In the parallelogram PQRS, QR = 12.7 cm and angle PQR = 50°.

1. Write the area of the parallelogram in terms of the given base and unknown side.

In parallelogram $PQRS$, opposite sides are equal, so $PQ=RS$.

The perpendicular height to base $QR$ is
$PQ\sin 50^\circ = RS\sin 50^\circ.$

Using $\text{Area of parallelogram}=\text{base}\times\text{height}$,
$52.6=12.7(RS\sin 50^\circ).$

2. Solve for RS and round to the required shown value.

$RS=\frac{52.6}{12.7\sin 50^\circ}$
$RS=5.407\ldots$

Hence, $RS=5.407\text{ cm}.$

1. Identify the longer diagonal and set up the cosine rule.

In parallelogram $PQRS$, $QP=RS=5.407\text{ cm}$.

The longer diagonal is $QS$, since it is across the acute angle between sides $QP$ and $QR$.

Using cosine rule in $\triangle PQS$ or the equivalent side-vector triangle,
$QS^2=12.7^2+5.407^2+2(12.7)(5.407)\cos 50^\circ.$

2. Calculate and round the length of the longer diagonal.

$QS=\sqrt{12.7^2+5.407^2+2(12.7)(5.407)\cos 50^\circ}$
$QS=16.697\ldots$
$QS\approx 16.7\text{ cm}.$