UPSC CDS 2 2019 Math Paper

Concept:Triangles formed by the intersection of lines connecting points on two parallel lines are similar, and the ratio of their areas equals the square of the ratio of corresponding sides.Explanation:Given two parallel lines $p$ and $q$.Points $B$ and $C$ lie on $p$, points $D$ and $E$ lie on $q$.Lines $BE$ and $CD$ intersect at $A$ between $p$ and $q$.In $\triangle ABC$ and $\triangle AED$:$\angle BAC = \angle EAD$ (vertically opposite angles at $A$).Since $p \parallel q$, $\angle ABC = \angle AED$ (alternate interior angles, with transversal $BE$).Thus, $\triangle ABC \sim \triangle AED$ by AA similarity criterion.The corresponding sides are:$AC$ corresponds to $AD$ (from $A$ to the second point on each line).Given $AC : AD = 4 : 9$.For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides.So area ratio $\frac{\text{area}(\triangle ABC)}{\text{area}(\triangle ADE)} = \left(\frac{AC}{AD}\right)^2 = \left(\frac{4}{9}\right)^2 = \frac{16}{81}$.Answer:The ratio is $16 : 81$. (Option C)