UPSC 2016 CDS I Math Paper

P is the mid-point of QR.∴ $\angle OPQ = 90^{\circ}$(Perpendicular from the centre of the circle to the chord bisects the chord)$PQ = PR = 1\,\text{cm}$Also, M is the mid-point of CD.∴$\angle OMP = 90^{\circ}$(Perpendicular from the centre of the circle to the chord bisects the chord)In right $\text{triangle } OPQ$ $OP=\sqrt{OQ^2-PQ^2}=\sqrt{7^2-1^2}$$=\sqrt{48}\,\text{cm}$In right $\text{triangle } OMP$$OM=\sqrt{OP^2-MP^2}=\sqrt{48-24}$$=\sqrt{24}$cm∴ $OM = MP$⇒ $\angle OPM = \angle POM = 45^{\circ}$∴ $\angle QPM = 90^{\circ} - 45^{\circ} = 45^{\circ}$⇒ $\angle QPD = 180^{\circ} - 45^{\circ} = 135^{\circ}$(Linear pair)Hence, statement 1 is correct. QR and CD are two chords of the circle that intersect at P.∴ $CP \times PD = QP \times PR \Rightarrow mn = 1$So, m and n are the roots of the quadratic equation$x^2-10x+1=0$,as the product of the roots is 1.Hence, statement 2 is correct.Area of $\text{triangle } OPR = \frac{1}{2} \times PR \times OP$$=\frac{1}{2} \times 1 \times \sqrt{48}=2\sqrt{3}\,\text{cm}^2$Area of $\text{triangle } OMP = \frac{1}{2} \times MP \times OM$$=\frac{1}{2} \times \sqrt{24} \times \sqrt{24}=12\,\text{cm}^2$Therefore, ratio of the area of $\text{triangle } OPR$ to the area of $\text{triangle } OMP$$=\frac{2\sqrt{3}}{12}=\frac{1}{2\sqrt{3}}=1:2\sqrt{3}$Hence, statement 3 is incorrect.