Group Event 10
$A B C D$ is a square of side length $20 \sqrt{5} x . P, Q$ are midpoints of $D C$ and $B C$ respectively.
G10.1 If $A P=a x$, find $a$.
G10.2 If $P Q=b \sqrt{10} x$, find $b$.
G10.3 If the distance from $A$ to $P Q$ is $c \sqrt{10} x$, find $c$.
G10.4 If $\sin \theta=\frac{d}{100}$, find $d$.
$\begin{array}{l}A P=\sqrt{A D^{2}+D P^{2}} \\ \quad=\sqrt{(20 \sqrt{5} x)^{2}+(10 \sqrt{5} x)^{2}}=50 x \\ \begin{aligned} \Rightarrow a & =50 \\ P Q & =\sqrt{C P^{2}+C Q^{2}}=10 \sqrt{10} x \\ \Rightarrow b & =10 \\ c \sqrt{10} x & =A C-\text { distance from } C \text { to } P Q \\ & =20 \sqrt{5} x \cdot \sqrt{2}-10 \sqrt{5} x \cdot\left(\frac{1}{\sqrt{2}}\right)\end{aligned} \\ \quad=15 \sqrt{10} x \\ \Rightarrow c=15\end{array}$
$\begin{array}{l}\text { Area of } \triangle A P Q=\frac{1}{2} \cdot A P \cdot A Q \sin \theta=\frac{1}{2} \cdot P Q \cdot(c \sqrt{10} x) \\ \Leftrightarrow \frac{1}{2} \cdot(50 x)^{2} \sin \theta=\frac{1}{2} \cdot 10 \sqrt{10} x \cdot 15 \sqrt{10} x \\ \sin \theta=\frac{d}{100}=\frac{3}{5} \\ \Rightarrow d=60\end{array}$