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

Let their be two isosceles triangles ABC and DEF.

∠A = ∠D (Given, vertical angles are equal)

Since, triangles are isosceles so,

∠B = ∠C = $\dfrac{180 - ∠A}{2}$ and ∠E = ∠F = $\dfrac{180 - ∠D}{2}$.

Since, ∠A = ∠D so, we can say

∠B = ∠E.

Hence, by AA axiom △ABC ~ △DEF.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes.

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △DEF}} = \dfrac{(\text{Height of △ABC})^2}{(\text{Height of △DEF})^2} \\[1em] \Rightarrow \dfrac{7}{16} = \Big(\dfrac{\text{Height of △ABC}}{\text{Height of △DEF}}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Height of △ABC}}{\text{Height of △DEF}} = \sqrt{\dfrac{7}{16}} \\[1em] \Rightarrow \dfrac{\text{Height of △ABC}}{\text{Height of △DEF}} = \dfrac{\sqrt{7}}{4}.$

Hence, the ratio of their corresponding heights is $\sqrt{7} : 4$.