UPSC CDS 1 2020 Math Paper

Concept:Properties of intersecting diagonals and triangle inequalities in a trapezium.Explanation:Opposite angles formed by intersecting lines are equal, so $\angle AOD = \angle BOC$.Statement 2 is correct.Let the heights of $\triangle AOD$ and $\triangle BOC$ from $O$ be $x$ and $y$ respectively, with total height $h = x + y$.By similarity of triangles, $\frac{x}{y} = \frac{AD}{BC} = \frac{a}{b}$.Thus $x = \frac{ah}{a+b}$ and $y = \frac{bh}{a+b}$.Area sum: $\frac{1}{2} a x + \frac{1}{2} b y = \frac{h(a^2+b^2)}{2(a+b)}$.Area of trapezium $= \frac{(a+b)h}{2}$.Area of $\triangle AOB + \triangle DOC = \frac{(a+b)h}{2} - \frac{h(a^2+b^2)}{2(a+b)} = \frac{abh}{a+b}$.These are not equal in general, so statement 1 is false.Using triangle inequality in four triangles:$AB + AD > BD$, $BC + CD > BD$, $AB + BC > AC$, $AD + CD > AC$.Adding: $2(AB+BC+CD+DA) > 2(AC+BD)$.Hence $AB+BC+CD+DA > AC+BD$; statement 3 is correct.Answer:Statements 2 and 3 only are correct (Option B).