $\triangle DQS$ is $\dfrac{6}{7}$ . So, $BQ=\dfrac{6}{7} DQ$ since they are corresponding sides. Note that $BD=5\sqrt{5}$ . So,

$DQ+\dfrac{6}{7} DQ =5\sqrt{5}$

$DQ=\dfrac{35\sqrt{5}}{13}$

Since $\triangle TCH$ and $\triangle TDF$ are similar, this proportion becomes true:

$\dfrac{CT}{CH}=\dfrac{DT}{DF}$

Substituting in known lengths,

$\dfrac{CT}{2}=\dfrac{CT+5}{6}$

$CT=\dfrac{5}{2}$

$\implies DT=\dfrac{15}{2}$

We also know that $\triangle BPR$ and $\triangle DPT$ are similar, where the ratio from $\triangle BPR$ to $\triangle DPT$ is $\dfrac{4}{3}$ . So, $BP =\dfrac{4}{3} DP$ since they are corresponding sides. So,

$DP+\dfrac{4}{3} DP=5\sqrt{5}$

$DP=\dfrac{15\sqrt{5}}{7}$

The length $PQ$ , which we are asked to find, is $PQ=DQ-DP=\dfrac{35\sqrt{5}}{13}-\dfrac{15\sqrt{5}}{7}=\dfrac{50\sqrt{5}}{91}$ . Therefore, $x+y+z=50+5+91=\boxed{146}$ .

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Note: This can also be done nicely using coordinate geometry .