Question

Two isosceles triangles have equal vertical angles and their areas are in the ratio $7:16$. Find the ratio of their corresponding height.

Answer 1

Consider the two isosceles triangle $PQR$ and $XYZ$,
$\angle P=\angle X$ … [from the question]
So, $\angle Q+\angle R=\angle Y+\angle Z$
$\angle Q=\angle R$ and $\angle Y=\angle Z$ [because opposite angles of equal sides]
Therefore, $\angle Q=\angle Y$ and $\angle R=\angle Z$
$△PQR\sim △XYZ$
Then, area of $△PQR$/area of $△XYZ=P{M}^2/X{N}^2$ … [from corollary of theorem]
$P{M}^2/X{N}^2=7/16$
$PM/XN=\sqrt{7}/\sqrt{16}$
$PM/XN=\sqrt{7}/4$
Therefore, ratio of $PM:DM=\sqrt{7}:4$.