O-Level E-Math 2024 Paper 2, Question 8: Congruence (RHS) with circle reasons

(a)(i) In \(\triangle BMO\) and \(\triangle CNO\): - \(\angle OMB = \angle ONC = 90^{\circ}\) (a line from the centre to the midpoint of a chord is perpendicular to the chord); - \(OB = OC\) (radii of the same circle); - \(BM = \tfrac12 AB = \tfrac12 CD = CN\) (\(M, N\) midpoints and \(AB = CD\)).

So \(\triangle BMO \equiv \triangle CNO\) by RHS (right angle, equal hypotenuses \(OB = OC\), equal sides \(BM = CN\)).

(ii) Equal chords subtend equal angles at the centre, so \(\angle AOB = \angle COD = 1.8\) rad. Major arc \(AB = r(2\pi - 1.8) = 4(2\pi - 1.8) = 17.9\) cm (3 s.f.).

(b) By symmetry about the line \(PQR\), the common chord through the two intersection points subtends, at \(P\), twice angle \(RPS\): \(80^{\circ}\). At \(R\) on the larger circle, \(\angle PRS = 15^{\circ}\), so the angle at the centre \(Q\) is \(\angle PQS = 2 \times 15^{\circ} = 30^{\circ}\) (angle at centre \(= 2\times\) angle at circumference); by symmetry the chord subtends \(60^{\circ}\) at \(Q\). The shaded lens is the sum of two circular segments: \[\text{shaded} = \underbrace{\left[\tfrac{80}{360}\pi(5^2) - \tfrac12(5^2)\sin 80°\right]}_{\text{small circle} = 5.14} + \underbrace{\left[\tfrac{60}{360}\pi(8^2) - \tfrac12(8^2)\sin 60°\right]}_{\text{large circle} = 5.80} = 10.9 \text{ cm}^2.\]