An equilateral triangle has an altitude of 15 m. What is the perimeter of the triangle?

we know that
In the equilateral triangle every sides and angles are equal
so
the measure of internal angles is equal to [tex] 60 [/tex] degrees
see the attached picture to better understand the problem
In the right triangle ABC
[tex] sin 60=BC/AB [/tex]
where
BC is the altitude (opposite side angle [tex] 60 [/tex] degrees
AB is the hypotenuse
Clear AB
[tex] AB=BC/sin 60\\ AB=15/(\frac{\sqrt{3}}{2} )\\ AB=30/\sqrt{3}m [/tex]
Perimeter of the triangle is equal to
[tex] P=AB+BD+AD [/tex]
but remember that
[tex] AB=BD=AD [/tex]
so
[tex] P=3*AB\\ P=3*\frac{30}{\sqrt{3}} \\ \\ P=\frac{90}{\sqrt{3}} \\ \\ P=30\sqrt{3} m [/tex]
therefore
the answer is
[tex] 30\sqrt{3} m [/tex]