A basic and conceptual question from similarity chapter in which we have been asked to find the ratio of their corresponding height if two isosceles triangles have equal vertical angles and their areas are in the ratio 7:16

ML Aggarwal Avichal Publication class 10, Similarity, chapter 13.3, Question no 16

## Consider the two isosceles triangle PQR and XYZ,

∠P=∠X … [from the question]

So, ∠Q+∠R=∠Y+∠Z

∠Q=∠R and ∠Y=∠Z [because opposite angles of equal sides]

Therefore, ∠Q=∠Y and ∠R=∠Z

△PQR∼△XYZ

Then, area of △PQR/area of △XYZ=PM2/XN2 … [from corollary of theorem]

PM2/XN2=7/16

PM/XN=√7/√16

PM/XN=√7/4

Therefore, ratio of PM:DM=√7:4