The coordinates of P and Q are (0, 3) and (15, 12) respectively.

1. Find the gradient of $PQ$ using the two coordinates.

For $P(0,3)$ and $Q(15,12)$,
$m=\frac{12-3}{15-0}=\frac{9}{15}=\frac{3}{5}.$

2. Use the y-intercept to write the equation of the line.

Since $P(0,3)$ lies on the y-axis, the y-intercept is $3$.

Using $y=mx+c$,
$y=\frac{3}{5}x+3.$

1. Find the coordinates of R after reflecting Q in the x-axis.

$Q=(15,12)$.

Reflection in the $x$-axis changes $(x,y)$ to $(x,-y)$.

So $R=(15,-12)$.

2. Find the lengths of the four sides of trapezium OPQR.

$O=(0,0)$ and $P=(0,3)$, so

$OP=3$.

$P=(0,3)$ and $Q=(15,12)$, so

$PQ=\sqrt{(15-0)^2+(12-3)^2}=\sqrt{15^2+9^2}=\sqrt{306}=3\sqrt{34}.$

$Q=(15,12)$ and $R=(15,-12)$, so

$QR=24$.

$R=(15,-12)$ and $O=(0,0)$, so

$RO=\sqrt{(15-0)^2+(-12-0)^2}=\sqrt{15^2+12^2}=\sqrt{369}=3\sqrt{41}.$

3. Add the side lengths to find the perimeter.

Perimeter of $OPQR$

$=OP+PQ+QR+RO$

$=3+3\sqrt{34}+24+3\sqrt{41}$

$=27+3\sqrt{34}+3\sqrt{41}$

$=63.7\text{ units }\text{(to 3 s.f.)}.$