Triangle ABC is similar to triangle PQR and AB : PQ= 2 : 3. AD is median to the side BC in triangle ABC and PS is the median to side QR in triangle PQR. What is the value of $${\left( {\frac{{{\text{BD}}}}{{{\text{QS}}}}} \right)^2}?$$

Solution (By Examveda Team)

$$\frac{{AB}}{{PQ}} = \frac{2}{3}\left( {{\text{Given}}} \right)$$

$$\eqalign{ & \frac{{{\text{area of }}\Delta ABC}}{{{\text{area of }}\Delta PQR}} = {\left( {\frac{{AB}}{{PQ}}} \right)^2} = \frac{4}{9} \cr & \left[ {{\text{Ratio of similar }}\Delta } \right] \cr & \frac{{{\text{area of }}\Delta ABD}}{{{\text{area of }}\Delta PQS}} = \frac{{\frac{1}{2} \times {\text{area of }}\Delta ABC}}{{\frac{1}{2} \times {\text{area of }}\Delta PQR}} \cr & = \frac{4}{9} = \frac{{B{D^2}}}{{Q{S^2}}} \cr & \left[ {\therefore AD\,\& \,PS\,{\text{are median}}} \right] \cr} $$