IMC 2012 Q21

In the diagram, \(T\) is the foot of the perpendicular from \(Q\) to \(SR\) produced. Angles \(PQR\) and \(QRT\) are alternate angles between parallel lines so \(\angle QRT = 60^\circ\). Triangle \(QRT\) has interior angles of 90°, 60°, 30° so it may be thought of as being half of an equilateral triangle of side 1 unit, since the length of \(QR\) is 1 unit. So the lengths of \(RT\) and \(QT\) are \(\tfrac{1}{2}\) and \(\tfrac{\sqrt{3}}{2}\) units respectively.
Applying Pythagoras' Theorem to \(\Delta QST\), \(SQ^2 = ST^2 + QT^2 = \left(\tfrac{5}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{25}{4} + \tfrac{3}{4} = 7\).
So the length of \(SQ\) is \(\sqrt{7}\) units.