$QR$ at $X$ and $PR$ at $Z.$
$OZ,$ $OX,$ $OY$ are perpendicular to the sides $PR,$ $QR,$ $PQ.$
Here $PQR$ is an isosceles triangle with sides $PQ = PR$ and also from the figure,
\[\Rightarrow ~PY\text{ }=\text{ }PZ\text{ }=\text{ }x\]
\[\Rightarrow ~YQ\text{ }=\text{ }QX\text{ }=\text{ }XR\text{ }=\text{ }RZ\text{ }=\text{ }y\]
From the figure we can see that,
\[\Rightarrow ~Area\left( \Delta PQR \right)=Area\left( \Delta POR \right)+Area\left( \Delta POQ \right)\]
\[+Area\left( \Delta QOR \right)\]
\[\Rightarrow ~PER\text{ }=\text{ }2\left( \surd 3r \right)\text{ }+\text{ }4\left( \surd 3r \right)\]
\[\Rightarrow ~PER\text{ }=\text{ }6\surd 3r\]
∴ Thus proved