Question

Two isosceles triangles have equal vertical angles and the ratios of the areas are 16 : 25. find ratio of their altitudes drawn from vertex to the opposite sides

Answer 1

$\frac{ \sqrt{3} }{4} {x}^{2} \div \frac{ \sqrt{3} }{4} {y}^{2} = \frac{16}{25}$
$\frac{ {x}^{2} }{ {y}^{2} } = \frac{16}{25}$
$\frac{x}{y} = \frac{4}{5}$
Altitudes ratio
$\frac{ \sqrt{3} }{2} x \div \frac{ \sqrt{3} }{2} y = \frac{4}{5}$
I.e. 4:5