UPSC CDS I 2019 Math Paper

Concept:The centroid divides any median in the ratio $2:1$, with the longer segment from the vertex.Explanation:Given triangle $\text{ABC}$, let $D$ be the midpoint of side $\text{BC}$.Then $BD = DC = \frac{BC}{2} = \frac{60}{2} = 30$ cm.In right triangle $\text{BOD}$ (where $O$ is the centroid), we have $\angle OBD = 30^{\circ}$.Using $\tan 30^{\circ} = \frac{OD}{BD}$, we get $\frac{1}{\sqrt{3}} = \frac{OD}{30}$.Thus $OD = \frac{30}{\sqrt{3}} = 10\sqrt{3}$ cm.The centroid $O$ lies on median $\text{AD}$ such that $AO : OD = 2 : 1$.Therefore $\frac{AO}{OD} = \frac{2}{1}$, so $\frac{AO}{10\sqrt{3}} = 2$.Hence $AO = 20\sqrt{3}$ cm.This is the distance $d$ from vertex $A$ to the centroid.Answer:$d = 20\sqrt{3}$ cm (Option C).