The cyclic quadrilateral $PQRS$ is such that the angle at $P$ is ${70}^{\circ },$ the length $QR$ is $4$ and the length $RS$ is $5.$

Find the area of the circumscribed circle.

Opposite angles in a cyclic quadrilateral sum to ${180}^{\circ }$ (that is to say, the sum of the angles at $R$ and $P$ is ${180}^{\circ }$)

Use the previous hint with the cosine rule on $\mathrm{△}QRS$ to find $QS.$