O-Level E-Math 2016 Paper 2, Question 6: Angle in a semicircle

(a) \(A\), \(O\), \(C\) lie on a straight line, so \(AC\) is a diameter; hence \(\angle ABC = 90^{\circ}\) (angle in a semicircle). The line \(ADB\) is a tangent to the smaller circle at \(D\), so the radius \(OD \perp AB\), giving \(\angle ADO = 90^{\circ}\). Therefore \(\angle ABC = \angle ADO\), and \(\angle BAC = \angle DAO\) (same angle, common to both triangles). With two pairs of equal angles, \(\triangle ABC \sim \triangle ADO\) (AA).

(b) In right-angled triangle \(ABC\), \(\angle BCA = 65^{\circ}\) so \(\angle BAC = 90^{\circ} - 65^{\circ} = 25^{\circ}\). Since \(O\) is the midpoint of diameter \(AC\), the line \(BO\) splits \(\triangle ABC\) into \(\triangle ABO\) and \(\triangle BOC\) of equal area (equal bases \(AO = OC\), same height from \(B\)), so area \(BOC = \tfrac{1}{2}\,\)area \(ABC\). From the similarity in (a), the linear scale factor of \(\triangle ADO\) to \(\triangle ABC\) is \(\dfrac{AO}{AC} = \dfrac{1}{2}\), so area \(ADO = \left(\tfrac{1}{2}\right)^2 \text{area } ABC = \tfrac{1}{4}\,\)area \(ABC\). Therefore area \(BOC\) : area \(ADO = \tfrac{1}{2} : \tfrac{1}{4} = 2 : 1\).

(c) The shaded region is the ring (annulus) between the two circles, area \(= \pi R^2 - \pi r^2\), where \(r = 3\) is the small radius and \(R\) the large radius. In right-angled triangle \(ODA\), \(\angle OAD = \angle BAC = 25^{\circ}\) and \(OA = R\), so \(OD = OA \sin 25^{\circ} \Rightarrow 3 = R \sin 25^{\circ} \Rightarrow R = \dfrac{3}{\sin 25^{\circ}} = 7.0986\) cm. Shaded area \(= \pi(R^2 - 3^2) = \pi(50.39 - 9) = \pi(41.39) = 130.0 \text{ cm}^2\) (to 3 s.f., \(\approx 130 \text{ cm}^2\)).